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a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
b) Ta có :
\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)
\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)
\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)
Vì a > 0,5 <=> a > 0 <=> | a | = a
Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a
<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )
= -1 + 2a = 2a - 1
Thay vào trên ta được :
\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)
Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)
b, \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}\)
\(=\frac{2\sqrt{5}a\left(2a-1\right)}{2a-1}=2\sqrt{5}a\)
a) 2x2−y2√3(x+y)222x2−y23(x+y)22
=2x2−y2√32.(x+y)2=2x2−y232.(x+y)2
=2x2−y2√(x+y)2.√32=2x2−y2(x+y)2.32
=2x2−y2
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a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2}{\left(x+y\right)\left(x-y\right)}.\dfrac{\sqrt{3}.\left(x+y\right)}{\sqrt{2}}=\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}.\sqrt{5}a.\left(2a-1\right)=2\sqrt{5}a\)
a)\(\dfrac{\sqrt{6}}{x-y}\)
b)\(2a\sqrt{5}\)
a, \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}với}x\ge0,y\ge0\)\(vàx\ne y\)
\(=\dfrac{\sqrt{6}}{(x-y)}\)
b, \(=2a\sqrt{5}\)
a) \dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y2223(x+y)2
=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y2223.(x+y)2
=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22(x+y)2.23
=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y
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a căn 6 trên (x-y)
b 2a căn 5
b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5\left(1-2.2a+\left(2a\right)^2\right)}=\dfrac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=\dfrac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)
a) \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\)
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{\left|x+y\right|}{x^2-y^2}\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\sqrt{6}=\dfrac{1}{x-y}\sqrt{6}\) b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2\left|a\right|}{2a-1}\sqrt{5.\left(1-2a\right)^2}=\dfrac{2a\left|1-2a\right|}{2a-1}\sqrt{5}=2a\sqrt{5}\)
2a căn5
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)=\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}\left(x+y\right)^2}\) =\(\dfrac{2}{x^2-y^2}\sqrt{\left(x+y\right)^2}.\sqrt{\dfrac{3}{2}}\)=\(\dfrac{2}{x^2-y^2}.\left|x+y\right|\sqrt{\dfrac{3}{2}}\)= \(\dfrac{\left|x+y\right|}{x^2-y^2}.\sqrt{\dfrac{3.2^2}{2}}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}.\sqrt{6}\) (do x\(\ge\)0 y\(\ge\)0 \(\Rightarrow x+y\ge\)nên\(\left|x+y\right|=x+y\)) b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)=\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-2.1.2a+\left(2a\right)^2\right)}\)
a)\(\dfrac{\sqrt{6}}{x-y}\)
b)\(2a\sqrt{5}\)
a= 2/(x-y) . (x+y) .\(\sqrt{ }\)3(x+y)^2.2/2.2 = 2/(x+y).(x-y) . \(\sqrt{ }\)6(x+y)^2/\(\sqrt{ }\)4 = 2/(x+y). (x-y) . \(\sqrt{ }\)6(x+y)/2 = \(\sqrt{ }\)6/x-y b = 2/2a-1 . \(\sqrt{ }\)5a^2 . (1-2a)^2 = 2/2a-1 . \(\sqrt{ }\)5 . \(\sqrt{ }\)a^2 . \(\sqrt{ }\)(1-2a)^2 = 2/2a-1 . \(\sqrt{ }\)5 . a . (2a-1) = 2\(\sqrt{ }\)5-a
a)\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\cdot\left(x+y\right)^2}{2}}\) (voi \(x\ge0;x\ne y\) ) =\(\dfrac{2}{x^2-y^2}\cdot\dfrac{\sqrt{3}\cdot\sqrt{\left(x+y\right)^2}}{\sqrt{2}}\) =\(\dfrac{2}{\left(x-y\right)\cdot\left(x+y\right)}\cdot\dfrac{\sqrt{3}\cdot\left|x+y\right|}{\sqrt{2}}\) =\(\dfrac{2\sqrt{3}}{\sqrt{2}\cdot\left(x-y\right)}\) =\(\sqrt{\dfrac{12}{2}}\cdot\dfrac{1}{x-y}\)
=\(\sqrt{6}\cdot\dfrac{1}{x-y}\)
b)\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-4a+4a^2\right)}\) ( Voi a>0,5 )
= \(\dfrac{2}{2a-1}\cdot\sqrt{5a^2\cdot\left(1-2a\right)^2}\)
=\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left|1-2a\right|\)
=\(\dfrac{2}{2a-1}\cdot\sqrt{5a^2}\cdot\left(2a-1\right)\)
= \(2\sqrt{5a^2}\)
= \(2a\sqrt{5}\)
a) 2x2−y23(x+y)22\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y2223(x+y)2
=2x2−y232.(x+y)2=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y2223.(x+y)2
=2x2−y2(x+y)2.32=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22(x+y)2
.23
=2x2−y2.∣x+y∣32=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y22.∣x+y∣23
=∣x+y∣x2−y23.222=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}=
a) \dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}x2−y2223(x+y)2
=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}=x2−y2223.(x+y)2
=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}=x2−y22(x+y)2.23
=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}=x2−y<...
