Ai giải giúp mình bài 1 với bài 4 trước đi

a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)

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

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

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

b, Thay x = -2 ta được : 

\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)

Vậy A nhận giá trị 1/3 

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

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

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

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

\(=\frac{x+1}{x-1}\)

Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)

b. Sử dụng các hằng đẳng thức

 \(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)

và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Do (a - b) + (b - c) + (c - a) =  0 nên áp dụng hđt  \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:

\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)

Bài 1 :

\(b,ax^2+3ax+9=a^2\) 

\(\Leftrightarrow a^2x+3ax+9-a^2=0\) 

\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\) 

\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)

Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\) 

\(\Leftrightarrow ax=a-3\) 

Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\) 

c.Ta có \(\left(\frac{1}{x}-\frac{1}{y}-\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xz}-\frac{2}{xy}+\frac{2}{yz}=1\)

Do x = y + z nên \(\frac{-2}{xz}-\frac{2}{xy}+\frac{2}{yz}=\frac{-2y-2z+2\left(y+z\right)}{\left(y+z\right)zy}=0\)

Vậy nên \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1.\)

a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)

ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)

    = \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

    = \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)

    = \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)

b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)

                       <=> \(\frac{x^2+1}{x+1}+1>0\)

                        <=> \(\frac{x^2+x+2}{x+1}>0\)

Vì x² + x + 2 >0 \(\forall x\)

=> A > 0 <=> x + 1 > 0 <=> x > -1

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)

Bài 1 

ĐK \(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)

 A =\(\left(\frac{x^2-x+7}{\left(x+2\right)\left(x-2\right)}+\frac{1}{x+2}\right):\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{2x}{\left(x+2\right)\left(x-2\right)}\right)\)

\(=\frac{x^2-x+7+x-2}{\left(x+2\right)\left(x-2\right)}:\frac{x^2+4x+4-x^2+4x-4-2x}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2+5}{\left(x+2\right)\left(x-2\right)}.\frac{\left(x+2\right)\left(x-2\right)}{6x}=\frac{x^2+5}{6x}\)

b , \(A=1\Rightarrow\frac{x^2+5}{6x}=1\Rightarrow x^2-6x+5=0\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}\left(tm\right)}\)

Vậy x=1 hoặc  x=5

Bài 2.

a. \(B=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2+x\right)\left(2-x\right)}:\frac{x+3}{2-x}\)

\(=\frac{4x^2+8x}{\left(2+x\right)\left(2-x\right)}.\frac{2-x}{x+3}=\frac{2x}{x+3}\)

b.  \(B=\frac{2x}{x+3}=2-\frac{6}{x+3}\)

B nguyên \(\Leftrightarrow x+3\inƯ\left(-6\right)\Rightarrow x+3\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

\(\Rightarrow x\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\)

Vậy \(x\in\left\{-9;-6;-5;-4;-2;-1;0;3\right\}\)thì B nguyên

a) ĐKXĐ : x \(\ne-2;x\ne1;x\ne0\)

\(A=\left(\frac{x}{x+2}-\frac{4}{x^2+2x}\right):\left(\frac{x^2-2x+1}{x^2-x}\right)=\left(\frac{x}{x+2}-\frac{4}{x\left(x+2\right)}\right):\left(\frac{\left(x-1\right)^2}{x\left(x-1\right)}\right)\)

\(=\frac{x^2-4}{x\left(x+2\right)}:\frac{x-1}{x}=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x}{x-1}=\frac{x-2}{x}.\frac{x}{x-1}=\frac{x-2}{x-1}\)

b) Để A > 1 

=> \(\frac{x-2}{x-1}>1\)

=> \(\frac{x-2}{x-1}-1>0\Rightarrow\frac{-1}{x-1}>0\Rightarrow x-1< 0\Rightarrow x< 1\)

Vậy để A > 1 thì x < 1 và x \(\ne-2;x\ne1;x\ne0\)

c) Ta có \(A=\frac{x-2}{x-1}=\frac{x-1-1}{x-1}=1-\frac{1}{x-1}\)

Để A \(\inℤ\Rightarrow\frac{1}{x-1}\inℤ\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)\Rightarrow x-1\in\left\{1;-1\right\}\)

Khi x - 1 = 1 => x = 2( tm)

Khi x - 1 =-1 => x = 0 (loại) 

Vậy x = 2 thì A nguyên