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

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

\(\Rightarrow x=3\)

đ) \(4.8^2=2^x\)

\(2^2.\left(2^3\right)^2=2^x\)

\(2^2.2^6=2^x\)

\(2^8=2^x\)

\(\Rightarrow x=8\)

d) \(\left|x+3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)

mấy câu trên dễ rồi tự làm em nhé 

a, 2x + 5 = 7

=> 2x = 2

=> x = 1

vậy____

\(\frac{2x-5}{6}:\frac{x-7}{5}=\frac{2x-5}{6}\times\frac{5}{x-7}=\frac{5\left(2x-5\right)}{6\left(x-7\right)}=\frac{10x-25}{6x-42}\)

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

\(\Leftrightarrow4\left(4x+1\right)=3\left(2x-3\right)\)

\(\Leftrightarrow16x+4=6x-9\)

\(\Leftrightarrow16x-6x=-9-4\)

\(\Leftrightarrow10x=-13\)

\(x=-\frac{13}{10}\)

\(x=-1,3\)

=> 3 - 2x = 4x - 5 

hoặc 3- 2x = - (4x - 5)

+) Trường hợp: 3 -2x = 4x - 5 => 3 + 5 = 4x + 2x => 8 = 6x => x = 4/3
+) Trường hợp: 3 - 2x = - (4x - 5) => 3 - 2x = - 4x + 5 => 4x - 2x = 5 - 3 => 2x = 2 => x = 1

Vậy x = 4/3 hoặc x = 1

nếu ai có cách khác thì giúp mình với

a) I 3x - 2 I > 1

\(\Rightarrow\orbr{\begin{cases}3x-2>1\\3x-2< -1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x>3\\3x< 1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>1\\x< \frac{1}{3}\end{cases}}\)

Vậy.....

b) I2x + 3I \(\le\)5

\(\Rightarrow\hept{\begin{cases}2x+3\le5\\2x+3\ge-5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x\le2\\2x\ge-8\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x\le1\\x\ge-4\end{cases}}\)

\(\Rightarrow-4\le x\le1\)

Vậy.....

**\(\dfrac{2x-5}{6}=\dfrac{x-7}{5}\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}=\dfrac{5\left(x-7\right)}{30}\)

\(\Leftrightarrow6\left(2x-5\right)=5\left(x-7\right)\Leftrightarrow12x-30=5x-35\)

\(\Leftrightarrow12x-5x=-35+30\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)

**\(\dfrac{4x+1}{3}=\dfrac{2x-3}{4}\Leftrightarrow\dfrac{4\left(x+1\right)}{12}=\dfrac{3\left(2x-3\right)}{12}\)

\(\Leftrightarrow4\left(x+1\right)=3\left(2x-3\right)\Leftrightarrow4x+4=6x-9\)

\(\Leftrightarrow4x-6x=-9-4\Leftrightarrow-2x=-13\Leftrightarrow x=\dfrac{13}{2}\)

ta có 4x-1 \(⋮x+1\)

        x+1\(⋮x+1\)=> \(4\left(x+1\right)⋮x+1\)

=>4(x+1) -(4x-1)\(⋮x+1\)

=> 4x+4-4x+1\(⋮x+1\)

=>\(5⋮x+1\)

=>\(x+1\inƯ\left(5\right)\)

=> \(x+1\in\left\{\pm1,\pm5\right\}\)=> \(x\in\left\{-6,-2,0,4\right\}\)

Vậy..........

các phần sau bn lm tương tự nhé

=>\(\frac{6\left(x-1\right)}{18}+\frac{9\left(3x-5\right)}{18}+\frac{2\left(2x\right)}{18}+\frac{2\left(-5x+3\right)}{18}=\frac{1}{2}\)

=>\(\frac{6x-6}{18}+\frac{27x-45}{18}+\frac{4x}{18}+\frac{-10x+6}{18}=\frac{1}{2}\)

=>\(\frac{\left(6x-6\right)+\left(27x-45\right)+4x+\left(-10x+6\right)}{18}=\frac{1}{2}\)

=>\(\frac{27x-45}{18}=\frac{1}{2}\)

=>54x-90=18

54x =108

=>x=2

quy đồng lên rồi nhân chéo rồi giải

x=2

dễ mà

a) -2(2x - 8) + 3(4 - 2x) = -72  - 5(3x - 7)

=> -4x + 18 + 12 - 6x = -72 - 15x + 35

=> -10x + 15x = -37 - 30

=> 5x = -37

=> x = -7,4

b) 3|2x² - 7| = 33

=> |2x² - 7| = 11

=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)

=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)