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\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)
\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)
\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)
\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)
\(i.\sqrt{x+2+2\sqrt{x+1}}=\sqrt{x+1+2\sqrt{x+1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=\sqrt{x+1}+1\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)
b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)
c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)
d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a) ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
b) ta có : \(\dfrac{x-\sqrt{3x}+3}{x\sqrt{x}+3\sqrt{3}}=\dfrac{x-\sqrt{3x}+3}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}+\sqrt{y}}\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
a, \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
b, Với a;b > 0
\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
c, Với x >= 0
\(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d, Với x >= 0 ; x khác 14
\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\frac{1}{\sqrt{3}}\)
b) \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)
c) \(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{\left(4\sqrt{x}+7\right)}=\sqrt{x}-1\)
d) \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\dfrac{\sqrt{2}}{\sqrt{6}}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)
c)\(\sqrt{x}-1\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a/ \(\dfrac{1}{\sqrt{3}}\)
b/\(\dfrac{\sqrt{ab}}{b}\)
c/\(\sqrt{x}-1\)
d/\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a)\(\dfrac{\sqrt{3}}{3}\)
b)\(\dfrac{\sqrt{ab}}{b}\)
c)\(\sqrt{x}-1\)
d)\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\sqrt{\dfrac{1}{3}}\)
b) \(\sqrt{\dfrac{a}{b}}\)
c) \(\sqrt{x-1}\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\dfrac{\sqrt{3}}{3}\)
b) \(\dfrac{\sqrt{a}+1}{\sqrt{b}+1}\)
c) \(\sqrt{x}-1\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a. \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)
b. \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt[]{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)
c. \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{4x-4\sqrt{x}+7\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d. \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\) = \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}\)= \(\dfrac{1}{\sqrt{3}}\)=\(\dfrac{\sqrt{3}}{3}\)
b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}\)
với a,b≥0 thì \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\)\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\) =\(\dfrac{\sqrt{ab}}{b}\)
c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)
với x≥0 thì \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\left(\sqrt{x}-4.\sqrt{x}+1\right)}{\left(\sqrt{x}-4.\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)( với x≥0; x≠14)
a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}\sqrt{18}}=\dfrac{1}{\sqrt{3}}\)
b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\sqrt{a}}{\sqrt{b}}\)
c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\sqrt{x}-1\)
d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}30+1810+6=6(5+3)2(5+3)=62=31
b) Với a,b>0a,b>0
\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\left(\sqrt{a}\right)^2+\sqrt{a}.\sqrt{b}}{\left(\sqrt{b}\right)^2+\sqrt{a}.\sqrt{b}}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}b+
a) \(\dfrac{1}{\sqrt{3}}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)
c) \(\dfrac{1}{\sqrt{x}-1}\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a) \(\dfrac{1}{\sqrt{3}}\)
b) \(\dfrac{\sqrt{a}}{\sqrt{b}}\)
c) \(\sqrt{x}-1\)
d) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a,
\(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}=\dfrac{1}{\sqrt{3}}\)
b,
với a,b>0
\(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}=\dfrac{\left(\sqrt{a}\right)^2+\sqrt{a}.\sqrt{b}}{\left(\sqrt{b}\right)^2+\sqrt{a}.\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\)
c, Với x\ge0x≥0
