+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)

\(=x^3+125-x^3+8x^2-16x+16x\)

\(=8x^2+125\)

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

a)

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

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

<=> x = - 1
Vậy S = {- 1}

b)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)

<=> \(\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)

<=> (x + 5)² - (x - 5)² = 20

<=> (x + 5 - x + 5)(x + 5 + x - 5) = 20

<=> 10 . 2x = 20

<=> x = 20 : 20

<=> x = 1

Vậy S = {1}

c)

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

<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x}{2\left(x-3\right)\left(x+1\right)}\)

<=> x(x + 1) + x(x - 3) = 2x

<=> x² + x + x² - 3x - 2x = 0

<=> 2x² - 4x = 0

<=> 2x(x - 2) = 0

<=> \(\left[\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy S = {0; 2}

\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

d, ĐKXĐ:\(x\ne-2,x\ne3\)

\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)

Câu 1 :

a, Ta có : \(x^2-10x=-25\)

=> \(x^2-10x+25=0\)

=> \(\left(x-5\right)^2=0\)

=> \(x-5=0\)

=> \(x=5\)

Vậy phương trình có nghiệm là x = 5 .

b, Ta có : \(5x\left(x-1\right)=x-1\)

=> \(5x\left(x-1\right)-x+1=0\)

=> \(5x\left(x-1\right)-\left(x-1\right)=0\)

=> \(\left(5x-1\right)\left(x-1\right)=0\)

=> \(\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{1}{5}\\x=1\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 1, x = \(\frac{1}{5}.\)

c, Ta có : \(2\left(x+5\right)-x^2-5x=0\)

=> \(2\left(x+5\right)-x\left(x+5\right)=0\)

=> \(\left(2-x\right)\left(x+5\right)=0\)

=> \(\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 2, x = -5 .

d, Ta có : \(x^2-2x-3=0\)

=> \(x^2-3x+x-3=0\)

=> \(x\left(x+1\right)-3\left(x+1\right)=0\)

=> \(\left(x-3\right)\left(x+1\right)=0\)

=> \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 3, x = -1 .

e, Ta có : \(2x^2+5x-3=0\)

=> \(2x^2+6x-x-3=0\)

=> \(x\left(2x-1\right)+3\left(2x-1\right)=0\)

=> \(\left(x+3\right)\left(2x-1\right)=0\)

=> \(\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = -3, x = \(\frac{1}{2}.\)

\(1.x^2-10x=-25\\ \Leftrightarrow x^2-10x+25=0\\\Leftrightarrow \left(x-5\right)^2=0\\\Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)

Vậy nghiệm của phương trình trên là \(5\)

\(2.5x\left(x-1\right)=x-1\\ \Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=1\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{1;\frac{1}{5}\right\}\)

\(3.2\left(x+5\right)-x^2-5x=0\\\Leftrightarrow 2x+10-x^2-5x=0\\ \Leftrightarrow-x^2-3x+10=0\\\Leftrightarrow x^2+3x-10=0\\\Leftrightarrow x^2-2x+5x-10=0\\\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\\\Leftrightarrow \left(x+5\right)\left(x-2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-5;2\right\}\)

\(4.x^2-2x-3=0\\\Leftrightarrow x^2+x-3x-3=0\\\Leftrightarrow \left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;3\right\}\)

\(5.2x^2+5x-3=0\\ \Leftrightarrow2\left(x^2+\frac{5}{2}x-\frac{3}{2}\right)=0\\ \Leftrightarrow x^2+\frac{5}{2}x-\frac{3}{2}=0\\ \Leftrightarrow x^2-\frac{1}{2}x+3x-\frac{3}{2}=0\\\Leftrightarrow x\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{2}\right)=0\\\Leftrightarrow \left(x+3\right)\left(x-\frac{1}{2}\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+3=0\\x-\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-3;\frac{1}{2}\right\}\)