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

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

\(\text{a. 246 : x + 34 : x = 5 }\\ \left(246+34\right):x=5\\ \\280:x=5\\ x=280:5\\ x=56\)

\(\text{ b. 360 : x – 126 : x = 6}\\ \left(360-126\right):x=6\\ 234:x=6\\ x=234:6\\ x=39\)

\(a,246:x+34:x=5\\ \Rightarrow\left(246+34\right):x=5\\ \Rightarrow280:x=5\\ \Rightarrow x=56\\ b,360:x-126:x=6\\ \Rightarrow\left(360-126\right):x=6\\ \Rightarrow234:x=6\\ \Rightarrow x=39\)

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=1/6

x=1/4

a) x = 1/6

b) x = 1/4

HT

a) 2/3 + x = 5/6

X = 5/6 - 2/3

X = 1/6

b)  x : 5/6 = 3/10

X = 5/6 x 3/10

X = 17/15

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

<=> \(x=\dfrac{17}{12}\)

5x - 7 = -21 - 2x

5x  + 2x =7-21 

7x=-14

x=-2

5 (x - 6) + 2 (x 3) = 4

5x-30+3x+6=4

5x+3x=4+30-6

8x=32

x=4

a x=340

b x=6800

a) (x + 52400) : 5 = (52400 + 340) : 5

x = 340

b) ( x – 5480) × 6 = (6800 – 5480) × 6

x = 6800.

6800 nhé !

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a: 12:x=144

=>\(x=\dfrac{12}{144}\)

=>\(x=\dfrac{1}{12}\)

b: \(x-17=\left(-2\right)\cdot27\)

=>\(x-17=-54\)

=>\(x=-54+17=-37\)

c: \(3x-125=145\)

=>\(3x=125+145=270\)

=>\(x=\dfrac{270}{3}=90\)

\(a,12:x=144\\ x=\dfrac{12}{144}=\dfrac{1}{12}\\ ---\\ b,x-17=\left(-2\right).27\\ x-17=-54\\ x=-54+17=-37\\ ----\\ 3x-125=145\\ 3x=145+125=270\\ x=\dfrac{270}{3}=90\)

a: 12:x=144

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

b: 

c: 

x=90

a) Ta có: \(\left(2x-5\right)^3=216\)

\(\Leftrightarrow2x-5=6\)

\(\Leftrightarrow2x=11\)

hay \(x=\dfrac{11}{2}\)

b) Ta có: \(2x-3⋮x+4\)

\(\Leftrightarrow-11⋮x+4\)

\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{-3;-5;7;-15\right\}\)

Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da