\(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{4x+7\sqrt{x}-4\sqrt{x}-7}{4\sqrt{x}+7}=\dfrac{\sqrt{x}\left(4\sqrt{x}+7\right)-\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d,với x≥0,x≠16
\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(d.=\dfrac{x+\sqrt{X}-4\sqrt{X}-4}{x+3\sqrt{X}-4\sqrt{X}-12}\\ \\ \\ =\dfrac{\sqrt{X}\left(\sqrt{X}+1\right)-4\left(\sqrt{X}+1\right)}{\sqrt{X}\left(\sqrt{X}+3\right)-4\left(\sqrt{X}+3\right)}=\dfrac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-4\right)}{\left(\sqrt{X}+3\right)\left(\sqrt{X}-4\right)}\\ \\ \\ \\ =\dfrac{\sqrt{X}+1}{\sqrt{X}+3}=\dfrac{\sqrt{X}+1}{\sqrt{X}+3}.\dfrac{\sqrt{X}-3}{\sqrt{X}-3}=\dfrac{X-3\sqrt{X}-\sqrt{X}-3}{x-9}=\dfrac{X-2\sqrt{X}-3}{x-9}\)\(c.\dfrac{4x+7\sqrt{X}-4\sqrt{X}-7}{4\sqrt{X}+7}\\ \\ \\ \\ =\dfrac{\sqrt{x}\left(4\sqrt{X}+7\right)-\left(4\sqrt{X}+7\right)}{4\sqrt{X}+7}\\ =\dfrac{\left(4\sqrt{X}+7\right)\left(\sqrt{X}-1\right)}{4\sqrt{X}+7}\\ \\ \\ =\sqrt{X}-1\)\(b.=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\dfrac{\sqrt{a}}{\sqrt{b}}\\ =\dfrac{\sqrt{a}}{\sqrt{b}}.\dfrac{\sqrt{b}}{\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{ab}}{b}\)\(a.\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{\sqrt{2}}{\sqrt{6}}\\ \\ =\sqrt{\dfrac{2}{6}}=\sqrt{\dfrac{1}{3}}\)
a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\) = \(\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{5}+\sqrt{3}\right)}\)= \(\dfrac{\sqrt{2}}{\sqrt{6}}\) = \(\dfrac{1}{\sqrt{3}}\)
b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}\) = \(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)=\(\dfrac{\sqrt{a}}{\sqrt{b}}\)
c) \(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{4x+7\sqrt{x}-4\sqrt{x}-7}{4\sqrt{x}+7}\)= \(\dfrac{\left(4\sqrt{x}+7\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}\)= \(\sqrt{x}-1\)
d)\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)= \(\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)= \(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}\)=\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a)\(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\)=\(\dfrac{\sqrt{10}+\sqrt{6}\left(\sqrt{30}\right)-\sqrt{18}}{30-18}\)=\(\dfrac{10\sqrt{3}-6\sqrt{3}}{12}\)=\(\dfrac{4\sqrt{3}}{12}\)=\(\dfrac{\sqrt{3}}{3}\)
b)\(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}\)=\(\dfrac{a+\sqrt{ab}.b-\sqrt{ab}}{b^2-ab}\)=\(\dfrac{a+\sqrt{ab}.\left(b-\sqrt{ab}\right)}{b^2-ab}\)=\(\dfrac{-a+\sqrt{ab}+b\sqrt{ab}}{b^2-ab}\)=\(\dfrac{\sqrt{ab}\left(a-b\right)}{b^2-ab}\)
c)\(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)=\(\dfrac{16x\sqrt{x}-16x-49\sqrt{x}+49}{16x-49}\)=\(\dfrac{16x\left(\sqrt{x}-1\right)-49\left(\sqrt{x}-1\right)}{16x-49}\)=\(\dfrac{\left(\sqrt{x}-1\right)\left(16x-49\right)}{16x-49}\)=\(\sqrt{x}\)-1
d)\(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
A)đkxđ x>=0
-x=1(t/m)
-x=4(t/m)
Vậy pt có 2 nghiệm là S={1;4}
B)đkxđ x>=1
x=2(tm)
Vậy pt có 1 nghiệm là S={2}
C)đkxđ x>=-1/2
x=-1/2(tm)
x=-4/9(tm)
a)=1/√3
b) Với a,b>0
=√a/√b
c)voi x>=0
=√x-1
d)Với x\ge0;x\ne16x≥0;x=16
=√x+1/√x+3
a) \(\dfrac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}\)\(=\dfrac{\left(\sqrt{10+\sqrt{6}}\right)\left(\sqrt{30}-\sqrt{18}\right)}{\left(\sqrt{30+\sqrt{18}}\right)\left(\sqrt{30}-\sqrt{18}\right)}\)\(=\dfrac{10\sqrt{3}-6\sqrt{5}+6\sqrt{5}-6\sqrt{3}}{30-6\sqrt{15}+6\sqrt{15}-18}\)\(=\dfrac{4\sqrt{3}}{12}=\dfrac{\sqrt{3}}{3}\)
b) \(\dfrac{a+\sqrt{ab}}{b+\sqrt{ab}}\)\(=\dfrac{\left(a+\sqrt{ab}\right)\left(a-\sqrt{ab}\right)}{\left(b+\sqrt{ab}\right)\left(b-\sqrt{ab}\right)}\)\(=\dfrac{ab-\sqrt{a^2b}+\sqrt{ab^2}-ab}{b^2-\sqrt{a^2b}+\sqrt{ab^2}-ab}\)\(=\dfrac{\sqrt{a^2b}+\sqrt{ab^2}}{b^2-ab}\)
c)\(\dfrac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}\)\(=\dfrac{4x-4\sqrt{x}+7\sqrt{x}-7}{4\sqrt{x}+7}\)\(=\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)-7\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}\)
\(=\dfrac{\left(4\sqrt{x}+7\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d) \(\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)\(=\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}\)\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-4\right)+3\left(\sqrt{x}-4\right)}\)\(=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)