a) Ta có : 

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

\(\Leftrightarrow3x=3x+2\)

\(\Leftrightarrow0=2\) ( vô lí )

Do đó pt đã cho vô nghiệm

b) Ta có  \(\left|x\right|=-x^2-2\) (1)

Nhân xét : VT (1) : \(\left|x\right|\ge0\forall x\)

VP (1) : \(-x^2\le0\Leftrightarrow-x^2-2\le-2\forall x\)

Do đó : \(VT\ne VP\)

Vì vậy pt đã cho vô nghiệm

Lời giải:

$M=\frac{2x^2-3x+3}{x-2}=\frac{(2x^2-4x)+(x-2)+5}{x-2}$

$=\frac{2x(x-2)+(x-2)+5}{x-2}=2x+1+\frac{5}{x-2}$

Với $x$ nguyên, để $M$ nguyên thì $\frac{5}{x-2}$ nguyên

$\Rightarrow x-2$ là ước của $5$ (do $x$ nguyên)

$\Rightarrow x-2\in\left\{5;-5;1;-1\right\}$

$\Rightarrow x\in\left\{7; -3; 3; 1\right\}$

cảm ơn cô

`a)ĐK:(x+1)(2x-6) ne 0`

`<=>(x+1)(x-3) ne 0`

`<=> x ne -1,x ne 3`

`b)C=(3x^2+3x)/((x+1)(2x-6))`

`=(3x(x+1))/((x+1)(2x-6))`

`=(3x)/(2x-6)`

`C=1`

`=>3x=2x-6`

`<=>x=-6(tm)`

Vậy `x=-6`

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

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

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

tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn

cảm ơn cậu giúp mk câu c với ạ

Câu 4 :

\(x^2+y^2-2\left(x-y-1\right)=0\)

\(\Leftrightarrow x^2+y^2-2x+2y+2=0\)

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

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

Do \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x;y\)

Dấu " = " xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Vậy \(x=1;y=-1\)

Làm c1,2,3 đi bạn

a: ĐKXĐ: x∉{2;-2;0;3}

b: \(P=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

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

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

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

c: |x-5|=2

=>x-5=2 hoặc x-5=-2

=>x=7(nhận) hoặc x=3(loại)

Thay x=7 vào P, ta được:

\(P=\frac{-4\cdot7^2\cdot\left(7-2\right)}{\left(7+2\right)\left(7-3\right)}=\frac{-4\cdot7^2\cdot5}{9\cdot4}=\frac{-7^2\cdot5}{9}=\frac{-245}{9}\)