a: \(\dfrac{10x^3y-5x^2y^2-25x^4y^3}{-5xy}=-2x^2+xy+5x^3y^2\)

c: \(\dfrac{27x^3-y^3}{3x-y}=9x^2+3xy+y^2\)

mình tl được rồi thôi ko cần đâu các bạn

ảo thật đấy

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

\(=\left(x-2\right)\left(4x^2-8x+3\right)\)

\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

\(=\left(x-2\right)\left(4x^2-8x+3\right)\)

\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)

\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

\(=\left(x-2\right)\left(4x^2-8x+3\right)\)

\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a:Ta có: \(x\left(x-1\right)+x=4\)

\(\Leftrightarrow x^2-x+x=4\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

5/2\(\times\)1=\(\frac{5}{2\times1}hay=\frac{5}{2}\times1\)

a: \(C\left(x\right)=A\left(x\right)+B\left(x\right)\)

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

=-2x+3

b: Đặt C(x)=0

=>-2x+3=0

hay x=3/2