Hồ Lê Phú Lộc bn vào câu trả lừoi tương tự nhé !! sẽ có nhiều bài giống như thế và  cách giải cũng nhiều hơn 

**** cho mình đi Hồ Lê Phú Lộc 

a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+7x-x^2-x+6=0\)

hay x=-1

b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

b. (x + 2)² - x² + 4 = 0

<=> (x + 2 - x)(x + 2 + x) + 4 = 0

<=> 2(2 + 2x) + 4 = 0

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

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

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

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

\(\text{1) -5x - (-3)= 13}\)

\(\Rightarrow-5x=10\)

\(x=10:-5\)

\(x=-2\)

\(\text{2) |x-3| - 7= 13}\)

\(\Rightarrow|x-3|=20\)

\(\Rightarrow\orbr{\begin{cases}x-3=20\\x-3=-20\end{cases}\Leftrightarrow\orbr{\begin{cases}x=23\\x=-17\end{cases}}}\)

\(\text{3) 17- (43 - |x|)= 45}\)

\(\Rightarrow43-|x|=-28\)

\(|x|=71\)

\(\Rightarrow\orbr{\begin{cases}x=71\\x=-71\end{cases}}\)

\(\text{5) (x-2).(x+15)= 0}\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-15\end{cases}}}\)

4,\(\text{4) (x-3).(x-5) < 0}\)\(\left(x-3\right).\left(x-5\right)< 0\)

\(\Rightarrow\left(x-3\right)\)và \(\left(x-5\right)\)trái dấu

Mà \(\left(x-3\right)>\left(x-5\right)\Rightarrow\left(x-3\right)>0\)và \(\left(x-5\right)< 0\)

\(+,x-3>0\Rightarrow x>3\)

\(+,x-5< 0\Rightarrow x< 5\)

\(\Rightarrow3< x< 5\)

\(\)Mà \(x\in Z\)

\(\Rightarrow x=4\)

học tốt

1<=>-5x+3=13

<=>-5x=10

<=>x=-2

2<=>|x-3|=20

th1:x-3=20

<=>x=23

th2:x-3=-20

<=>x=-17

3,<=>17-43+|x|=45

<=>|x|=71

th1:x=71

th2:x=-71

4<=>x-3<0                  x-5>0

<=>x<3                       x>5(loại vì ko có số naod vừa lớn hơn 5 và nhỏ hơn 3)

<=>x-3>0                   x-5<0

<=>x>3                      x<5

=>3<x<5

5,<=>x-2=0                  x+15=0

<=>x=2                       x=-15

https://www.youtube.com/channel/UCb2H-q6FmW61PgcsL1OGPfw ủng hộ bạn t:))

1) -5x- (-3)= 13

=>-5x +3= 13   => -5x = 10  => x = 10: (-5)   => x =-2

2) |x-3|- 7=  13

=>|x-3|= 20  =>  \(\orbr{\begin{cases}x-3=20\\x-3=-20\end{cases}}\Rightarrow\orbr{\begin{cases}x=20+3\\x=3-20\end{cases}}\Rightarrow\orbr{\begin{cases}x=23\\x=-17\end{cases}}\)

4)

 \(17-\left(43-\left|x\right|\right)=45\\ \Rightarrow43-\left|x\right|=-28\Rightarrow\left|x\right|=71\Rightarrow\orbr{\begin{cases}x=71\\x=-71\end{cases}}\)

5)\(\left(x-2\right).\left(x+15\right)=0\Rightarrow\orbr{\begin{cases}x-2=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-15\end{cases}}\)

b) Ta có: \(\dfrac{x-2}{4}=\dfrac{2x+1}{3}\)

\(\Leftrightarrow3\left(x-2\right)=4\left(2x+1\right)\)

\(\Leftrightarrow3x-6=8x+4\)

\(\Leftrightarrow3x-8x=4+6\)

\(\Leftrightarrow-5x=10\)

hay x=-2

Vậy: x=-2

a) x³-9x²-4x-36=0

⇔ x²(x-9)-4(x-9)=0

⇔ (x-9)(x²-4)=0

⇒ Xảy ra 2 trường hợp:

