\(ĐK:x\ne\pm1\)

\(\Leftrightarrow\frac{ax^2-x+ax-1+bx-b}{x^2-1}=\frac{a\left(x^2+1\right)}{x^2-1}\)

\(\Leftrightarrow\frac{ax^2+x\left(a-1+b\right)-b-1}{x^2-1}=\frac{ax^2+a}{x^2-1}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ne\pm1\\a+b-1=0\\-b-1=a\end{cases}}\)

Giải ra :D

đề bài là gì??!!!!

tính ớ bạn

A = \(\frac{1}{x}\left(\frac{x^2-xy}{x+y}\right)^2\left[\frac{x+y}{\left(x-y\right)^2}+\frac{x+y}{xy-y^2}\right]-\frac{x}{x+y}\)

A = \(\frac{1}{x}\left(\frac{x^2-xy}{x+y}\right)^2\left[\frac{x+y}{\left(x-y\right)^2}+\frac{x+y}{y\left(x-y\right)}\right]-\frac{x}{x+y}\)

A = \(\frac{1}{x}\left[\frac{x\left(x-y\right)}{x+y}\right]^2\left[\frac{y\left(x+y\right)+\left(x-y\right)\left(x+y\right)}{y\left(x-y\right)^2}\right]-\frac{x}{x+y}\)

A = \(\frac{1}{x}\cdot\frac{x^2\left(x-y\right)^2}{\left(x+y\right)^2}\left[\frac{xy+y^2+x^2-y^2}{y\left(x-y\right)^2}\right]-\frac{x}{x+y}\)

A = \(\frac{x\left(x-y\right)^2}{\left(x+y\right)^2}\cdot\frac{x\left(x+y\right)}{y\left(x-y\right)^2}-\frac{x}{x+y}\)

A = \(\frac{x^2}{y\left(x+y\right)}-\frac{x}{x+y}\)

A = \(\frac{x^2-xy}{y\left(x+y\right)}\)

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                                                   Bài giải

\(\left(x-3\right)2-\left(x-3\right)\left(x+3\right)=0\)

\(\left(x-3\right)\left(x+3-2\right)=0\)

\(\left(x-3\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{3\text{ ; }-1\right\}\)

\(\left(x-3\right).2-\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left(x-3\right)\left[2-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(2-x-3\right)=0\Leftrightarrow\left(x-3\right)\left[\left(-1\right)-x\right]\). Xét 2 trường hợp

Xét 2 trường hợp. \(TH1:x-3=0\Leftrightarrow x=0+3=3\)

\(TH2:\left(-1\right)-x=0\Leftrightarrow x=\left(-1\right)-0=-1\). Vậy \(x\in\left\{-1;3\right\}\)

\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)

<=> \(\frac{4}{\left(x-1\right)\left(x-2\right)}-\frac{3}{2x^2-6x+1}+1=0\)

<=> 4(2x² - 6x + 1) - 3(x - 1)(x - 2) + (x - 1)(x - 2)(2x² - 6x + 1) = 0

<=> 28x² - 30x + 2x⁴ - 12x³ = 0

<=> 2x(14x - 15 + x² - 6x²) = 0

<=> 2x(x² - 3x + 5)(x - 3) = 0

vì x² - 3x + 5 khác 0 nên:

<=> 2x = 0 hoặc x - 3 = 0

<=> x = 0 hoặc x = 3

\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\)

\(\Leftrightarrow\frac{2x^4-12x^3+28x^2-30x}{2x^4-12x^3+28x^2-15x+2}=0\)

\(\Leftrightarrow2x^4-12x^3+28x^2-30x=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x^2-3x+5\right)=0\)

mà  \(x^2-3x+5\) khác 0 

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vì a<b => 3a < 3b => 3a +4 < 3b+4 < 3b+4+1 = 3b+5

nên 3a+4 < 3b+5

Ta có:a<b =>3a<3b     (1)

                     4<5        (2)

Cộng vế với vế của (1) vs (2) ta được: 3a+4<3b+5 (đpcm)

\(\left(2x-1\right)^2=2x\left(2x+3\right)\)

=>\(4x^2-4x+1=4x^2+6x\)

=>\(4x^2-4x+1-4x^2-6x=0\)

=>\(-10x+1=0=>x=\frac{1}{10}\)

                                                      Bài giải

\(\left(2x-1\right)^2=2x\left(2x+3\right)\)

\(\left(2x-1\right)^2=2x\left(2x-1+4\right)\)

\(\left(2x-1\right)^2=2x\left(2x-1\right)+8x\)

\(\left(2x-1\right)^2-2x\left(2x-1\right)=8x\)

\(\left(2x-1-2x\right)\left(2x-1\right)=8x\)

\(-2x+1=8x\)

\(8x+2x=1\)

\(10x=1\)

\(x=\frac{1}{10}\)