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

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

\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)

<=>x=1

vậy ...

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

\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

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

vậy ...

Trong Th này bn nên dùng dấu ''hoặc''

a,\(\left(x^3-1\right)\left(x^2+1\right)=0\)

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

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

\(\left[{}\begin{matrix}x=1\\x=\pm1\end{matrix}\right.\)

b, \(\left(2x+6\right)\left(3x^2-12\right)=0\)

\(\left[{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-3\\x=\pm2\end{matrix}\right.\)

hk hừng đẳng thưc chưa

chưa bn

(3x-2)^2=(2-3x)^4

(3x-2)(3x-2)=(2-3x)^4

-(2-3x)x-(2-3x)=(2-3x)(2-3x)(2-3x)(2-3x)

0=(2-3x)^2

0=2-3x

x=3/2

hok tôt

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

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

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

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

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

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

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

Vậy pt có tập nghiệm S = { 0;3 }

khỉ nghĩ như này..

x³-3x²=0

(=)x² (x-3)=0

(=)x²=0,hoac x-3=0

(=)x=3

5: \(=-\left(x^2+3x+5\right)\)

\(=-\left(x^2+3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)

\(=-\left(x+\dfrac{3}{2}\right)^2-\dfrac{11}{4}< 0\)

6: \(=-3\left(x^2+2x+\dfrac{4}{3}\right)=-3\left(x^2+2x+1+\dfrac{1}{3}\right)\)

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

Có   \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow x-5=1-3x\)

\(\Leftrightarrow x+3x=1+5\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=6:4\)

\(\Leftrightarrow x=\frac{3}{2}\)

Giải:

a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)

\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)

\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)

\(\Leftrightarrow4x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

Vậy ...

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)

\(\Leftrightarrow30x-37=4\)

\(\Leftrightarrow30x=41\)

\(\Leftrightarrow x=\dfrac{41}{30}\)

Vậy ...

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)

\(\Leftrightarrow x^3+8-x^3-3=14x\)

\(\Leftrightarrow5=14x\)

\(\Leftrightarrow x=\dfrac{5}{14}\)

Vậy ...

d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)

\(\Leftrightarrow x^3+1-x^3-3x=2\)

\(\Leftrightarrow1-3x=2\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Vậy ...

a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)

=> \(x^2-2x-x-3x+7+9x=6\)

=> \(x^2-2x-x^2-3x+7+9x=6\)

=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)

=> \(4x=-1\)

Vậy \(x=\dfrac{-1}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)

=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)

=> \(42x=43\)

Vậy \(x=\dfrac{43}{42}\)

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)

=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)

=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)

=> \(5=14x\)

Vậy \(x=\dfrac{5}{14}\)

d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)

=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)

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

=> \(-3x=1\)

Vậy \(x=\dfrac{-1}{3}\)

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)