a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)

\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)

\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)

\(=\frac{x^2+4x+4}{x^2}\)

\(\left(\frac{x+2}{x}\right)^2\)

=>phép chia = 1 với mọi x # 0 và x#-1

b)Cm tương tự

khó quá

Ta thấy A gồm có 99 số hạng nên ta nhóm mỗi nhóm 3 số hạng.

Ta có: A = 1 + 5 + 5² + 5³ + 5⁴ + 5⁵ +...+ 5⁹⁷ + 5⁹⁸ + 5⁹⁹

             = (1 + 5 + 5² )+ (5³ + 5⁴ + 5⁵ )+...+( 5⁹⁷ + 5⁹⁸ + 5⁹⁹ )

             =(1 + 5 + 5² )+ 5³(1 + 5 + 5² ) +...+ 5⁹⁷(1 + 5 + 5² )

             = ( 1 + 5 + 5²)(1 + 5³+....+5⁹⁷)

             = 31(1 + 5³+....+5⁹⁷)

Vì có một thừa số là 31 nên A chia hết cho 31.

 P/s Đừng để ý câu trả lời của mình

Bài làm

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

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

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

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

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

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

b) Thay \(x=\frac{1}{2}\)vào P ta được:

\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}\)

\(P=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{2}{2}}\)

\(P=\frac{3}{2}:\frac{-1}{2}\)

\(P=\frac{3}{2}.\left(-2\right)\)

\(P=-3\)

Vậy giá trị của \(P=-3\) tại \(x=\frac{1}{2}\)

a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)

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

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

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

\(\Leftrightarrow P=\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{x+1}\)

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

Vậy \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)

b) Ta có \(P=\frac{x+1}{x-2}\left(x\ne-1;x\ne\pm2\right)\)

Thay x=\(\frac{1}{2}\left(tm\right)\)vào P ta có:

\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}-\frac{4}{2}}=\frac{\frac{3}{2}}{\frac{-3}{2}}=\frac{3}{2}:\frac{-3}{2}=-1\)

Vậy \(P=-1\)khi x=\(\frac{1}{2}\)

😎😎😎😎😎😎😎

Câu 1 : 

a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)

b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)

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

Bài 2 : 

a, Với \(x\ne\pm2\)

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

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

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)

b, Thay x = -4 vào biểu thức trên ta được : 

\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)

c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

b) Ta có :

\(\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x}=-1\)

\(\frac{1}{x}=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{4x^2-11x+10+x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{5x^2-11x+6}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{5x^2-5x-6x+6}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{5x\left(x-1\right)-6\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\frac{1}{x}=\frac{\left(x-1\right)\left(5x-6\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=x\left(x-1\right)\left(5x-6\right)\)

\(\Leftrightarrow x=\frac{\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(5x-6\right)}\)

a) \(\frac{1}{x+2}+\frac{2}{x+3}=\frac{6}{x+4}\)

ĐKXĐ \(x\ne-2,-3,-4\)

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

=> \(\frac{3x+7}{\left(x+2\right)\left(x+3\right)}-\frac{6}{x+4}=0\)

=> \(\frac{\left(3x+7\right)\left(x+4\right)-6\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x+4\right)}=0\)

=> (3x + 7)(x + 4) - 6(x² + 5x + 6) = 0

=> 3x² + 19x + 28 - 6x² - 30x - 36 = 0

=> -3x² - 11x - 8 = 0

=> -3x² - 3x - 8x - 8 = 0

=> -3x(x + 1) - 8(x + 1) = 0

=> (x + 1)(-3x - 8) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-\frac{8}{3}\end{cases}}\)

Vậy ...

b) Thiếu dữ liệu cuả đề 

c) \(\frac{6x+22}{x+2}-\frac{2x+7}{x+3}=\frac{x+4}{x^2+5x+6}\)

ĐKXĐ \(x\ne-2;-3\)

=> \(\frac{\left(6x+22\right)\left(x+3\right)-\left(x+2\right)\left(2x+7\right)}{\left(x+2\right)\left(x+3\right)}=\frac{x+4}{\left(x+2\right)\left(x+3\right)}\)

=> \(6x^2+40x+66-x\left(2x+7\right)-2\left(2x+7\right)=x+4\)

=> \(6x^2+40x+66-2x^2-7x-4x-14=x+4\)

=> 4x² + 29x + 52 = x + 4

=> 4x² + 29x + 52 - x - 4 = 0

=> 4x² + 28x + 48 = 0

=> 4(x² + 7x + 12) = 0

=> x² + 7x +12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 3)(x + 4) = 0

=> \(\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\) 

Mà \(x\ne-2,-3\)nên x = -3 loại

Vậy x = -4

\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)

\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow27x-2x-4x-27+2=0\)

\(\Leftrightarrow21x=25\)

\(\Leftrightarrow x=\frac{25}{21}\)

Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !

\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)

\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)

\(\Leftrightarrow-20x-12=56\)

\(\Leftrightarrow-20x=68\)

\(\Leftrightarrow x=-\frac{17}{5}\)

Tự check lại nhá

\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

Ta dễ thấy \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}< 0\) nên

\(x+95=0\Leftrightarrow x=-95\)