\(\Leftrightarrow\left(3x+1\right)\left(4x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

c.

ĐLXĐ: \(x\ge-\dfrac{1}{3}\)

\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)

Đặt \(\sqrt{3x+1}=t\ge0\)

\(\Rightarrow-t^2+t+4x^2-10x+6=0\)

\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{61}{12}\)

\(\Leftrightarrow36x^2+12x-58-2\sqrt{12x+61}=0\)

\(\Leftrightarrow\left(36x^2+24x+4\right)-\left(12x+61+2\sqrt{12x+61}+1\right)=0\)

\(\Leftrightarrow\left(6x+2\right)^2-\left(\sqrt{12x+61}+1\right)^2=0\)

\(\Leftrightarrow\left(6x+1-\sqrt{12x+61}\right)\left(6x+3+\sqrt{12x+61}\right)=0\)

\(\Leftrightarrow...\) tương tự câu a

a.

ĐKXĐ: \(x\ge-\dfrac{5}{4}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{7};5\right\}\)

\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)

\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

Giải tiếp tương tự

Sau đó giải tiếp câu còn lại

a, \(\left|2+3x\right|=\left|4x-3\right|\)

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}}\)

Câu b tương tự

c, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Rightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{5}{8}x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}x+\frac{5}{8}x=-\frac{3}{5}+\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}}\)

d, \(\left|\frac{7}{5}x+\frac{3}{2}\right|-\left|\frac{4}{3}-\frac{1}{4}\right|=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|-\frac{13}{12}=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|=\frac{13}{12}\)

Đến đây dễ rồi, tự làm tiếp :)

P/s: Ko chắc, sai ib với t :v

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

=>3x-2=0 hoặc 4x+5=0

=>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

=>x=3 hoặc x=-20

c: =>(x-3)(2x+5)=0

=>x-3=0 hoặc 2x+5=0

=>x=3 hoặc x=-5/2

\(\Leftrightarrow\left|4x-7\right|\left(\left|4x-7\right|-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left|4x-7\right|=0\\\left|4x-7\right|=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-7=0\\4x-7=5\\7-4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 1. a) 4x - 3 = 0

⇔ x = \(\dfrac{3}{4}\)

KL.....

b) - x + 2 = 6

⇔ x = - 4

KL...

c) -5 + 4x = 10

⇔ 4x = 15

⇔ x = \(\dfrac{15}{4}\)

KL....

d) 4x - 5 = 6

⇔ 4x = 11

⇔ x = \(\dfrac{11}{4}\)

KL....

h) 1 - 2x = 3

⇔ -2x = 2

⇔ x = -1

KL...

Bài 2. a) ( x - 2)( 4 + 3x ) = 0

⇔ x = 2 hoặc x = \(\dfrac{-4}{3}\)

KL......

b) ( 4x - 1)3x = 0

⇔ x = 0 hoặc x = \(\dfrac{1}{4}\)

KL.....

c) ( x - 5)( 1 + 2x) = 0

⇔ x = 5 hoặc x = \(\dfrac{-1}{2}\)

KL.....

d) 3x( x + 2) = 0

⇔ x = 0 hoặc x = -2

KL.....

Bài 3.a) 3( x - 4) - 2( x - 1) ≥ 0

⇔ x - 10 ≥ 0

⇔ x ≥ 10

b) 3 - 2( 2x + 3) ≤ 9x - 4

⇔ - 4x - 3 ≤ 9x - 4

⇔ 13x ≥1

⇔ x ≥ \(\dfrac{1}{13}\)