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25 tháng 8 2020

a) \(A=x^2y+y+xy^2-x\) (hẳn đề là vậy)

\(A=xy\left(x+y\right)+\left(y-x\right)\)

\(A=\left(-5\right).2\left(-5+2\right)+2+5\)

\(A=30+7=37\)

b) \(B=3x^3-2y^3-6x^2y^2+xy\)

\(B=3.\left(\frac{2}{3}\right)^3-2.\left(\frac{1}{2}\right)^3-6.\left(\frac{2}{3}\right)^2.\left(\frac{1}{2}\right)^2+\frac{2}{3}.\frac{1}{2}\)

\(B=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}\)

\(B=\frac{11}{36}\)

c) \(C=2x+xy^2-x^2y-2y\)

\(C=2.\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right).\left(-\frac{1}{3}\right)^2-\left(-\frac{1}{2}\right)^2.\left(-\frac{1}{3}\right)-2.\left(-\frac{1}{3}\right)\)

\(C=-1-\frac{1}{18}+\frac{1}{12}+\frac{2}{3}\)

\(C=-\frac{11}{36}\)

\(A=2x+xy^2-x^2y-2y\)

\(=2\left(x-y\right)-xy\left(x-y\right)\)

\(=\left(x-y\right)\left(2-xy\right)\)

\(=\left(-\dfrac{1}{2}-\dfrac{-1}{3}\right)\left(2-\dfrac{-1}{2}\cdot\dfrac{-1}{3}\right)\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\left(2-\dfrac{1}{6}\right)\)

\(=\dfrac{-1}{6}\cdot\dfrac{11}{6}=-\dfrac{11}{36}\)

Bài 3: 

a: Ta có: C=A+B

\(=x^2-2y+xy+1+x^2+y-x^2y^2-1\)

\(=2x^2-y+xy-x^2y^2\)

b: Ta có: C+A=B

\(\Leftrightarrow C=B-A\)

\(=x^2+y-x^2y^2-1-x^2+2y-xy-1\)

\(=-x^2y^2+3y-xy-2\)

5 tháng 8 2021

a, `(8x^3-4x^2): 4x -(4x^2-5x) : 2x + (2x)^2`

`=4x (2x^2-x) : 4x - 2x(2x-5/2 ) :2x + 4x^2`

`=2x^2-x-2x+5/2+4x^2`

`=6x^2-3x+5/2`

b, `(3x^3-x^2y) :x^2 -(xy^2+x^2y) :xy + 2x(x+1)`

`=x^2 (3x-y) :x^2 -xy(y+x) + (2x^2+2x)`

`=3x-y-y-x+2x^2+2x`

`=2x^2+4x-2y`

29 tháng 7 2019

a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)

 

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1\)

Ta có ĐT tương đương

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

Thay \(x=9\) ; \(y=10\) ; \(z=11\) vào BT có :

\(\left(9-1\right)\left(10-1\right)\left(11-1\right)=720\)

Vậy .........

17 tháng 7 2018

C = xyz - xy - yz - xz + x + y +z- 1

= xy(z-1) - y(z-1) - x(z-1) + 1(z-1)

(xy-y-x+1)(z-1)

Ta có: \(\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)

\(=\left[x^2\left(x-y\right)+y^2\left(x-y\right)\right]\left(x+y\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=x^4-y^4=2^4-\left(\dfrac{1}{2}\right)^4=16-\dfrac{1}{16}=\dfrac{255}{16}\)