a) \(\dfrac{2}{x^2-y^2}\)\(\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)
= \(\dfrac{2}{\left(x-y\right).\left(x+y\right)}\)\(\sqrt{\dfrac{3}{2}.\left(x+y\right)^2}\)
= \(\dfrac{2}{\left(x+y\right).\left(x-y\right)}\).|x+y| \(\sqrt{\dfrac{3}{2}}\)
= \(\dfrac{x+y}{\left(x-y\right).\left(x+y\right)}\) .\(\sqrt{\dfrac{3.2^2}{2}}\)
= \(\dfrac{\sqrt{6}}{x-y}\)
b) \(\dfrac{2}{2a-1}\) \(\sqrt{5a^2\left(1-4a+4a^2\right)}\)
= \(\dfrac{2}{2a-1}\)\(\sqrt{5a^2\left(1-2a\right)^2}\)
=\(\dfrac{2}{2a-1}\)\(\sqrt{5}\).\(\sqrt{a^2}\).\(\sqrt{\left(1-2a\right)^2}\)
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\). |a| . |1-2a|
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\) . |a| .|1-2a|
= \(\dfrac{2}{2a-1}\)\(\sqrt{5}\).a.( 2a-1) ( vì a > 0,5 )
= 2a\(\sqrt{5}\)
=
a) 2x2−y2√3(x+y)222x2−y23(x+y)22
=2x2−y2√32.(x+y)2=2x2−y232.(x+y)2
=2x2−y2√(x+y)2.√32=2x2−y2(x+y)2.32
=2x2−y2
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Bài 30 (trang 19 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau:
a) $\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}$ với $x>0,y \ne 0$ ; b) $2y^2.\sqrt{\dfrac{x^4}{4y^2}}$ với $y<0$ ;
c) $5xy.\sqrt{\dfrac{25x^2}{y^6}}$ với $x<0$,$y>0$; d) $0,2x^3y^3.\sqrt{\dfrac{16}{x^4y^8}}$ với $x \ne 0, y\ne 0$.
(Vì x > 0 nên |x| = x; y2 > 0 với mọi y ≠ 0)
(Vì x2 ≥ 0 với mọi x; và vì y < 0 nên |2y| = – 2y)
(Vì x < 0 nên |5x| = – 5x; y > 0 nên |y3| = y3)
(Vì x2y4 = (xy2)2 > 0 với mọi x ≠ 0, y ≠ 0)
a) 1/y
b) - x^2 y
c) -25x^2 / y^2
d) 4x/5y
Bài 1: Cho A=\(\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)với x≥0; y≥0; x≠y
a) Rút gọn A
b) Chứng minh A≥0
Bài 2:Cho A= \(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
với x>0; x≠1
a) Rút gọn A
b)Tìm x để A=6
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
1.giải hệ phương trình:
\(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)
2.Rút gọn biểu thức
\(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\) với x\(\ge\)0;x\(\ne\)4
1) Ta có: \(\left\{{}\begin{matrix}2x-y=3\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=3\\x=-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x=-1\end{matrix}\right.\)
Vậy: (x,y)=(1;-1)
2) Ta có: \(A=\dfrac{x+20}{x-4}+\dfrac{2}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}-2}\)
\(=\dfrac{x+20+2\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+20+2\sqrt{x}-4-6\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
Bài 61 (trang 33 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\dfrac{3}{2} \sqrt{6}+2 \sqrt{\dfrac{2}{3}}-4 \sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}$;
b) $\left(x \sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2 x}{3}}+\sqrt{6 x}\right): \sqrt{6 x}=2 \dfrac{1}{3} $ với $x>0$.
a) -17√3/3 b) 11√6
c) 21 d) 11
a) a) Biến đổi vế trái thành 32√6+23√6−42√6326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (√6x+13√6x+√6x):√6x(6x+136x+6x):6x và làm tiếp
bài 2: rút gọn
a) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) (x\(\ge\) 0
c) \(\dfrac{x-y}{\sqrt{y}-1}\sqrt{\dfrac{\left(y-2\sqrt{x}+1\right)^2}{\left(x-1\right)^4}}\) x \(\ne\) 1; y \(\ne\)1 ; y \(\ge\)0
Đề câu c co bị sai ko vậy bạn? (y - 2\(\sqrt{x}\) +1)
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
Cho biểu thức A= \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)( với x≥0, x≠4)
a. Rút gọn A
b. Tìm x để A>0
a/ \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{-1}{\sqrt{x}-2}\)
b/ \(A>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0\)
Ta thấy: - 1 < 0 nên để A > 0 thì:
\(\sqrt{x}-2< 0\)\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
kết hợp với đkxđ: => \(0\le x< 4\)
A=\(2\sqrt{12}-\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
B=\(\dfrac{x}{x-16}+\dfrac{2}{\sqrt{x}-4}+\dfrac{2}{\sqrt{x}+4}\)( Với x\(\ge\)0; x\(\ne\)16)
a) Rút gọn 2 biểu thức A, B
b) Tìm giá trị của x để B\(-\dfrac{1}{2}\)A=0
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???
Bài tập: Chứng minh
a,\(\left(\sqrt{3}-\sqrt{2}\right).\sqrt{5+2\sqrt{6}}=1\)
b,\(\left[\dfrac{\sqrt{x}-\sqrt{y}}{x-y}+\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right].\dfrac{\sqrt{xy}+1}{\sqrt{x+\sqrt{y}}}\) (với x\(\ge\) 0; y\(\ge\) 0; x\(\ne\)y)
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
Rút gọn : a) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
b)\(\dfrac{x+4y-4\sqrt{xy}}{\sqrt{x}-2\sqrt{y}}+\dfrac{y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x\ge0;y\ge0;x\ne4y\right)\)
c)\(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
d)\(\dfrac{9-x}{\sqrt{3x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}\)
e)\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\)
g)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)với\) a, b \(\ge\)0 , a \(\ne\)9; b\(\ne\)25
Mọi người giúp tớ với , cảm ơn nhiều nhiều ạ !!
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: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)