1. (3x - 5)² - (3x + 1)² = 8

=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8

=> -6(6x - 4) = 8

=> 6x - 4 = \(\dfrac{-4}{3}\)

\(\Rightarrow x=\dfrac{4}{9}\)

2) 2x(8x - 3) - (4x - 3)² = 27

=> 16x² - 6x - 16x² + 24x - 9 = 27

=> 18x - 9 = 27

=> x = 2

3) (2x - 3)² - (2x + 1)² = 3

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

=> -4(4x - 2) = 3

=> 4x - 2 = \(\dfrac{-3}{4}\)

\(\Rightarrow x=\dfrac{5}{16}\)

4) (x + 5)² - x² = 45

=> (x + 5 - x)(x + 5 + x) = 45

=> 5(2x + 5) = 45

=> 2x + 5 = 9

=> x = 2

5) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x + 1)² = 18

=> x³ - 9x² + 27x - 27 - x³ + 27 + 9(x² + 2x + 1) = 18

=> -9x² + 27x + 9x² + 18x + 9 = 18

=> 45x + 9 = 18

=> 45x = 9

=> x = \(\dfrac{1}{5}\)

6) x(x - 4)(x + 4) - (x - 5)(x² + 5x + 25) = 13

=> x (x² - 16) - (x³ - 125) = 13

=> x³ - 16x - x³ + 125 = 13

=> -16x = -112

=> x = 7.

a)\(\left(2x-3\right)\left(x+1\right)< 0\)

\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\)  hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại)  hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)

\(\Leftrightarrow-1< x< \frac{3}{2}\)

b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)

\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)

c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)

Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)

\(\Leftrightarrow x=4\)

đề câu c sai rầu kìa má

(x - 2/3)³ = -1/27

=> (x - 2/3)³ = (-1/3)³

=> x - 2/3 = -1/3

=> x = -1/3 + 2/3

=> x = 1/3

Từ bài ra ta có \(\left(x-\frac{2}{3}\right)^3=\left(\frac{-1}{3}\right)^3\)

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

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

\(\Rightarrow x=\frac{1}{3}\)

Vậy ... nếu đúng thì k nha

a) \(\left(x-10\right)^2-x\left(x+8\right)=-12x+100=-11,76+100=88,24\)

b) \(x^3-9x^2+27x-27=\left(x-3\right)^3=\left(5-3\right)^3=8\)

c) \(6x\left(2x-7\right)-\left(3x-5\right)\left(4x+7\right)=-43x+35=121\)

\(a)\) \(\left(x-10\right)^{^2}-x.\left(x+8\right)\) \(với\) \(x=0,98\)

\(=-12x+100\)

\(=-11,76+100\)

\(=88,24\)

\(b)\) \(x^3-9x^2+27.x-27\) \(với\) \(x=5\)

\(=\left(x-3\right)^3\)

\(=\left(5-3\right)^3\)

\(=8\)

\(c)\)\(6x.\left(2x-7\right)-\left(3x-5\right).\left(4x+7\right)\) \(tại\) \(x=-2\)

\(=-43+35\)

\(=121\)

Chúc bạn hôc tốt nha ❤

a) \(\left|2x-3\right|=7\)

\(\Rightarrow2x-3=7\) hoặc \(2x-3=-7\)

+) \(2x-3=7\Rightarrow x=5\)

+) \(2x-3=-7\Rightarrow x=-2\)

Vậy x = 5 hoặc x = -2

b) \(\left|5x-3\right|-2=9\)

\(\Rightarrow\left|5x-3\right|=11\)

\(\Rightarrow5x-3=11\) hoặc \(5x-3=-11\)

+) \(5x-3=11\Rightarrow x=\frac{14}{5}\)

+) \(5x-3=-11\Rightarrow x=\frac{-8}{5}\)

Vậy \(x=\frac{14}{5}\) hoặc \(x=\frac{-8}{5}\)

Tìm x bik 

\(a,\left|2x-3\right|=7\)

\(b,\left|5x-3\right|-2=9\)

a) 

\(\left|2x-3\right|=7\)

\(\Rightarrow\begin{cases}2x-3=7\\2x-3=-7\end{cases}\)

\(\Rightarrow\begin{cases}2x=10\\2x=4\end{cases}\)

\(\Rightarrow\begin{cases}x=5\\x=2\end{cases}\)

Vậy \(x=\begin{cases}5\\2\end{cases}\)

a) \(\left|2x-3\right|=7\)

\(\Leftrightarrow\begin{cases}2x-3\ge0\\2x-3=7\end{cases}\)  hoặc \(\begin{cases}2x-3< 0\\2x-3=-7\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge\frac{3}{2}\\x=5\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< \frac{3}{2}\\x=-2\left(tm\right)\end{cases}\)

Vậy x={-2;5}

b) \(\left|5x-3\right|-2=9\Leftrightarrow\left|5x-3\right|=11\)

\(\Leftrightarrow\begin{cases}x\ge\frac{3}{5}\\5x-3=11\end{cases}\) hoặc \(\begin{cases}x< \frac{3}{5}\\5x-3=-11\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge\frac{3}{5}\\x=\frac{14}{5}\left(tm\right)\end{cases}\) hoặc \(\begin{cases}x< \frac{3}{5}\\x=-\frac{8}{5}\left(tm\right)\end{cases}\)

Vậy \(x=\left\{\frac{14}{5};-\frac{8}{5}\right\}\)

=> 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

\(\left(x+3\right)^3-3\cdot\left(3x+1\right)^2+\left(2x+1\right)\cdot\left(4x^2-2x+1\right)=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-3\cdot\left(9x^2+6x+1\right)+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-27x^2-18x-3+8x^3-4x^2+2x+4x^2-2x+1=54\)

\(\Leftrightarrow9x^3-18x^2+9x-29=0\)

\(\Leftrightarrow x=2,208024627\)