Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1
a) \(-\frac{1}{3}xy\).(3\(x^2yz^2\))
=\(\left(-\frac{1}{3}.3\right)\).\(\left(x.x^2\right)\).(y.y).\(z^2\)
=\(-x^3\).\(y^2z^2\)
b)-54\(y^2\).b.x
=(-54.b).\(y^2x\)
=-54b\(y^2x\)
c) -2.\(x^2y.\left(\frac{1}{2}\right)^2.x.\left(y^2.x\right)^3\)
=\(-2x^2y.\frac{1}{4}.x.y^6.x^3\)
=\(\left(-2.\frac{1}{4}\right).\left(x^2.x.x^3\right).\left(y.y^2\right)\)
=\(\frac{-1}{2}x^6y^3\)
Bài 3:
a) \(f\left(x\right)=-15x^2+5x^4-4x^2+8x^2-9x^3-x^4+15-7x^3\)
\(f\left(x\right)=\left(5x^4-x^4\right)-\left(9x^3+7x^3\right)-\left(15x^2+4x^2-8x^2\right)+15\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
b)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=4\cdot1^4-16\cdot1^3-11\cdot1^2+15\)
\(f\left(1\right)=-8\)
\(f\left(x\right)=4x^4-16x^3-11x^2+15\)
\(f\left(-1\right)=4\cdot\left(-1\right)^4-16\cdot\left(-1\right)^3-11\cdot\left(-1\right)^2+15\)
\(f\left(-1\right)=24\)
Bài 3:
a: \(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)
b: \(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)
Bài 2:
\(A+B=4x^4-5xy+5y^2+3x^2+2xy-y=4x^4+3x^2-3xy+5y^2-y\)
\(A-B=4x^4-5xy+5y^2-3x^2-2xy+y=4x^4-3x^2+5y^2-7xy+y\)
\(B-A=-\left(A-B\right)=-4x^4+3x^2-5y^2+7xy-y\)
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
Bài 1 :
\(A=x^2-2xy^2+y^4=\left(x-y^2\right)^2=-\left(y^2-x\right)^2\)
Mà \(B=-\left(y^2-x\right)^2\)
Nên ta có : đpcm
Bài 2
Đặt \(\left(x+1\right)\left(x-2\right)\left(2x-1\right)=0\)
TH1 : x = -1
TH2 : x = 2
TH3 : x = 1/2
Bài 4 :
a, \(\left(2x+3\right)\left(5-x\right)=0\Leftrightarrow x=-\frac{3}{2};5\)
b, \(\left(x-\frac{1}{2}\right)\left(3x+1\right)\left(2-x\right)=0\Leftrightarrow x=\frac{1}{2};-\frac{1}{3};2\)
c, \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;-2\)
d, \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)