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12 tháng 10 2020

a) 2x (x-5) -(x2-10x +25)=0

\(\Leftrightarrow\)2x(x-5)-(x-5)2=0

\(\Leftrightarrow\)(x-5)(2x-x+5)=0

\(\Leftrightarrow\)(x-5)(x+5)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)

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

b) x2 - 9 +3x(x+3) = 0

\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0

\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0

\(\Leftrightarrow\)(x+3)(x-3+3x)=0

\(\Leftrightarrow\)(x+3)(4x-3)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)

c) x3 - 16x = 0

\(\Leftrightarrow\)x(x2-16)=0

\(\Leftrightarrow\)x(x-4)(x+4)=0

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

d) (2x+3)(x-2) - (x2 -4x+4) = 0

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

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

\(\Leftrightarrow\)(x-2)(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

e) 9x2 -(x2 -2x +1)=0

\(\Leftrightarrow\)(3x)2-(x-1)2=0

\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0

\(\Leftrightarrow\)(2x+1)(4x-1)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

f)x3-4x2 -9x +36 = 0

\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0

\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0

\(\Leftrightarrow\)(x-4)(x2-9)=0

\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0

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

\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)

g) 3x - 6 = (x-1).(x-2)

\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)

\(\Leftrightarrow\)x-1=3

\(\Leftrightarrow\)x=4

i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)

k) x2 -1 = (x-1).(2x+3)

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

\(\Leftrightarrow\)x+1=2x+3

\(\Leftrightarrow\)x-2x=3-1

\(\Leftrightarrow\)-x=2

\(\Leftrightarrow\)x=-2

l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6

\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6

\(\Leftrightarrow\)6x-8=6

\(\Leftrightarrow\)6x=14

\(\Leftrightarrow\)x=\(\frac{7}{3}\)

29 tháng 10 2021

a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

9 tháng 12 2021

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

25 tháng 3 2020

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

13 tháng 12 2020

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

9 tháng 12 2023

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

= (x + y - 3x)(x + y + 3x)

= (y - 2x)(4x + y)

e) (x + 5)² - (2x - 1)²

= (x + 5 - 2x + 1)(x + 5 + 2x - 1)

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

9 tháng 12 2023

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

(x - 10)(x + 2) = 0

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2

11 tháng 9 2019

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023

Đề bài yêu cầu gì vậy em.

1 tháng 11 2021

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)