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

. Ai đó giúp tôi đi mà ._.

bài khó quá bạn ạ

(x+2)(x+3)-(x-2)(x+5)=0

=> x²+5x+6-x²-3x+10=0

=>2x+16=0 

 =>2x=-16

=>x=-8

Answer:

\(\left(2x-3\right).\left(x+1\right)-x.\left(2x+3\right)-9=0\)

\(\Rightarrow\left(2x^2+2x-3x-3\right)-2x^2-3x-9=0\)

\(\Rightarrow\left(2x^2-x-3\right)-2x^2-3x-9=0\)

\(\Rightarrow2x^2-x-3-2x^2-3x-9=0\)

\(\Rightarrow\left(2x^2-2x^2\right)-\left(x+3x\right)-\left(3+9\right)=0\)

\(\Rightarrow-4x-12=0\)

\(\Rightarrow x+3=0\)

\(\Rightarrow x=-3\)

\(2x.\left(x-3\right)-x+3=0\) (Sửa đề)

\(\Rightarrow2x.\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right).\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)

\(2x.\left(x^2-4\right)+6.\left(4-x^2\right)=0\)

\(\Rightarrow2x.\left(x^2-4\right)-6.\left(x^2-4\right)=0\)

\(\Rightarrow2.\left(x-3\right).\left(x+2\right).\left(x-2\right)=0\)

Trường hợp 1: \(x-3=0\Rightarrow x=3\)

Trường hợp 2: \(x+2=0\Rightarrow x=-2\)

Trường hợp 3: \(x-2=0\Rightarrow x=2\)

Nhiều thế vậy ...!!

Sao làm nổi?

a) Ta có: \(\left(5x-15\right)\left(4+6x\right)=0\)

\(\Leftrightarrow5\left(x-3\right)\cdot2\cdot\left(2+3x\right)=0\)

\(\Leftrightarrow10\left(x-3\right)\left(2+3x\right)=0\)

Vì 10\(\ne\)0 nên

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

Vậy: \(x\in\left\{3;\frac{-2}{3}\right\}\)

b) Ta có: \(\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\5x=6\\\frac{1}{2}x=\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=\frac{3}{4}:\frac{1}{2}=\frac{3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{6}{5};\frac{3}{2}\right\}\)

c) Ta có: \(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\)

\(\Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=3\\x=\frac{25}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{12}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{4};\frac{25}{12}\right\}\)

d) Ta có: \(\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5\left(x-1\right)-\frac{3}{2}-\frac{\left(2-3\right)\left(x-1\right)}{3}\right]=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5x-5-\frac{3}{2}-\frac{-1\left(x-1\right)}{3}\right]=0\)

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

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(5x-\frac{13}{2}-\frac{1}{3}+\frac{x}{3}\right)=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{15x}{3}-\frac{41}{6}+\frac{x}{3}\right)=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{16x}{3}-\frac{41}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{1}{6}=0\\\frac{16x}{3}-\frac{41}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{1}{6}\\\frac{16}{3}\cdot x=\frac{41}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}:\frac{2}{3}\\x=\frac{41}{6}:\frac{16}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{41}{32}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{4};\frac{41}{32}\right\}\)

\(a.\left(5x-15\right)\left(4+6x\right)=0\\ \left[{}\begin{matrix}5x-15=0\\4+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)

\(b.\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}=0\right)\\ \left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=-\frac{3}{2}\end{matrix}\right.\)

c.

\(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\\ \Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\\ \left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{2}\end{matrix}\right.\)

bạn ơi !!!

đăng từng câu thôi thế này nhìn loạn cả mắt luôn á

rối mắt quá

a, 2/3+1/3:x=-7

1/3:x=-7-2/3

1/3:x=-23/3

x=1/3:-23/3

x=-1/23

Vậy x=-1/23

c, (2x-3)(6-2x)=0

*TH1: 2x-3=0

2x=3

x=3/2

*TH2: 6-2x=0

2x=6

x=6/2

x=3

Vậy x=3/2 hoặc x=3

d,x:3/4+1/4=-2/3

x:3/4=-2/3-1/4

x:3/4=-11/12

x=-11/12*3/4

x=-11/16

\(a.2x-3=4x+6\) 

\(\Leftrightarrow2x-3-4x-6=0\) 

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow x=\dfrac{9}{2}\)

\(S=\left\{\dfrac{9}{2}\right\}\) 

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

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

\(\Leftrightarrow2x^2+2=0\)

\(\Leftrightarrow x\left(2x+2\right)=0\) 

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

\(S=\left\{0,-1\right\}\) 

Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm

a: =>-2x=9

=>x=-9/2

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=0 hoặc x=-1

\(a,2x-3=4x+6\)

\(\Leftrightarrow2x-4x=6+3\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

\(b,\) Ghi vậy mình không làm được.

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

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

\(\Leftrightarrow x\left(2x+2\right)=0\)

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

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

\(d,\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}=0\left(dkxd:x\ne-1;x\ne3\right)\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x-3\right)-2.2}{2\left(x+1\right)\left(x-3\right)}=0\)

\(\Leftrightarrow x^2+x-x^2+3x-4=0\)
\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow x=1\left(tmdk\right)\)

Vậy \(S=\left\{1\right\}\)