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Rút gọn:
\(M=\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
M= \(\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
M=\(\dfrac{(x+2+\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}+\dfrac{(x+2-\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}\)
M=\(\dfrac{\left(x+2\right)^2+2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}+\dfrac{\left(x+2\right)^2-2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{\left(x+2\right)^2+x^2-4+\left(x+2\right)^2+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{x^2+4x+4+x^2-4+x^2+4x+4+x^2-4}{4x+8}\)
M= \(\dfrac{4x^2+8}{4x+8}\)
M= \(\dfrac{x.\left(4x+8\right)}{4x+8}\)
M= x
OK nha bn !!
a: ĐKXĐ: x>=0; x<>4
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\cdot\dfrac{\sqrt{x}-2+2}{2}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}}{2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)
TH1: M>=căn M
=>M^2>=M
=>M^2-M>=0
=>5*căn x-1>=0
=>x>=1/25 và x<>4
TH2: M<căn M
=>5căn x-1<0
=>x<1/25
Kết hợp ĐKXĐ, ta được: 0<=x<1/25
a: Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right)\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}:\dfrac{x+4\sqrt{x}+4+x-2\sqrt{x}-x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2\sqrt{x}+4}{1}\cdot\dfrac{\sqrt{x}+2}{x+2\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+4}{\sqrt{x}}\)
b: \(P-2=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ
nên P>2
Ta có: \(P=\left(\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{-\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+4}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
a) ĐK:\(x\ge0,x\ne4\)
\(P=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}+2\right)-2-5\sqrt{x}}{x-4}\)
\(=\dfrac{x\sqrt{x}+4x}{x-4}\)
b) ĐK: \(x\ge0,x\ne1\)
\(A=\dfrac{\sqrt{x}\left(x-1\right)+3\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+4-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+3x-\sqrt{x}-5}{\left(\sqrt{x}+3\right)\left(x-1\right)}\)
E = \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) = \(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
E = \(\sqrt{x}+1+\sqrt{x}\) = \(2\sqrt{x}+1\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}-4}{4-x}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{x-4}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+2\sqrt{x}+3\sqrt{x}+6-\left(x-2\sqrt{x}-\sqrt{x}+2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+5\sqrt{x}+6-x+2\sqrt{x}+\sqrt{x}-2-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{4\sqrt{x}+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4}{\sqrt{x}-2}\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4x+2\sqrt{x}-4}{\sqrt{x}^2-2^2}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(\sqrt{x}-2\right)-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-4-4\sqrt{x}-x+x-2\sqrt{x}-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\right)\)
\(P=\dfrac{-4x\left(\sqrt{x}\left(2-\sqrt{x}\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4x\left(-\sqrt{x}\left(\sqrt{x}-2\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{\sqrt{16x^3}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
Có j bạn xem lại coi có sai xót chỗ nào ko nhé, mk ko chắc là đúng 100% đâu.
cảm ơn c nhiều
điều kiện \(\left[{}\begin{matrix}x\ge2\\x< -2\end{matrix}\right.\)
ta có : \(M=\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
\(=\dfrac{\left(x+2+\sqrt{x^2-4}\right)^2+\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2-\sqrt{x^2-4}\right)\left(x+2+\sqrt{x^2-4}\right)}\)
\(=\dfrac{4x^2+8x}{4x+8}=x\)Mysterious Person giúp mk