Ai giúp em với ạ em cần rất gấp bài 4 thôi ạ
Gọi O là tâm đường tròn \(\Rightarrow\) O là trung điểm BC
\(\stackrel\frown{BE}=\stackrel\frown{ED}=\stackrel\frown{DC}\Rightarrow\widehat{BOE}=\widehat{EOD}=\widehat{DOC}=\dfrac{180^0}{3}=60^0\)
Mà \(OD=OE=R\Rightarrow\Delta ODE\) đều
\(\Rightarrow ED=R\)
\(BN=NM=MC=\dfrac{2R}{3}\Rightarrow\dfrac{NM}{ED}=\dfrac{2}{3}\)
\(\stackrel\frown{BE}=\stackrel\frown{DC}\Rightarrow ED||BC\)
Áp dụng định lý talet:
\(\dfrac{AN}{AE}=\dfrac{MN}{ED}=\dfrac{2}{3}\Rightarrow\dfrac{EN}{AN}=\dfrac{1}{2}\)
\(\dfrac{ON}{BN}=\dfrac{OB-BN}{BN}=\dfrac{R-\dfrac{2R}{3}}{\dfrac{2R}{3}}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{EN}{AN}=\dfrac{ON}{BN}=\dfrac{1}{2}\) và \(\widehat{ENO}=\widehat{ANB}\) (đối đỉnh)
\(\Rightarrow\Delta ENO\sim ANB\left(c.g.c\right)\)
\(\Rightarrow\widehat{NBA}=\widehat{NOE}=60^0\)
Hoàn toàn tương tự, ta có \(\Delta MDO\sim\Delta MAC\Rightarrow\widehat{MCA}=\widehat{MOD}=60^0\)
\(\Rightarrow\Delta ABC\) đều
Mọi người giúp em bài này với ạ.EM cần gấp ạ
Mọi người giúp em bài này với ạ.EM cần gấp ạ
dạng này dễ mà bạn
bạn tìm ĐK, đối chiếu giá trị với ĐK thấy thỏa mãn rồi thay vô
toàn SCP nên tính cũng đơn giản:)
1) Thay x = 64 (TMĐK ) vào A, có :
A = \(\frac{\sqrt{64}}{\sqrt{64}-2}\)=\(\frac{4}{3}\)
Vậy A = \(\frac{4}{3}\)khi x = 64
2) Thay x = 36 ( TMĐK ) vào A, có
A =\(\frac{\sqrt{36}+4}{\sqrt{36}+2}\)=\(\frac{5}{4}\)
Vậy A =\(\frac{5}{4}\)khi x = 36
3) Thay x=9 (TMĐK ) vào A, có :
A= \(\frac{\sqrt{9}-5}{\sqrt{9}+5}\)= \(\frac{-1}{4}\)
Vậy A=\(\frac{-1}{4}\)khi x = 9
4) Thay x = 25( TMĐK ) vào A có:
A =\(\frac{2+\sqrt{25}}{\sqrt{25}}\)=\(\frac{7}{5}\)
Vậy A=\(\frac{7}{5}\) khi x = 25
P1 = (\(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)) : \(\frac{\sqrt{x}}{x+\sqrt{x}}\)= \(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\):\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\).
(\(\sqrt{x}+1\)) =\(\frac{x+\sqrt{x}+1}{\sqrt{x}}\)(ĐKXĐ : x > 0 )
P2 =\(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)= \(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
(ĐKXĐ: x\(\ge\)0, x\(\ne\)1)
a, \(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)ĐK : \(x\ge0;x\ne1\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b, \(B=\frac{3x-4}{x-2\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{\sqrt{x}-1}{2-\sqrt{x}}\)ĐK : \(x>0;x\ne4\)
\(=\frac{3x-4-\left(x-4\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
c, \(Q=\frac{3}{\sqrt{a}-3}+\frac{2}{\sqrt{a}+3}+\frac{a-5\sqrt{a}-3}{a-9}\)ĐK : \(a\ge0;a\ne9\)
\(=\frac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\frac{a}{a-9}\)
d, \(B=\frac{x}{x-4}-\frac{1}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\)ĐK : \(x\ge0;x\ne4\)
\(=\frac{x}{x-4}+\frac{\sqrt{x}+2}{x-4}+\frac{\sqrt{x}-2}{x-4}=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)
Bài 4:
a: \(\sqrt{1,6}\cdot\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}\)
