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.

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

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

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

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

Đặt x/3=y/5=k

=>x=3k; y=5k

\(A=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{5\cdot9+3\cdot25}{10\cdot9-3\cdot25}=8\)

a) \(5x^2\)\(\left(x-2y\right)\)\(-\)\(15x\)\(\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

\(=5x\left(x-2y\right)\left(x-3\right)\)

b)  \(3\left(x-y\right)\)\(-\)\(5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(3+5x\right)\)

c)  \(10x\left(x-y\right)\)\(-\)\(8y\left(y-x\right)\)

\(=\)\(10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(5x+4\right)\left(x-y\right)\)

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

\(=x^2\)\(\left(x-5\right)\)\(-\)\(4\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-4\right)\)

\(=\left(x-5\right)\left(x-2\right)\left(x-2\right)\)

a) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

\(=\left(x-2y\right)\left(x-3\right)5x\)

b)\(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(3+5x\right)\left(x-y\right)\)

c)\(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(10x+8y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

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

\(=x^2\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x^2-4\right)\left(x-5\right)\)

\(=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)

câu 1/ 5x(\(4x^2\)-2x+1) - 2x(\(10x^2\)-5x-2)

= 5x.\(4x^2\)-5x.2x+ 5x.1 - ( 2x.\(10x^2\)-2x.5x-2x.2)

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

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

= \(-11x^3\) + 9x

à cj ơi, e 2k6, đọc phần lí thuyết r lm, nên có lỗi sai j mong cj thông cảm

5x.( 4x²-2x+1)- 2x( 10x²-5x-2)

= 20x³- 10x²+5x - 20x³+ 10x²+ 4xx

= (5x+4x)= 9x

Thay x=3 vào 9x, ta có:

9×3=27

Vậy biểu thức trên bằng 27

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

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

\(5x^4y+10x^3y+10x^2y^3+5xy^4\)

\(=5xy.x^3+5xy.2x^3+5xy.2xy^3+5xy.y^3\)

\(=5xy\left(x^3+2x^3+2xy^3+y^3\right)\)

Ht pt

Lời giải:
Đặt $\frac{x}{3}=\frac{y}{5}=t(t\neq 0)$

$\Rightarrow x=3t; y=5t$

Khi đó:

$\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5(3t)^2+3(5t)^2}{10(3t)^2-3(5t)^2}=\frac{120t^2}{15t^2}=8$