\(a,\Leftrightarrow x^2+2x+1-x^2+3x-2x=3\\ \Leftrightarrow3x=2\Leftrightarrow x=\dfrac{3}{2}\\ b,\Leftrightarrow x^2-x-6-x^2+6x-9=15\\ \Leftrightarrow5x=30\Leftrightarrow x=6\\ c,\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=0\\ \Leftrightarrow x=-4\)

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

\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)

b) \(\left(x+2\right)\left(x-3\right)-\left(x-3\right)^2=15\Rightarrow x^2-3x+2x-6-\left(x^2-6x+9\right)=15\)

\(\Rightarrow5x=30\Rightarrow x=6\)

`P(x)=\(4x^2+x^3-2x+3-x-x^3+3x-2x^2\)

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

`= 2x^2+3`

`Q(x)=`\(3x^2-3x+2-x^3+2x-x^2\)

`= -x^3+(3x^2-x^2)+(-3x+2x)+2`

`= -x^3+2x^2-x+2`

`P(x)-Q(x)-R(x)=0`

`-> P(X)-Q(x)=R(x)`

`-> R(x)=P(x)-Q(x)`

`-> R(x)=(2x^2+3)-(-x^3+2x^2-x+2)`

`-> R(x)=2x^2+3+x^3-2x^2+x-2`

`= x^3+(2x^2-2x^2)+x+(3-2)`

`= x^3+x+1`

`@`\(\text{dn inactive.}\)

a: P(x)-Q(x)-R(x)=0

=>R(x)=P(x)-Q(x)

=2x^2+3+x^3-2x^2+x-2

=x^3+x+1

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

hay \(x\in\left\{1;-1\right\}\)

plss

a,

\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0

\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)

=>\(x=\left\{3,-3\right\}\)

b,

\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)

\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)

=>\(x=\left\{-4,0,4\right\}\)

d,

\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)

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

=>\(x=\left\{-3,0,3\right\}\)

e,

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

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

=>\(x=\left\{-2,-1,0\right\}\)

f,

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

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

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

=>x=0 hoặc x=-1

4: =>(2x-3)(x+2)=0

=>x=3/2 hoặc x=-2

6: =>6x=7 hoặc 6x=-7

=>x=7/6 hoặc x==7/6

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

ảnh lỗi r ạ

Lỗi rùi

lưu ảnh rồi gửi vào đây nhé

Lời giải:

a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$

$\Leftrightarrow -4x.6=8$

$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$

b.

$9x^5-72x^2=0$

$\Leftrightarrow 9x^2(x^3-8)=0$

$\Leftrightarrow x^2=0$ hoặc $x^3=8$

$\Leftrightarrow x=0$ hoặc $x=2$

c.

$5x^4-8x^2-4=0$

$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$

$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$

$\Leftrightarrow (5x^2+2)(x^2-2)=0$

$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)

$\Leftrightarrow x=\pm \sqrt{2}$

d.

PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$

$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$

$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$

$\Leftrightarrow x+2=0$ hoặc $x+1=0$

$\Leftrightarrow x=-2$ hoặc $x=-1$

a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)

\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)

\(\Leftrightarrow-24x=8\)

hay \(x=-\dfrac{1}{3}\)

b: Ta có: \(9x^5-72x^2=0\)

\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)

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

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

c: Ta có: \(5x^4-8x^2-4=0\)

\(\Leftrightarrow5x^4-10x^2+2x^2-4=0\)

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

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

a) x³-1-(x²+2x)(x-2)=5

⇔ x³-1-x³+4x=5

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)