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$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
a) \(5x^2y^4:10x^2y=\dfrac{5x^2y^4}{10x^2y}=\dfrac{5.x^2.y.y^3}{5.2.x^2.y}=\dfrac{y^3}{2}\)
Các câu khác tương tự mà làm
b) \(\dfrac{3}{4}x^3y^3:\left(-\dfrac{1}{2}x^2y^2\right)=\left[\dfrac{3}{4}:\left(-\dfrac{1}{2}\right)\right].\left(x^3:x^2\right).\left(y^3:y^2\right)\)
\(=-\dfrac{3}{2}xy\)
c)\(\left(-xy\right)^{10}:\left(-xy\right)^5=\left(-xy\right)^{10-5}=\left(-xy\right)^5\)
a ) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b ) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
c ) \(\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\left(x-3y\right)\)
\(=\left(x^2y^2-\dfrac{1}{3}xy+3y\right)x-3y\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\)
\(=x^3y^2-\dfrac{1}{3}x^2y+3xy-3x^2y^3+xy^2-9y^2\)
d ) \(\left(\dfrac{1}{5}x-1\right)\left(x^2-5x+2\right)\)
\(=\dfrac{1}{5}x\left(x^2-5x+2\right)-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-x^2+\dfrac{2}{5}x-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-2x^2+\dfrac{27}{5}x-2\)
\(a,\left(4x^3-3xy^2+2xy\right).\left(-\dfrac{1}{3}x^2y\right)\)
\(=\dfrac{-x}{3y}+\dfrac{y}{x}-\dfrac{2}{3x}\)
\(b,\left(5xy-x^2+y\right)\left(\dfrac{2}{5}xy^2\right)\)
\(=\dfrac{2}{y}-\dfrac{2x}{5y^2}+\dfrac{2}{5xy}\)
c,=\(\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\) b,=\(2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)
a)
\(A=x^2y-y+xy^2-x\)
\(A=\left(x^2y-x\right)-\left(y-xy^2\right)\)
\(A=x.\left(xy-1\right)-y.\left(1-xy\right)\)
\(A=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(A=\left(xy-1\right).\left(x+y\right)\)
Thay \(x=-5\) và \(y=2\) vào biểu thức A, ta được:
\(A=\left[\left(-5\right).2-1\right].\left[\left(-5\right)+2\right]\)
\(A=\left(-11\right).\left(-3\right)\)
\(A=33.\)
Vậy giá trị của biểu thức A tại \(x=-5\) và \(y=2\) là \(33.\)
Chúc bạn học tốt!
a) \(A=\left(3x+2\right)^2-9x\left(x+1\right)\)
\(A=9x^2+12x+4-9x^2-9x\)
\(A=3x+4\)
\(B=\left(2x-1\right)^2-2\left(2x-1\right)\left(5x-1\right)+\left(5x-1\right)^2\)
\(B=\left[2x-1-\left(5x-1\right)\right]^2\)
\(B=\left(2x-1-5x+1\right)^2\)
\(B=\left(-3x\right)^2\)
\(B=9x^2\)
a: \(=-2x^2y-\dfrac{5}{4}x^2y^2-\dfrac{4}{3}xy^4\)
b: \(=4x^{n-1+n+1}-6x^{n-1+1}=4x^{2n}-6x^n\)
`(xy-5)(xy+2)+3(xy-2)(xy+2)-(3xy-1/2)^2 +5x^2y^2`
`=(xy+2)(xy-5+3xy-6)-(9x^2y^2-3xy+1/4)+5x^2y^2`
`=(xy+2)(4xy-11)-9x^2y^2+3xy-1/4+5x^2y^2`
`=4x^2y^2-11xy+8xy-22-4x^2y^2+3xy-1/4`
`=-89/4`
Ta có: \(\left(xy-5\right)\left(xy+2\right)+3\left(xy-2\right)\left(xy+2\right)-\left(3xy-\dfrac{1}{2}\right)^2+5x^2y^2\)
\(=x^2y^2-3xy-10+3x^2y^2-12-\left(9x^2y^2-3xy+\dfrac{1}{4}\right)+5x^2y^2\)
\(=9x^2y^2-3xy-22-9x^2y^2+3xy-\dfrac{1}{4}\)
\(=\dfrac{-89}{4}\)