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b: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Leftrightarrow12x=12\)
hay x=2
d: Ta có: \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow9x=-2\)
hay \(x=-\dfrac{2}{9}\)
a) \(=3\left(x-2\right)\)
b) \(=2\left(x+5\right)\)
c) \(=x\left(x-3\right)\)
d) \(=\left(x-3\right)\left(2x+7\right)\)
e) \(=\left(x-1\right)\left(3x+2\right)\)
f) \(=x\left(x+2\right)\left(x-5\right)\)
1) \(x\left(x-1\right)+\left(1-x\right)^2\)
\(=x\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x-1\right)\left(x+x-1\right)\)
\(=\left(x-1\right)\left(2x-1\right)\)
2) \(2x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left[2x-\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
3) \(3x\left(x-1\right)^2-\left(1-x\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^3\)
\(=\left(x-1\right)^2\left(3x+x-1\right)\)
\(=\left(x-1\right)^2\left(4x-1\right)\)
4) \(3x\left(x+2\right)-5\left(x+2\right)^2\)
\(=\left(x+2\right)\left[3x-5\left(x+2\right)\right]\)
\(=\left(x+2\right)\left(3x-5x-10\right)\)
\(=\left(x+2\right)\left(-2x-10\right)\)
\(=-2\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow2.\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)
nha
2.(x+3)-x^2-3x=0
<=> 2(x+3) -x.(x+3) = 0
<=> (2-x).(x+3) = 0
<=> 2-x = 0 hoặc x+3 = 0
<=> x= 2 hoặc x= -3
Vậy _____________
Bài 1.
8: \(=\left(x-2y\right)\cdot x\cdot\left(x+3\right)\)
9: \(=\left(5x+2\right)\left(x-3\right)-x\left(x-3\right)\)
\(=\left(x-3\right)\left(4x+2\right)\)
=2(2x+1)(x-3)
3: \(=2\left(x+2\right)\left(25x-15-x\right)\)
\(=2\left(x+2\right)\left(24x-15\right)\)
=6(x+2)(8x-5)
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
1: \(x\left(x-1\right)+\left(1+x\right)^2\)
\(=x^2-x+x^2+2x+1\)
\(=2x^2+x+1\)
Đa thức này ko phân tích được nha bạn
2: \(\left(x+1\right)^2-3\left(x+1\right)\)
\(=\left(x+1\right)\cdot\left(x+1\right)-\left(x+1\right)\cdot3\)
\(=\left(x+1\right)\left(x+1-3\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
3: \(2x\cdot\left(x-2\right)-\left(x-2\right)^2\)
\(=2x\left(x-2\right)-\left(x-2\right)\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(2x-x+2\right)\)
\(=\left(x-2\right)\left(x+2\right)\)
4: \(3x\left(x-1\right)^2-\left(1-x\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^3\)
\(=3x\left(x-1\right)^2+\left(x-1\right)^2\cdot\left(x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(3x+x-1\right)\)
\(=\left(x-1\right)^2\cdot\left(4x-1\right)\)
5: \(3x\left(x+2\right)-5\left(x+2\right)^2\)
\(=\left(x+2\right)\cdot3x-\left(x+2\right)\cdot\left(5x+10\right)\)
\(=\left(x+2\right)\left(3x-5x-10\right)\)
\(=\left(-2x-10\right)\left(x+2\right)\)
\(=-2\left(x+5\right)\left(x+2\right)\)
6: \(4x\left(x-y\right)+3\left(y-x\right)^2\)
\(=4x\left(x-y\right)+3\left(x-y\right)^2\)
\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)
\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)
\(=\left(x-y\right)\left(7x-3y\right)\)
a,3x.(x-1)+x-1=0
3x(x-1) +(x-1) =0
(3x+1)(x-1) =0
Th1: 3x+1 =0
3x = -1
x= -1/3
Th2: x-1 =0
x=1
Vậy x= -1/3 và x=1
a,3x.(x-1)+x-1=0
=> (x - 1) (3x + 1) = 0
=> x - 1 = 0 hoặc 3x + 1 = 0
=> x = 1 hoặc x = \(\frac{-1}{3}\)
b,2.(x+3)-x2-3x=0
=> 2. (x + 3) - x (x-3) = 0
=> (x - 3) (2 - x) = 0
=> x - 3 = 0 hoặc 2 - x = 0
=> x = 3 hoặc x = 2
P/s: Mỗi chữ hoặc bạn thay = dấu [ nhé (Thay dấu này chắc bạn biết cách trình bày rồi nha)
- Nhớ tick [Nếu đúng] nha