- TH1: x-9=0 ⇔ x=9

- TH2: x²-4=0 ⇔ x=2 hoặc x=-2

Vậy x=9 hoặc x=2 hoặc x=-2.

a: Ta có: \(2x^3-18x=0\)

\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)

\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)

\(\Leftrightarrow-13x=13\)

hay x=-1

c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)

\(\Leftrightarrow3x=12\)

hay x=4

a) 2x³-18x=0

⇔ 2x(x²-9)=0

⇔ 2x(x-3)(x+3)=0

⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

b)(3x-1)(2x+1)-6x(x+2)=11

⇔ 6x²+x-1-6x²-12x=11

⇔ -11x=12

\(\Leftrightarrow x=-\dfrac{12}{11}\)

c) (x-1)³-(x+2).(x²-2x+4)=3.(1-x²)

⇔ x³-3x²+3x-1-x³-8-3+3x²=0

⇔ 3x=12

⇔   x=4

c. (x - 1)³ - (x + 2)(x² - 2x + 4) = 3(1 - x²)

<=> (x³ - 3x² + 3x - 1) - (x³ - 2x² + 4x + 2x² - 4x + 8) = 3 - 3x²

<=> x³ - 3x² + 3x - 1 - x³ + 2x² - 4x - 2x² + 4x - 8 = 3 - 3x²

<=> x³ - x³ - 3x² + 2x² - 2x² + 3x² + 3x - 4x + 4x = 3 + 1 + 8

<=> 3x = 12

<=> x = 4

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) (x + 3)^2 - 4x - 12 = 0

<=> (x + 3)^2 - 4(x + 3) = 0

<=> (x + 3)(x - 1) = 0

<=> x = -3 hoặc x = 1

b) x(x + 5)(x- 5) - (x - 3)(x^2 + 3x + 9) = 7

<=> x^3 - 25x - x^3 + 27 = 7

<=> -25x + 27 = 7

<=> x = 4/5

a/ \(\left(x+3\right)^2-4x-12=0\)

\(\left(x+3\right)^2-4\left(x+3\right)=0\)

\(\left(x+3\right)\left(x+3-4\right)=0\)

\(\left[{}\begin{matrix}x+3=0\Rightarrow x=-3\\x+3-4=0\Rightarrow x=1\end{matrix}\right.\)

---

b/ \(x\left(x+5\right)\left(x-5\right)-\left(x-3\right)\left(x^2+3x+9\right)=7\)

\(x\left(x^2-25\right)-\left(x^3-27\right)=7\)

\(x^3-25x-x^3+27=7\)

\(-25x=-20\)

\(x=\dfrac{20}{25}=\dfrac{4}{5}\)

a, <=>x² +6x+9-4x-12=0

<=> x² +2x -3=0

<=> x² +3x -x-3=0

<=> x.(x+3) - (x+3) =0

<=> (x-1)(x+3)=0

<=> x=1 hoặc x=-3

b, <=> x(x² -25) - (x-3)(x+3)² -7=0 

<=> x³ -25x + (9-x²) (x+3) -7=0

<=> x³ -25x+ 9x+27-x³ -3x² -7=0

<=> -3x² -16x +20=0

<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)

<=> x= 10/3 hoặc x=2

Chúc bạn học tốt nha!

a. `4x^2-20x+25=0`

`<=>(2x)^2-2.2x.5 +5^2=0`

`<=>(2x-5)^2=0`

`<=>2x-5=0`

`<=>x=5/2`

b. `(x-5)(x+5)-(x-3)^2=2(x-7)`

`<=>x^2-25-x^2+6x-9=2x-14`

`<=>6x-34=2x-14`

`<=>4x=20`

`<=>x=5`

\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)

\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)

b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)

\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)

a) Có: (2x)² - 2.2.5.x + 5² = 0

          ⇒ (2x - 5)² = 0 ⇒ 2x - 5 = 0

          ⇒ 2x = 5 ⇒ x = \(\dfrac{5}{2}\)

b) Có: x² - 25 - x² + 6x - 9 = 2x - 14

           ⇒ 6x - 36 = 2x - 14

           ⇒ 4x = 22

           ⇒ x = \(\dfrac{11}{2}\)