\(a,=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\\ b,=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\\ c,=\left(x-5\right)^2\\ d,=\left(x-8\right)\left(x+8\right)\)

\(5^x-2-3^2=2^4-\left(2^8x2^4-2^{10}x2^2\right)\)

\(5^x-2-3^2=2^4-\left(2^{8+4}-2^{10+2}\right)\)

\(5^x-2-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(5^x-2-3^2=2^4-0\)

\(5^x-2-3^2=2^4\)

\(5^x-2-9=16\)

\(5^x-2=16+9\)

\(5^x-2=25\)

\(5^x=25+2\)

\(5^x=27\)

Bởi vì 27 không phân tích được 1 số có số mũ là 2

\(\Rightarrow\) Không tồn tại x

\(5^{x-2}-9=16-\left(256.16-1024.4\right)\)

\(\Rightarrow5^{x-2}-9=16-\left(4096-4096\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}-9=16\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow x-2=25:5\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=5\)

\(5^{x-2}-3^2=2^4-\left(2^8\cdot2^4-2^{10}\cdot2^2\right)\)

\(\Rightarrow5^{x-2}-3^2=2^4-\left(2^{12}-2^{12}\right)\)

\(\Rightarrow5^{x-2}-9=16-0\)

\(\Rightarrow5^{x-2}=16+9\)

\(\Rightarrow5^{x-2}=25\)

\(\Rightarrow5^{x-2}=5^2\)

\(\Rightarrow x-2=2\)

\(\Rightarrow x=2+2\)

\(\Rightarrow x=4\)

Vậy: x=4

d. 2x²(x - y) + 2y(y - x)

= 2x²(x - y) - 2y(x - y)

= (2x² - 2y)(x - y)

= 2(x² - y)(x - y)

e. 5a²b(a - 2b) - 2a(2b - a)

= 5a²b(a - 2b) + 2a(a - 2b)

= (5a²b + 2a)(a - 2b)

= a(5ab + 2)(a - 2b)

f. 4x²y(x - y) + 9xy²(x - y)

= (4x²y + 9xy²)(x - y)

= xy(4x + 9y)(x - y)

g. 50x²(x - y)² - 8y²(y - x)²

= 50x²(x² - 2xy + y²) - 8y²(y² - 2xy + x²)

= 50x²(x² - 2xy + y²) - 8y²(x² - 2xy + y²)

= 50x²(x - y)² - 8y²(x - y)²

= (50x² - 8y²)(x - y)²

= 2(25x² - 4y²)(x - y)².

g: Ta có: \(50x^2\left(x-y\right)^2-8y^2\left(y-x\right)^2\)

\(=2\left(x-y\right)^2\left(25x^2-4y^2\right)\)

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

$D\,=2x(10x^2-5x-2)-5x(4x^2-2x-1)\\\quad =20x^3-10x^2-4x-20x^3+10x^2+5x\\\quad =(20x^3-20x^3)+(-10x^2+10x^2)+(-4x+5x)\\\quad =x$

Thay $x=-5$ vào $D=x$

$\Rightarrow D=-5$

Vậy $D=-5$ với $x=-5$

Ta có: \(D=2x\left(10x^2-5x-2\right)-5x\left(4x^2-2x-1\right)\)

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

=x=-5

\(4x-9.3x-10.2x=0\)

\(=>4x-27x-20x=0\)

\(=>x\left(4-27-20\right)=0\)

\(=>x.\left(-43\right)=0\)

\(=>x=0\)

(25x5 – 5x4 + 10x2) : 5x2

= 25x5 : 5x2 + (-5x4) : 5x2 + 10x2 : 5x2

= (25 : 5).(x5 : x2) + (-5 : 5).(x4 : x2) + (10 : 5).(x2 : x2)

= 5.x5 – 2 + (-1).x4 – 2 + 2.1

= 5x3 – x2 + 2

\(\dfrac{x}{2}=\dfrac{y}{5}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{25}=\dfrac{10x^2-y^2}{10\cdot4-25}=\dfrac{x^2+y^2}{4+25}\\ \Leftrightarrow A=\dfrac{40-25}{29}=\dfrac{15}{29}\)

a) Ta có: \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)

\(\Leftrightarrow M=x^2+11xy-y^2\)

Vậy: \(M=x^2+11xy-y^2\)

b) Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)

\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)

\(\Leftrightarrow N=-x^2+10xy-12y^2\)

Vậy: \(N=-x^2+10xy-12y^2\)