\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

xin lỗi ,em mới lớp7 , khó wá đối zới em rồi

Em moi lop 9bang anh ,nhung kho qua

\(\left|x-5\right|=2x\)ĐK : x>=0 

TH1 : x - 5 = 2x <=> x = -5 ( loại )

TH2 : x - 5 = -2x <=> 3x = 5 <=> x = 5/3 ( tm )

Vậy tập nghiệm pt là S = { 5/3 } 

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

\(\Leftrightarrow x^2-4x+4+2x-2-x^2-4\le0\)

\(\Leftrightarrow-2x-2\le0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)

Vậy tập nghiệm bft là S = { x | x > = -1 } 

Ta có: \(\left|x-5\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x\left(x\ge5\right)\\x-5=-2x\left(x< 5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=5\\x+2x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=5\\3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

`sqrt{x^{3}+1}=(2x^{2}+4)/5`

`dk:x>=-1`

`pt<=>5sqrt{(x+1)(x^{2}-x+1)}=2x^{2}-2x+2+2x+2`

Đặt `a=sqrt{x+1},b=sqrt{x^{2}-x+1}`

`pt<=>5ab=2a^{2}+2b^{2}`

`<=>2a^{2}-5ab+2b^{2}=0`

`<=>(a-2b)(b-2a)=0`

Đến đây bạn chia 2 th rồi giải thôi!

`(3x-1)/(x-1)-(2x+5)/(x+3)+4/(x^2+2x-3)=1(x ne 1,-3)`

`<=>((3x-1)(x+3))/(x^2+2x-3)-((2x+5)(x-1))/(x^2+2x-3)+4/(x^2+2x-3)=(x^2+2x-3)/(x^2+2x-3)`

`<=>(3x-1)(x+3)-(2x+5)(x-1)+4=x^2+2x-3`

`<=>3x^2+8x-3-2x^2-3x+5+4=x^2+2x-3`

`<=>x^2+5x+6=x^2+2x-3`

`<=>3x=-9`

`<=>x=-3(loại)`

Vậy `S={cancel0}`

ĐKXĐ: \(x\notin\left\{1;-3\right\}\)

Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

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

\(\Leftrightarrow\dfrac{3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{3x^2+8x-3-\left(2x^2+3x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{3x^2+8x+1-2x^2-3x+5}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

Suy ra: \(x^2+5x+6-x^2-2x+3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\)

hay x=-3(Không nhận)

Vậy: \(S=\varnothing\)

ĐKXĐ: ` x ne 1 ; x ne 4`

`(2x+1)/(x^2-5x+4) + 5/(x-1) = 2/(x-4)`

`<=> (2x+1)/[(x-1)(x-4)] + [5(x-4)]/[(x-1)(x-4)] = [2(x-1)]/[(x-1)(x-4)]`

`=> 2x+1 + 5x -20 = 2x-2`

`<=> 5x = 17`

`<=> x= 17/5`(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của phương trình là `S={ 17/5}`

\(ĐK:x\ne0;x\ne-1\)

\(\dfrac{2x-5}{x}+\dfrac{4}{x+1}=2\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x+1\right)+4x}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)

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

\(\Leftrightarrow2x^2+2x-5x-5+4x=2x^2+2x\)

\(\Leftrightarrow-5x+4x+2x-2x=5\)

\(\Leftrightarrow-x=5\Leftrightarrow x=-5\left(tm\right)\)

\(\dfrac{2x-5}{x}+\dfrac{4}{x+1}=2\)  đkxđ : x khác 0 , x khác -1

\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{4x}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{2x^2+2x-5x-5}{......}+\dfrac{4x}{.....}-\dfrac{2x^2+2x}{....}=0\)

\(\Leftrightarrow2x^2+2x-5x-5+4x-2x^2-2x=0\)

\(\Leftrightarrow-5x-5=0\)

\(\Leftrightarrow x=\left(0+5\right):-5=-1\left(ktm\right)\)

Vậy pt....

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)