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\(a,ĐK:x\ne0;x\ne-5\\ b,A=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\\ A=\dfrac{x^2+4x-5}{2\left(x+5\right)}=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}=\dfrac{x-1}{2}\\ c,A=1\Leftrightarrow x-1=2\Leftrightarrow x=3\left(tm\right)\)

\(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\\ ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

a) ĐKXĐ: \(x^2-2x+1\ge0\Rightarrow\left(x-1\right)^2\ge0\left(luônđúng\right)\)

\(Q=2x-\sqrt{x^2-2x+1}=2x-\sqrt{\left(x-1\right)^2}=2x-x+1=x+1\)

b) \(Q=x+1\\ \Rightarrow7=x+1\\ \Rightarrow x=6\)

a: Ta có: \(Q=2x-\sqrt{x^2-2x+1}\)

\(=2x-\left|x-1\right|\)

\(=\left[{}\begin{matrix}2x-x+1=x+1\left(x\ge1\right)\\2x+x-1=3x-1\left(x< 1\right)\end{matrix}\right.\)

b: Ta có: Q=7

nên \(\left[{}\begin{matrix}x+1=7\left(x\ge1\right)\\3x-1=7\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{8}{3}\left(loại\right)\end{matrix}\right.\)

a) \(Q^2=2x-\sqrt{\left(x^2+2x+1\right)}\)

\(Q^2=2x-\sqrt{\left(x+1\right)^2}\)

\(Q^2=2x-x-1\)

\(Q^2=x-1\)

\(Q=\sqrt{x-1}\)

b) Để Q=9 thì \(\sqrt{x-1}=9\)  => \(x=82\)

Mình lộn câu b) Để Q=7 =>\(\sqrt{x-1}\)=7 => x=50

a: Ta có: \(Q=2x-\sqrt{x^2+2x+1}\)

\(=2x-\left|x+1\right|\)

\(=\left[{}\begin{matrix}2x-x-1=x-1\left(x\ge-1\right)\\2x+x+1=3x+1\left(x< -1\right)\end{matrix}\right.\)

b: Ta có: Q=7

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\left(x\ge-1\right)\\3x+1=7\left(x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(loại\right)\\x=2\left(loại\right)\end{matrix}\right.\)

giúp mình mọi người ơi

Nhìn mãi mới hiểu cái đề bài @-@

`a)đk:` $\begin{cases}\sqrt{x^2-2x} \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} ne 0\\\end{cases}$
`<=>` $\begin{cases}x \ge 2\,or\,x<0\\x \ne 0\end{cases}$
`b)A=(x+sqrt{x^2-2x})/(x-sqrt{x^2-2x})-(x-sqrt{x^2-2x})/(x+sqrt{x^2+2x})`
`=((x+sqrt{x^2-2x})^2-(x-sqrt{x^2-2x})^2)/((x+sqrt{x^2-2x})(x-sqrt{x^2-2x}))`
`=(x^2+x^2-2x+2sqrt{x^2-2x}-x^2-x^2+2x+2sqrt{x^2-2x})/(x^2-x^2+2x)`
`=(4sqrt{x^2-2x})/(2x)`
`=(2sqrt{x^2-2x})/x`
`c)A<2`
`<=>2sqrt{x^2-2x}<2x`
`<=>sqrt{x^2-2x}<x(x>=2)`(BP 2 vế thì x>=2)
`<=>x^2-2x<x^2`
`<=>2x>0`
`<=>x>0`
`<=>x>=2`
Vậy `x>=2` thì `A<2`.

1.

a. ĐKXĐ : x lớn hơn hoặc bằng 1/2 

b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)

= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)

=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)

= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)

Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)

\(\Rightarrow A=2\)

Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)

Do đó : A= \(\sqrt{4x-2}\)

Vậy ............

2. 

a. \(x\ge2\)hoặc x<0

b. A= \(2\sqrt{x^2-2x}\)

c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)

\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)

Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)

Vậy...........

a: \(M=\dfrac{x^2\left(x-2\right)}{x-2}+\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2+x+1\)

b: Để M=7 thì (x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(loại)

Vậy: x=-3