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: \(x\cdot3\cdot5=2,7\)

=>\(x\cdot15=2,7\)

=>\(x=\dfrac{2,7}{15}=0,18\)

b: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)

a: x=2/3-4/5=10/15-12/15=-2/15

b: 1/2-x=7/12

=>x=1/2-7/12=-1/12

c: =>7/2:x=-7/2

=>x=-1

d: =>1/6x=3/8-5/2=3/8-20/8=-17/8

=>x=-17/8*6=-102/8=-51/4

e: =>1,5x=-1,5

=>x=-1

a, x( x - 3) - ( x + 2)( x - 1) = 3

<=> x² - 3x - x² + x - 2x + 2 = 3

<=>  x² - 3x - x² + x - 2x = 3 - 2

<=> -4x = 1

<=> x =\(-\frac{1}{4}\)

Vậy_

b,  \(x-\frac{x-1}{5}+\frac{x+2}{6}=4+\frac{x+1}{3}\)

\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)

=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10

<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10

<=> 19x = 114

<=> x = 6

Vậy _

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 \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

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

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

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

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) (3x-15)⁷ = 0

3x-15 = 0

3x = 0+15

3x = 15

x = 15:3

x = 5

b) 4^2x-6 = 1

 2x-6 = 0

2x = 0+6

2x = 6

x = 6:2

x = 3

c) Tớ ko bít 

d) (x - 6)³ = (x - 6)²

Th₁:

x - 6 = 1

x = 1 + 6

x = 7

Th₂:

x - 6 = 0

x = 6

Vậy x = 7

      x = 6

--thodagbun--

a, (3x-15)^7=0 <=> 3x-15=0 <=> x=5

b, 4²x+6=1 <=> 16x=-5 <=>x=-5/16

c, \(\dfrac{\left(3-x\right)^{10x}}{\left(3-x\right)^{20}}=1\Leftrightarrow\left(3-x\right)^{10x-20}=1\)

TH1: 10x-20 = 0 <=> x=2

TH2: 3-x=1 <=> x=2

Vậy x=2

d, (x-6)^3 = (x-6)^2

<=> (x-6)^2.[(x-6)-1]=0

<=> (x-6)^2=0 hoặc (x-6)-1=0

<=> x=6 hoặc x=7

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)