a: Ta có: \(x\left(2-x\right)+x^2+x=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)

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

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

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,

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

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

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

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

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

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

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

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

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

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

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

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

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

a: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

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

b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)

\(\Leftrightarrow3x=12\)

hay x=4

a) 2x³-18x=0

⇔ 2x(x²-9)=0

⇔ 2x(x-3)(x+3)=0

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

b)(3x-1)(2x+1)-6x(x+2)=11

⇔ 6x²+x-1-6x²-12x=11

⇔ -11x=12

\(\Leftrightarrow x=-\dfrac{12}{11}\)

c) (x-1)³-(x+2).(x²-2x+4)=3.(1-x²)

⇔ x³-3x²+3x-1-x³-8-3+3x²=0

⇔ 3x=12

⇔   x=4

c. (x - 1)³ - (x + 2)(x² - 2x + 4) = 3(1 - x²)

<=> (x³ - 3x² + 3x - 1) - (x³ - 2x² + 4x + 2x² - 4x + 8) = 3 - 3x²

<=> x³ - 3x² + 3x - 1 - x³ + 2x² - 4x - 2x² + 4x - 8 = 3 - 3x²

<=> x³ - x³ - 3x² + 2x² - 2x² + 3x² + 3x - 4x + 4x = 3 + 1 + 8

<=> 3x = 12

<=> x = 4

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

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

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

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

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

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a) Ta có: \(2-x=2\left(x-2\right)^3\)

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

\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

b) Ta có: \(8x^3-72x=0\)

\(\Leftrightarrow8x\left(x^2-9\right)=0\)

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

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

Vậy: S={0;3;-3}

c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)

\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

d) Ta có: \(2x^3+3x^2+3+2x=0\)

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

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow2x=-3\)

hay \(x=-\dfrac{3}{2}\)

e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)

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

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

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

Vậy: S={0;1;-2}

f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

Vậy: S={0;2;12}

a: Ta có: \(40x^4+5x=0\)

\(\Leftrightarrow5x\left(8x^3+1\right)=0\)

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

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

b: Ta có: \(8x^2-2x-1=0\)

\(\Leftrightarrow8x^2-4x+2x-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)