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

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

1.

Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)

Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)

\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}m=-x^2+3x\\m=-x^2+x+1\end{matrix}\right.\)

Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):

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

\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)

Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\): 

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)

Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)

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

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

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

3)(x²+2x+4)(2-x)+x(x-3)(x+4)-x²+24=0

\(\Rightarrow\)8-x³+x(x²+4x-3x-12)-x²+24=0

\(\Rightarrow\)8-x³+x³+4x²-3x²-12x-x²+21=0

\(\Rightarrow\)-12x+29=0

\(\Rightarrow\)-12x=-29

\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)

a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_#