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a) \(3x^2-3y^2=3\left(x^2-y^2\right)=3\left(x-y\right)\left(x+y\right)\)
b) \(x^2-xy+7x-7y=\left(x^2+7x\right)-\left(xy+7y\right)\)
\(=x\left(x+7\right)-x\left(y+7\right)=x\left(x+7-y-7\right)=x\left(x-y\right)\)
c)\(x^2-3x+2=x^2-2x-x+2=\left(x^2-x\right)-\left(2x-2\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)=\left(x-2\right)\left(x-1\right)\)
d) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)\)
\(=x\left[\left(x+y\right)^2-16\right]=x\left(x+y-4\right)\left(x+y+4\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
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)
a) 7(x-y)
b) (x-3y)2
c) xy(x+y)-(x+y)=(xy-1)(x+y)
d) x2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)
a) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )
a) Ta có : x2 + 7x + 12
= x2 + 3x + 4x + 12
= (x2 + 3x) + (4x + 12)
= x(x + 3) + 4(x + 3)
= (x + 4)(x + 3)
Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha
c) \(\left(x^2-x+6\right)^2+\left(x-3\right)^2\)
\(=-\left(\left(x^2+x-6\right)-\left(x+3\right)^2\right)\)
\(=-\left(x^2+x-6-x-3\right)\left(x^2+x+6+x+3\right)\)
\(=-\left(x^2-9\right)\left(x^2+2x+9\right)\)
\(=-\left(\left(x-9\right)\left(x+9\right)\left(x^2+2x+9\right)\right)\)
a) 7x2 - 14xy + 7y2
= 7(x2 - 2xy + y2)
= 7(x - y)2
b) x2 + 3x - 10
= - ( x2 - 3x + 10 )
= - ( x2 - 2 . 3/2x + 9/4 + 31/4 )
= - (( x - 3/2 )2 + 31/4
= (x - 3/2)2 - \(\left(\sqrt{\frac{31}{4}}\right)^2\)
= \(\left(x-\frac{3}{2}-\sqrt{\frac{31}{4}}\right)\left(x-\frac{3}{2}+\sqrt{\frac{31}{4}}\right)\)