a) `(x-8)(x^3+8)=0`

`<=>(x-8)(x+2)(x^2-2x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\) (Vì `x^2-2x+4 \ne 0 forall x)`

Vậy `A={8;-2}`.

b) `(4x-3)-(x+5)=3(10-x)`

`,=>4x-3-x-5=30-3x`

`<=>3x-8=30-3x`

`<=>6x=38`

`<=>x=19/3`

Vậy `S={19/3}`.

a) Ta có: \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)

b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

Vậy: \(x\in\left\{-8;6\right\}\)

a) \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

b) \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a.|4-5.x|=24

TH1: 4-5.x=24

        ⇔5x=-20

        ⇔x=-4

TH1: 4-5.x=-24

        ⇔5x=28

        ⇔x=28/5

b.(8+x).(6-x)=0

\(\left\{{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a) x = 5/8 : 3/4

x = 5/6

b) x = 5/6 - 1/12

x = 3/4

a)x=\(\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{6}\)

b) x=\(\dfrac{5}{6}-\dfrac{1}{12}=\dfrac{3}{4}\)

a \(x=\dfrac{5}{8}:\dfrac{3}{4}\)

\(x=\dfrac{5}{8}\times\dfrac{4}{3}\)

\(x=\dfrac{5}{6}\)

b \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

\(x=\dfrac{10}{12}-\dfrac{1}{12}\)

\(x=\dfrac{9}{12}\)

\(x=\dfrac{3}{4}\)

x      = 9/3 - 5/8

x      =  57/24

x+5/8=9/3

x       -= 9/3 - 5/8

x        =  57/24

\(x+\dfrac{5}{8}=\dfrac{9}{3}\)

\(x=\dfrac{9}{3}-\dfrac{5}{8}=\dfrac{19}{8}\)

\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)

\(< =>20x-5x^2+5x^2-12-x-6=0\)

\(< =>19x-18=0\)

\(< =>x=\dfrac{18}{19}\)

\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)

\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)

\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)

a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)

b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)

a)x=18/19

b)x=15/2

\(a,x-\dfrac{5}{7}=\dfrac{3}{5}\times\dfrac{1}{2}\)

\(x-\dfrac{5}{7}=\dfrac{3}{10}\)

\(x=\dfrac{3}{10}+\dfrac{5}{7}\)

\(x=\dfrac{21}{70}+\dfrac{50}{70}\)

\(x=\dfrac{71}{70}\)

\(Vayx=\dfrac{71}{70}\)

\(b,x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{3}{4}\)

\(x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{6}{8}\)

\(x\times\dfrac{2}{3}=\dfrac{11}{8}\)

\(x=\dfrac{11}{8}:\dfrac{2}{3}\)

\(x=\dfrac{11}{8}\times\dfrac{3}{2}\)

\(x=\dfrac{33}{16}\)

\(Vayx=\dfrac{33}{16}\)

Câu 2:Đáp số là 8/17 nhé

(5/9+4/9)x8/17

=>1x8/17=8/17

Câu 1: Tìm x:

a)  2/5× x =(thiếu đề)

 8/7: x =4/5 

x=8/7:4/5

x=10/7

Câu 2: Tính bằng cách thuận tiện nhất

5/9× 8/17 + 4/9 × 8/17

=(5/9+4/9)x8/17

=1x8/17=8/17

a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)

\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)

\(\Leftrightarrow x=15\)

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

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)

\(\Leftrightarrow x^2-36=x^2\)(vô lý)

a. (x - 5)(x + 3) = x(x - 3)

<=> x² + 3x - 5x - 15 = x² - 3x

<=> x² - x² + 3x - 5x + 3x - 15 = 0

<=> x = 15

b. (x + 2)² = (x - 1)(x + 2)

<=> x² + 4x + 4 = x² + 2x - x - 2

<=> x² - x² + 4x - 2x + x = -2 - 4

<=> 3x = -5

<=> \(x=\dfrac{-5}{3}\)

c. (x - 6)(x + 6) = x²

<=> x² - 36 - x² = 0

<=> x² - x² = 36

<=> 0 = 36 (vô lí)

Vậy nghiệm của PT là \(S=\varnothing\)

d. (2x - 3)² = 4x² - 8 

<=> 4x² - 12x + 9 - 4x² + 8 = 0

<=> 4x² - 4x² - 12x = -8 - 9

<=> -12x = -17

<=> \(x=\dfrac{17}{12}\)

a) \(x+\dfrac{4}{15}=\dfrac{4}{12}\)

\(x=\dfrac{4}{12}-\dfrac{4}{15}\)

\(x=\dfrac{20}{60}-\dfrac{16}{60}\)

\(x=\dfrac{1}{15}\)

b) \(x-\dfrac{5}{8}=1\dfrac{2}{3}\)

\(x-\dfrac{5}{8}=\dfrac{5}{3}\)

\(x=\dfrac{5}{3}+\dfrac{5}{8}\)

\(x=\dfrac{40}{24}+\dfrac{15}{24}\)

\(x=\dfrac{55}{24}\)