a: ĐKXĐ: x>=-2

\(\sqrt{5x+10}=8-x\)

=>\(\begin{cases}8-x\ge0\\ \left(8-x\right)^2=5x+10\end{cases}\Rightarrow\begin{cases}x\le8\\ x^2-16x+64=5x+10\end{cases}\)

=>\(\begin{cases}-2\le x\le8\\ x^2-21x+54=0\end{cases}\Rightarrow\begin{cases}-2\le x\le8\\ \left(x-3\right)\left(x-18\right)=0\end{cases}\)

=>x=3

b: ĐKXĐ: \(4x^2+x-12\ge0\)

=>\(x^2+\frac14x-3\ge0\)

=>\(x^2+2\cdot x\cdot\frac18+\frac{1}{64}-\frac{193}{64}\ge0\)

=>\(\left(x+\frac18\right)^2\ge\frac{193}{64}\)

=>\(\left[\begin{array}{l}x+\frac18\ge\frac{\sqrt{193}}{8}\\ x+\frac18\le-\frac{\sqrt{193}}{8}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{193}-1}{8}\\ x\le\frac{-\sqrt{193}-1}{8}\end{array}\right.\)

\(\sqrt{4x^2+x-12}=3x-5\)

=>\(\begin{cases}3x-5\ge0\\ \left(3x-5\right)^2=4x^2+x-12\end{cases}\Rightarrow\begin{cases}3x\ge5\\ 9x^2-30x+25-4x^2-x+12=0\end{cases}\)

=>\(\begin{cases}x\ge\frac53\\ 5x^2-31x+37=0\end{cases}\)

\(\Delta=\left(-31\right)^2-4\cdot5\cdot37=221\) >0

=>Phương trình có hai nghiệm phân biệt là

\(\left[\begin{array}{l}x=\frac{31-\sqrt{221}}{2\cdot5}=\frac{31-\sqrt{221}}{10}\left(loại\right)\\ x=\frac{31+\sqrt{221}}{10}\left(nhận\right)\end{array}\right.\)

a) 3x² + 2x - 1 = 0

<=> 3x² + 3x - x - 1 = 0

<=> 3x( x + 1 ) - ( x + 1 ) = 0

<=> ( x + 1 )( 3x - 1 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

<=> x( x - 2 ) - 3( x - 2 ) = 0

<=> ( x - 2 )( x - 3 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

c) x² - 3x + 2 = 0

<=> x² - x - 2x + 2 = 0

<=> x( x - 1 ) - 2( x - 1 ) = 0

<=> ( x - 1 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

d) 2x² - 6x + 1 = 0

<=> 2( x² - 3x + 9/4 ) - 7/2 = 0

<=> 2( x - 3/2 )² = 7/2

<=> ( x - 3/2 )² = 7/4

<=> \(\left(x-\frac{3}{2}\right)=\left(\pm\sqrt{\frac{7}{4}}\right)^2=\left(\pm\frac{\sqrt{7}}{2}\right)^2\)

<=> \(\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{7}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{7}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{7}}{2}\\x=\frac{3-\sqrt{7}}{2}\end{cases}}\)

a) Trường hợp 1. Xét 4 - 5x = 5 - 6x.

Tìm được x = 1.

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a)

\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)

b)

\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)

c)

\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)

d)

\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

e)

\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)

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

\(\left(3x+1\right)\left(x-5\right)\left(-4x+5\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{3}\\\frac{5}{4}\le x\le5\end{matrix}\right.\)

\(\frac{x+2}{x-2}\le\frac{3x+1}{2x-1}\Leftrightarrow\frac{3x+1}{2x-1}-\frac{x+2}{x-2}\ge0\)

\(\Leftrightarrow\frac{x^2-8x}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\frac{x\left(x-8\right)}{\left(2x-1\right)\left(x-2\right)}\ge0\Leftrightarrow\left[{}\begin{matrix}x\le0\\\frac{1}{2}< x< 2\\x\ge8\end{matrix}\right.\)