\(A=6xy\left(xy-y^2\right)-8x^2\left(x-y\right)^2+5y\left(x^2-xy\right)\)

\(=6xy^2\left(x-y\right)-8x^2\left(x-y\right)\left(x-y\right)+5xy\left(x-y\right)\)

\(=x\left(x-y\right)\left(6y^2-8x\left(x-y\right)+5y\right)\)

\(=x\left(x-y\right)\left(6y^2-8x^2+8xy+5y\right)\)

\(=x\left(x-y\right)\left[2\left(3y+2x\right)\left(y-2x\right)+16xy+5y\right]\)

Thay x=1/2; y =2 ta được

\(A=\frac{1}{2}\left(\frac{1}{2}-2\right)\left[0+16\cdot\frac{1}{2}\cdot2+5\cdot2\right]=-\frac{1}{2}\cdot\frac{3}{2}\cdot26=-\frac{39}{2}\).

6xy ( xy - y² ) - 8x² ( x - y )² + 5y ( x² - xy ) 

= 6x²y² - 6xy³ - 8x³ + 8x²y + 5yx² - 5xy²

= xy ( 6xy - 6y² + 8x + 5x - 5y ) - 8x³

Thay x= \(\frac{1}{2}\)     ; y = 2

= 6 - 6.4 + 8. \(\frac{1}{2}\)      + 5. \(\frac{1}{2}\)       -  5.2 - 8.8

=> 6 - 24 + 4 + 2,5 - 10 - 64

= - 85,5

\(6xy\left(xy-ỳ^2\right)-8x^2\left(x-y\right)^2+5y\left(x^2-xy\right)\)

\(=6x^2y^2-6xy^3-8x^2\cdot\left(x^2-y^2\right)+5x^2y-5xy^2\)

\(=6x^2y^2-6xy^3-8x^4+8x^2y^2+5x^2y-5xy^2\)

\(=-8x^4+14x^2y^2+5x^2y-6xy^3-5xy^2\)

 Thay \(x=\frac{1}{2};y=2\) vào đa thức \(-8x^4+14x^2y^2+5x^2y-6xy^3-5xy^2\)

\(-8\left(\frac{1}{2}\right)^4+14\left(\frac{1}{2}\right)^2\cdot2^2+5\left(\frac{1}{2}\right)^2\cdot2-6\cdot\frac{1}{2}\cdot2^3-5\cdot\frac{1}{2}\cdot2^2\)

\(=-\frac{1}{2}+14+\frac{5}{2}-24-10\)

\(=-18\)

Giải:

\(6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(=6xy^2\left(x-y\right)-8x^2\left(x-y^2\right)+5xy^2\left(x-y\right)\)

\(=11xy^2\left(x-y\right)-8x^2\left(x-y^2\right)\)

Thay x và y vào ta được:

\(11.\dfrac{1}{2}.2^2\left(\dfrac{1}{2}-2\right)-8\left(\dfrac{1}{2}\right)^2\left(\dfrac{1}{2}-2^2\right)\)

\(=22\left(\dfrac{-3}{2}\right)-2\left(\dfrac{-7}{2}\right)\)

\(=-33+14=-19\)

Vậy ...

\(5x\left(4x^2+2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(=20x^3+10x^2+5x-20x^3+10x^2+4x\)

\(=20x^2+9x\)

thay x = 15 ta được

\(20.15^2+9.15=4635\)

câu b tương tự

THay vào và tính bạn nhé

VICTOR_Nobita Kun ns như thật

Lời giải:
a.

$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$

$=9x=9.15=135$

b.

$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$

$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$

c.

$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$

$=-8x^3+9x^2y^2-xy^3$

$=(-2x)^3+(3xy)^2-xy^3$

$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$

a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)

Thay x = 15 vào bt A ta có

A = 9 . 15 = 135

b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)

Thay x = -1/5 ; y = - 1/2 vào bt B ta có

\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(=9x^2y^2-xy^3-8x^3\)

Thay x = 1/2 ; y = 2 vào bt C ta có

\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)

d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)

\(=12x^2+12x-3\)

\(\left|x\right|=2\Rightarrow x=\pm2\)

Thay x = 2 vào bt D có

\(D=12.4+12.2-3=69\)

Thay x = - 2 vào bt D ta có

\(D=12.4-12.2-3=21\)

Ta có : 6xy.( xy - y2 ) - 8x2 .( x - y2 ) + 5y2 ( x2 - xy )

=6x²y²-6xy³-8x³+8x²y²+5x²y²-5xy³

⁼ (6x²y²+8x²y²+5x²y²) +(-6xy³-5xy³)-8x³

=19x²y²-11xy³-8x³

Thay x= 1/2, y=2 ta đc :

19.1/2².2²-11.1/2.2³-8.1/2³

=19.1/4.4-11.1/2.8-8.1/8

=19-11.4-1

=-36