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

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

b)\(m=\frac{x+2-\left(x-2\right)+x^2+4x}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)

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

Để m nguyên thì \(4⋮x-2\)

\(\Rightarrow x-2\in\left\{1,2,4,-1,-2,-4\right\}\)

\(\Leftrightarrow x\in\left\{3,4,6,1,0,-2\right\}\)

\(M=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4x}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow M=\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow M=\frac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow M=\frac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{x-2}\)

Để M có giá trị nguyên thì x+2 chia hết cho x-2

Ta có x+2=x-2+4

=> x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

Vì x nguyên => x-2 nguyên

=> x-2 thuộc Ư (4)={-4;-2;-1;1;2;4}

Ta có bảng

\(\Leftrightarrow y^2-3y+7-10+y=0\)

\(\Leftrightarrow y^2-2y+1-4=0\)

\(\Leftrightarrow\left(y-1\right)^2-4=0\)

\(\Leftrightarrow\left(y-1-4\right)\left(y-1+4\right)=0\)

\(\Leftrightarrow\left(y-5\right)\left(y+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=5\\y=-3\end{cases}}\)

Vậy ....

\(y^2-3y+7=10-y\)

\(y^2-3y+7-10+y=0\)

\(y^2-2y-3=0\)

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

\(y=3;-1\)

ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

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

\(=\frac{x-3}{2x}\left(1-\frac{6}{x-3}\right)\)

\(=\frac{x-3}{2x}.\frac{x-9}{x-3}=\frac{x-9}{2x}\)

\(M=\frac{\left(x-3\right)^2}{2x^2-6x}\left(1-\frac{6x+18}{x^2-9}\right)\left(x\ne\pm3;x\ne0\right)\)

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

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\left(1-\frac{6}{x-3}\right)\)

\(\Leftrightarrow M=\frac{x-3}{2x}\cdot\frac{x-9}{x-3}\)

\(\Leftrightarrow M=\frac{x-9}{2x}\)

Vậy với \(x\ne\pm3;x\ne0\)thì \(M=\frac{x-9}{2x}\)

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

a) Để A có nghĩa \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\2-2x^2\ne0\end{cases}}\Leftrightarrow x\ne\pm1\)

b) Ta có \(A=\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)

\(\Rightarrow2A=\frac{x}{x-1}+\frac{x^2+1}{1-x^2}=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{\left(x-1\right)\left(x+1\right)}\)

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

\(\Rightarrow A=\frac{1}{2x+2}\)

KL...

c) Để \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{1}{2x+2}=\frac{1}{2}\)

\(\Leftrightarrow2x+2=2\Leftrightarrow2x=0\Leftrightarrow x=0\)(t/m ĐKXĐ)

KL...

mk giải từng nha == tại vì mk sợ nhiều qus bị troll 

\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)

\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)

\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)

\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)

\(-7+18x^2-6x=x-4\)

\(3-18x^2+7x=0\)

\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)

\(9\left(2x+1\right)=4\left(x-5\right)^2\)

\(18x+9=4x^2-40x+100\)

\(18x+9-4x^2+40x-100=0\)

\(58x-91-4x^2=0\)

\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)

Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath

Tại sao lại có a?

tại đề bài gốc có a. ai biết được

a) \(\frac{8xy}{3x-1}:\frac{12xy^3}{5-15x}\)

\(=\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

\(=\frac{2}{3x-1}.\frac{5\left(1-3x\right)}{3y^2}\)

\(=\frac{10}{3y^2}\)

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

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

=.=, làm nhanh lẫn

a) \(=\frac{8xy}{3x-1}.\frac{5\left(3x-1\right)}{-12xy^3}\)

\(=\frac{-10}{3y^2}\)

Cô si hết lên!

\(a+1\ge2\sqrt{a}\)

\(b+1\ge2\sqrt{b}\)

\(c+1\ge2\sqrt{c}\)

\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)

Dấu "=" khi a = b = c = 1

Áp dụng bdt AM-GM cho 2 số dương a và 1 ta được:

\(a+1\ge2\sqrt{a}\)

tương tự ta có: \(b+1\ge2\sqrt{b}\);\(c+1\ge2\sqrt{c}\)

Suy ra \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(do \(abc=1\Rightarrow\sqrt{abc}=1\))

Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)(đpcm)

Ta có: \(x^2y-xy^2+y^2z-yz^2+xz^2-x^2z=xy\left(x-y\right)-z\left(x^2-y^2\right)+z^2\left(x-y\right)\)

\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)

\(=\left(x-y\right)\left(x\left(y-z\right)-z\left(y-z\right)\right)=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)