\(=\sqrt{1,6\cdot250}+\sqrt4\)
\(=\sqrt{400}+2=20+2=22\)
b: \(\sqrt{1\frac34\cdot2\frac27\cdot5\frac49}\)
\(=\sqrt{\frac74\cdot\frac{16}{7}\cdot\frac{49}{9}}=\sqrt{\frac{16}{4}\cdot\frac{49}{9}}=2\cdot\frac73=\frac{14}{3}\)
c: \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
\(=20\sqrt{20}-15\sqrt{45}+5\sqrt5\)
\(=40\sqrt5-45\sqrt5+5\sqrt5=0\)
d: \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)
\(=\sqrt{25}-\sqrt9+2\cdot\sqrt{16}\)
\(=5-3+2\cdot4\)
=2+8
=10
e: \(\frac{2\sqrt8-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt5+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\frac{4\sqrt2-2\sqrt3}{\sqrt6\left(\sqrt3-2\sqrt2\right)}-\frac{\sqrt5+\sqrt{27}}{\sqrt6\left(\sqrt5+\sqrt{27}\right)}\)
\(=\frac{2\left(2\sqrt2-\sqrt3\right)}{-\sqrt6\left(2\sqrt2-\sqrt3\right)}-\frac{1}{\sqrt6}=-\frac{2}{\sqrt6}-\frac{1}{\sqrt6}=-\frac{3}{\sqrt6}=\frac{-3\sqrt6}{6}=-\frac{\sqrt6}{2}\)
f: \(\frac{3+2\sqrt3}{\sqrt3}+\frac{2+\sqrt2}{\sqrt2+1}-\left(\sqrt2+\sqrt3\right)\)
\(=2+\sqrt3+\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}-\sqrt2-\sqrt3\)
\(=2-\sqrt2+\sqrt2\)
=2
bài 1:
\(a.\sqrt{25 . 144}=\sqrt{25}.\sqrt{144}=5.12=60\)
\(b.\sqrt{45 . 80}=\sqrt{9 . 5 . 5 . 16}=\sqrt{9 . 25 . 16}=\sqrt{9}.\sqrt{25}.\sqrt{16}=3.5.4=60\)
\(c.\sqrt{52}.\sqrt{13}=\sqrt{52 . 13}=\sqrt{4 . 13 . 13}=\sqrt{4 . 13^2}=\sqrt{4}.\sqrt{13^2}=2.13=26\)
\(d.\sqrt{7}.\sqrt{28}=\sqrt{7 . 28}=\sqrt{7 . 7 . 4}=\sqrt{7^2 . 4}=\sqrt{7^2}.\sqrt{4}=7.2=14\)
\(e.\sqrt{1 \frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{\sqrt{25}}{\sqrt{16}}=\frac{5}{4}\)
\(f.\sqrt{\frac{25}{64}}=\frac{\sqrt{25}}{\sqrt{64}}=\frac{5}{8}\)
\(g.\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=5\)
\(h.\frac{\sqrt{230}}{\sqrt{2,3}}=\sqrt{\frac{230}{2,3}}=\sqrt{100}=10\)
bài 2:
\(a.\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\sqrt{6}\)
\(= \sqrt{\frac{2}{3}} . \sqrt{6} + \sqrt{\frac{50}{3}} . \sqrt{6} - \sqrt{24} . \sqrt{6}\)
\(= \sqrt{\frac{2}{3} . 6} + \sqrt{\frac{50}{3} . 6} - \sqrt{24 . 6}\)
\(= \sqrt{4} + \sqrt{100} - \sqrt{144}\)
\(=2+10-12=0\)
b. \(\sqrt{3 + \sqrt{5}}.\sqrt{2}=\sqrt{(3 + \sqrt{5}) . 2}\)
\(=\sqrt{6 + 2\sqrt{5}}=\sqrt{5 + 2\sqrt{5} + 1}\)
\(=\sqrt{(\sqrt{5} + 1)^2}=\vert{}\sqrt{5}+1\vert{}=\sqrt{5}+1\)
\(c.\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right).\sqrt{12}\)
\(= \sqrt{\frac{3}{4}} . \sqrt{12} - \sqrt{3} . \sqrt{12} + 5\sqrt{\frac{4}{3}} . \sqrt{12}\)
\(= \sqrt{\frac{3}{4} . 12} - \sqrt{3 . 12} + 5\sqrt{\frac{4}{3} . 12}\)
\(= \sqrt{9} - \sqrt{36} + 5\sqrt{16}\)
\(= 3 - 6 + 5 . 4\)
\(=3-6+20=17\)
\(d.\sqrt{3 - \sqrt{5}}.\sqrt{8}=\sqrt{3 - \sqrt{5}}.\sqrt{2}.\sqrt{4}\)
\(=\sqrt{(3 - \sqrt{5}) . 2}.2=2\sqrt{6 - 2\sqrt{5}}\)
\(=2\sqrt{5 - 2\sqrt{5} + 1}=2\sqrt{(\sqrt{5} - 1)^2}\)
\(=2\vert{}\sqrt{5}-1\vert{}=2(\sqrt{5}-1)=2\sqrt{5}-2\)
bài 3:
\(a.\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
