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Ta có:\(7x^3-5x^2=x^2\left(7x-5\right)\)
\(x^2-10x+25=\left(x-5\right)^2\)
\(x^2-10x+16=x^2-8x-2x+16=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)
\(x^2-2x-15=x^2-5x+3x-15=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
\(2x^2+7x+3=2x^2+x+6x+3=x\left(2x+1\right)+3\left(2x+1\right)=\left(x+3\right)\left(2x+1\right)\)
a) \(x^2-10x+16=x^2-8x-2x+16=\left(x^2-8x\right)-\left(2x-16\right)=x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x-2\right)\)b) \(x^2-2x-15=x^2+3x-5x-15=\left(x^2+3x\right)-\left(5x+15\right)=x\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-5\right)\)c) \(2x^2+7x+3=2x^2+x+6x+3=\left(2x^2+x\right)+\left(6x+3\right)=x\left(2x+1\right)+3\left(2x+1\right)=\left(2x+1\right)\left(x+3\right)\)
\(a,\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
\(b,\left(x^2+1\right)^2-4x^2\)
\(=x^4+2x^2+1-4x^2\)
\(=x^4-2x^2+1\)
\(\left(x^2-1\right)^2\)
\(c\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-8y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-8y+4+y-2\right)\)
\(=\left(y+2\right)\left(y^2-7y+2\right)\)
a) ( a3 - b3) + ( a - b)2
= (a-b) (a2 + ab + b2 ) + (a-b)2
= (a-b) (a2 + ab + b2 +a -b )
hok tốt
1) \(6x^2-15x+6\)
\(=6x^2-3x-12x+6\)
\(=3x\left(2x-1\right)-6\left(2x-1\right)\)
\(=\left(3x-6\right)\left(2x-1\right)\)
\(=3\left(x-2\right)\left(2x-1\right)\)
b) \(6x^2-20x+6\)
\(=6x^2-2x-18x+6\)
\(=2x\left(3x-1\right)-6\left(3x-1\right)\)
\(=\left(2x-6\right)\left(3x-1\right)\)
\(=2\left(x-3\right)\left(3x-1\right)\)
c) \(-10x^2-7x+6\)
\(=5x-10x^2+6-12x\)
\(=5x\left(1-2x\right)+6\left(1-2x\right)\)
\(=\left(5x+6\right)\left(1-2x\right)\)
d) \(10x^2-7x-12\)
\(=10x^2-15x+8x-12\)
\(=5x\left(2x-3\right)+4\left(2x-3\right)\)
\(=\left(5x+4\right)\left(2x-3\right)\)
1) 6x^2 - 15x + 6
= 6x^2 - 12x - 3x +6
= 6x (x - 2) - 3 (x - 2)
= (x - 2) (6x - 3)
= 3 (x - 2) (x - 1)
2) 6x^2 - 20x + 6
= 6x^2 - 18x - 2x + 6
= 6x (x - 3) - 2 (x - 3)
= (x - 3) (6x - 2)
= 2 (x - 3) (3x - 1)
3) -10x^2 - 7x + 6
= -10x^2 + 5x - 12x + 6
= 5x (1 - 2x) + 6 (1 - 2x)
= (1 - 2x) (5x + 6)
4) 10x^2 - 7x - 12
= 10x^2 - 15x + 8x - 12
= 5x (2x - 3) + 4 (2x - 3)
= (2x - 3) (5x + 4)
good luck !
1 )
=x3-2x2+6x2-12x+5x-10
=x2(x-2)+6x(x-2)+5(x-2)
=(x-2)(x2+6x+5)
=(x-2)(x2+x+5x+5)
=(x-2)[x(x+1)+5(x+1)]
=(x-2)(x+1)(x+5)
toàn mũ lớn hơn 3 khó làm quá!!!! >.<
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Bài 1 :
a) \(A=x^2-6x+11\)
\(A=x^2-2\cdot x\cdot3+3^2+2\)
\(A=\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(B=2\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
c) \(C=5x-x^2\)
\(C=-\left(x^2-5x\right)\)
\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)
\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Bài 2 :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-x^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)
\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
1)x(x2 - 19 - 30)
2)x(x2 - 7 - 6)
3)x(x2 + 4x - 7 - 10)
( 4 tích mình làm tiếp 3 câu cuối)
a) 7x3 - 5x2
= x2( 7x - 5 )
b) x2 - 10x + 25
= x2 - 2.5.x + 52
= ( x - 5 )2