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

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

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

thay x = 15 ta được

\(20.15^2+9.15=4635\)

câu b tương tự

THay vào và tính bạn nhé

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.

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

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

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

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

khó thế

P = x(x - y) - x + y²(x - y) - y² + 5

P = x - x + y² - y² + 5

P = 5
 

Q = x²(x - y) - x² + y²(x - y) - y² + 5(x - y) - 2015

Q = 5 - 2015

Q = -2010

\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y+x^2}{xy^2}\)

\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3}{x^2y^2}-\dfrac{5xy+x^3}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{\left(5xy-5xy\right)+x^3+y^3}{x^2y^2}\)

\(=\dfrac{x^3+y^3}{x^2y^2}\)

\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}=\dfrac{y\left(5x+y^2\right)}{y\cdot x^2y}-\dfrac{x\left(5y-x^2\right)}{x\cdot xy^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\) \(\left(x,y\ne0\right)\)

\(\frac{5x+y^2}{x^2y}-\frac{5y-x^2}{xy^2}\)

\(=\frac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)

\(=\frac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\frac{x^3+y^3}{x^2y^2}\)

Đề bài là j vậy ?O?