\(= \sqrt{\frac{1}{7}} : \sqrt{7} - \sqrt{\frac{16}{7}} : \sqrt{7} + \sqrt{7} : \sqrt{7}\)
\(= \sqrt{\frac{1}{7} : 7} - \sqrt{\frac{16}{7} : 7} + 1\)
\(= \sqrt{\frac{1}{49}} - \sqrt{\frac{16}{49}} + 1\)
\(=\frac{1}{7}-\frac{4}{7}+1=-\frac{3}{7}+1=\frac{4}{7}\)
\(b.\sqrt{36 - 12\sqrt{5}}:\sqrt{6}=\sqrt{\frac{36 - 12\sqrt{5}}{6}}\)
\(=\sqrt{6 - 2\sqrt{5}}=\sqrt{5 - 2\sqrt{5} + 1}\)
\(=\sqrt{(\sqrt{5} - 1)^2}=\vert{}\sqrt{5}-1\vert{}=\sqrt{5}-1\)
\(c.\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(= \sqrt{\frac{1}{3}} : \sqrt{3} - \sqrt{\frac{4}{3}} : \sqrt{3} + \sqrt{3} : \sqrt{3}\)
\(= \sqrt{\frac{1}{3} : 3} - \sqrt{\frac{4}{3} : 3} + 1\)
\(=\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+1=\frac{1}{3}-\frac{2}{3}+1\)
\(=-\frac{1}{3}+1=\frac{2}{3}\)
\(e.\sqrt{3 - \sqrt{5}}:\sqrt{2}=\sqrt{\frac{3 - \sqrt{5}}{2}}\)
\(=\sqrt{\frac{6 - 2\sqrt{5}}{4}}=\frac{\sqrt{6 - 2\sqrt{5}}}{\sqrt{4}}\)
\(=\frac{\sqrt{5 - 2\sqrt{5} + 1}}{2}=\frac{\sqrt{(\sqrt{5} - 1)^2}}{2}\)
\(=\frac{\vert{}\sqrt{5} - 1\vert{}}{2}=\frac{\sqrt{5} - 1}{2}\)
bài 4:
\(a.\sqrt{1,6}.\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}=\sqrt{1,6 . 250}+\sqrt{\frac{19,6}{4,9}}\)
\(=\sqrt{400}+\sqrt{4}=20+2=22\)
\(b.\sqrt{1 \frac{3}{4}}.\sqrt{2 \frac{2}{7}}.\sqrt{5 \frac{4}{9}}=\sqrt{\frac{7}{4}}.\sqrt{\frac{16}{7}}.\sqrt{\frac{49}{9}}\)
\(=\sqrt{\frac{7}{4} . \frac{16}{7} . \frac{49}{9}}=\sqrt{\frac{16 . 49}{4 . 9}}=\sqrt{\frac{4 . 49}{9}}\)
\(=\frac{\sqrt{4} . \sqrt{49}}{\sqrt{9}}=\frac{2 . 7}{3}=\frac{14}{3}\)
\(c.\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)
\(= 20\sqrt{300} : \sqrt{15} - 15\sqrt{675} : \sqrt{15} + 5\sqrt{75} : \sqrt{15}\)
\(= 20\sqrt{\frac{300}{15}} - 15\sqrt{\frac{675}{15}} + 5\sqrt{\frac{75}{15}}\)
\(= 20\sqrt{20} - 15\sqrt{45} + 5\sqrt{5}\)
\(= 20\sqrt{4 . 5} - 15\sqrt{9 . 5} + 5\sqrt{5}\)
\(= 20 . 2\sqrt{5} - 15 . 3\sqrt{5} + 5\sqrt{5}\)
\(= 40\sqrt{5} - 45\sqrt{5} + 5\sqrt{5}\)
\(= (40 - 45 + 5)\sqrt{5}\)
\(=0\sqrt{5}=0\)
d. \(\left( \sqrt{325} - \sqrt{117} + 2\sqrt{208} \right) : \sqrt{13}\)
\(= \sqrt{325} : \sqrt{13} - \sqrt{117} : \sqrt{13} + 2\sqrt{208} : \sqrt{13}\)
\(= \sqrt{\frac{325}{13}} - \sqrt{\frac{117}{13}} + 2\sqrt{\frac{208}{13}}\)
\(= \sqrt{25} - \sqrt{9} + 2\sqrt{16}\)
\(=5-3+2.4=10\)
\(e.\frac{2\sqrt{8} - \sqrt{12}}{\sqrt{18} - \sqrt{48}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{30} + \sqrt{162}}=\frac{2\sqrt{4 . 2} - \sqrt{4 . 3}}{\sqrt{9 . 2} - \sqrt{16 . 3}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{6 . 5} + \sqrt{81 . 2}}\)
\(=\frac{2 . 2\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}.\sqrt{5} + 9\sqrt{2}}=\frac{4\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} - 4\sqrt{3}}.\frac{\sqrt{5} + 3\sqrt{3}}{\sqrt{6}(\sqrt{5} + 3\sqrt{3})}\)
\(=\frac{2(2\sqrt{2} - \sqrt{3})}{3\sqrt{2} - 4\sqrt{3}}.\frac{1}{\sqrt{6}}=\frac{2(2\sqrt{2} - \sqrt{3})}{\sqrt{6}(3\sqrt{2} - 4\sqrt{3})}\)
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