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Bài 1:
a) 2x^2 -3x + 1 = 2x^2 -2x -x +1 = 2x.(x-1) - (x-1) = (x-1).(2x-1)
b) 2x^3y - 2xy^3 - 4xy^2 - 2xy = 2xy.(x^2 - y^2 - 2y -1) = 2xy.[ x^2 - (y^2 + 2y+1)] = 2xy.[x^2 - (y+1)^2]
= 2xy.(x-y-1).(x+y+1)
c) (x^2 + x+3).(x^2 + x +5) - 8 = (x^2+x+4-1).(x^2+x+4+1) - 8 = (x^2+x+4)^2 - 1 - 8 = (x^2+x+4)^2 - 3^2
= (x^2+x+4-3).(x^2+x+4+3) = (x^2+x+1).(x^2+x+7)
Bài 2:
a) (x+2).(x^2-2x+4) - (x^3+2x) = 0
x^3 + 8 - x^3 - 2x = 0
8 - 2x = 0
x = 4
b) x^2 - 2x - 8 = 0
x^2 +2x - 4x - 8 = 0
x.(x+2) - 4.(x+2) = 0
(x+2).(x-4) = 0
...
bn tự làm tiếp nha
1) bạn ktra lại đề
2) \(x^6+2x^5+x^4-2x^3-2x^2+1=\left(x^3+x^2-1\right)^2\)
3)
a) \(x^2+x-2=0\)
<=> \(\left(x-1\right)\left(x+2\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Vậy...
b) \(3x^2+5x-8=0\)
<=> \(\left(x-1\right)\left(3x+8\right)=0\)
<=> \(\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
Vậy...
a) \(3x^2-3y^2-12x+12y\)
\(=\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-3y-12\right)\)
\(=\left(x-y\right).3.\left(x-y-4\right)\)
b) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
c) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Bài 1
\(x^5+x^4+1=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)+\left(-x^3-x^2-x\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
Bài 2
Ta có: \(\left(ax+b\right)\left(x^2+cx+1\right)=ax^3+bx^2+acx^2+bcx+ax+b\)
\(=ax^3+\left(b+ac\right)x^2+\left(bc+a\right)x+b=x^3-3x-2\)
\(\Rightarrow a=1\)
\(\Rightarrow b+ac=0\)
\(\Rightarrow bc+a=-3\)
\(\Rightarrow b=-2\)
Thay giá trị của \(a=1;b=-2\)vào \(b+ac=0\)ta được
\(\Leftrightarrow-2+c=0\Rightarrow c=2\)
Vậy \(a=1;b=-2;c=2\)
Bài 3
Ta có \(\left(x^4-3x^3+2x^2-5x\right)\div\left(x^2-3x+1\right)=x^2+1\left(dư-2x+1\right)\)
\(\Rightarrow b=2x-1\)
Bài 4 (cũng làm tương tự như bài 3 nhé )
Bài 5(bài nãy dễ nên bạn tự làm đi nhé)
Bài 6
\(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\)
\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\)\(\Rightarrow a-b=0\Rightarrow a=b\)
Bài 7
\(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2+a^2+b^2+b^2+c^2+c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a-b=0\Rightarrow a=b\)
\(\Rightarrow b-c=0\Rightarrow b=c\)
\(\Rightarrow a-c=0\Rightarrow a=c\)
Vậy \(a=b=c\)