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

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

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

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

Bài 1:

a: \(\Leftrightarrow3^x\cdot10=810\)

\(\Leftrightarrow3^x=81\)

hay x=4

b: \(\Leftrightarrow2x+6+3x+12+4x+20=8x+2345\)

=>9x+38=8x+2345

=>x=2307

a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)

\(\Rightarrow\left(3x-1\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)

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

Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)

b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)

c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)

d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

a, x=-3/8

b,x=3/100

c,x=0

d,x=-1/3 hoặc x=1

\(a,\sqrt{9x^2}=2x+1\\ \Leftrightarrow\left[{}\begin{matrix}3x=2x+1,\forall x\ge0\\-3x=2x+1,\forall x< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1,\forall x\ge0\left(N\right)\\x=-1,\forall x< 0\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-1,\forall x+3\ge0\\x+3=1-3x,\forall x+3< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2,\forall x\ge-3\left(N\right)\\x=-\dfrac{1}{2},\forall x< -3\left(L\right)\end{matrix}\right.\Leftrightarrow x=2\)

\(c,\sqrt{x^2-2x+4}=2x-3\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=\left(2x-3\right)^2\\ \Leftrightarrow x^2-2x+4=4x^2-12x+9\\ \Leftrightarrow3x^2-10x+5=0\\ \Delta=100-4\cdot3\cdot5=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10-\sqrt{40}}{6}\\x=\dfrac{10+\sqrt{40}}{6}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{10}}{3}\\x=\dfrac{5+\sqrt{10}}{3}\end{matrix}\right.\)

\(a.\sqrt{9x^2}=2x+1\)

<=> \(\sqrt{9}x=2x+1\)

<=> 3x = 2x + 1

<=> 3x - 2x = 1

<=> x = 1

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

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

$a)ĐK:8x+2\ge 0$

$\to 8x \ge -2$

$\to x \ge -\dfrac14$

$b)ĐK:\dfrac{-5}{6-3x} \ge 0(x \ne 2)$

Mà $-5<0$

$\to 6-3x<0$

$\to 6<3x$

$\to x>2$

$*A=x-2\sqrt{x-2}+3(x \ge 2)$

$=x-2-2\sqrt{x-2}+1+4$

$=(\sqrt{x-2}-1)^2+4 \ge 4$

Dấu "=" xảy ra khi $\sqrt{x-2}-1=0 \Leftrightarrow \sqrt{x-2}=1\Leftrightarrow x=3$

a) \(x\ge-\dfrac{1}{4}\)

b) x<2

Bài 2:

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

\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow5x=22\Rightarrow x=\frac{22}{5}\)

c)\((8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)\)

\(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)

Suy ra x=3;x=-11/10