Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
a: \(=a^2+2ab+b^2-a^2-2ab-b^2=0\)
b: \(=x^3+27-54-x^3=-27\)
Câu 4:
\(\Leftrightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)
\(\Leftrightarrow3x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{0;1\right\}\)
Bài 2:
a: \(\dfrac{5xy-4y}{2x^2y^3}+\dfrac{3xy+4y}{2x^2y^3}\)
\(=\dfrac{5xy-4y+3xy+4y}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)
b: \(\dfrac{1-2x}{2x}+\dfrac{2x}{2x-1}+\dfrac{1}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)\left(2x-1\right)+2x\cdot2x-1}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1+4x^2-1}{2x\left(2x-1\right)}=\dfrac{4x-2}{2x\left(2x-1\right)}=\dfrac{1}{x}\)
c: \(\dfrac{3}{x+y}+\dfrac{1}{x-y}-\dfrac{3x}{x^2-y^2}\)
\(=\dfrac{3x-3y+x+y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-2y}{\left(x-y\right)\left(x+y\right)}\)
Bài 1 :
a) \(3x^2+4x-7\)
\(=3x^2-3x+7x-7\)
\(=3x\left(x-1\right)+7\left(x-1\right)\)
\(\left(x-1\right)\left(3x+7\right)\)
b) \(3x^2+48+24x-12y^2\)
\(=3\left(x^2+16+8x-4y^2\right)\)
\(=3\left[\left(x+4\right)^2-\left(2y\right)^2\right]\)
\(=3\left(x-2y+4\right)\left(x+2y+4\right)\)
Bài 2 :
a) Phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x-3y\ne0\\2xy-1\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne3y\\2xy\ne1\\x\ne-2\end{cases}}}\)
b) \(A=\left(\frac{x+2y}{x-3y}+\frac{5y}{3y-x}-2xy\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)
\(A=\left(\frac{x+2y}{x-3y}-\frac{5y}{x-3y}-\frac{2xy\left(x-3y\right)}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)
\(A=\left(\frac{x+2y-5y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)
\(A=\left(\frac{x-3y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)
\(A=\frac{\left(x-3y\right)-2xy\left(x-3y\right)}{x-3y}\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)
\(A=\frac{-\left(x-3y\right)\left(2xy-1\right)\left(x+2\right)}{\left(x-3y\right)\left(2xy-1\right)}+\frac{x^2-3}{x+2}\)
\(A=\frac{-\left(x+2\right)\left(x+2\right)}{\left(x+2\right)}+\frac{x^2-3}{x+2}\)
\(A=\frac{-x^2-4x-4+x^2-3}{x+2}\)
\(A=\frac{-4x-7}{x+2}\)
c) Thay x = 3 ( vì y bị triệt tiêu hết nên ko xét đến đỡ mệt ng :) )
\(A=\frac{-4\cdot3-7}{3+2}=\frac{-19}{5}\)
Câu 1 :
a. \(7x^2y-7xy^2=7xy\left(x-y\right)\)
b. \(x^2-5x+xy-5y=x\left(x-5\right)+y\left(x-5\right)\)
\(=\left(x+y\right)\left(x-5\right)\)
c. \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)\)
\(\left(x-3\right)\left(x-1\right)\)
Câu 2 :
a. \(\dfrac{x}{x+6}+\dfrac{6}{x+6}=\dfrac{x+6}{x+6}=1\)
b. \(\dfrac{10x^2}{x^2-4x+4}:\dfrac{2x}{\left(x-2\right)^3}=\dfrac{10x^2}{\left(x-2\right)^2}.\dfrac{\left(x-2\right)^3}{2x}\)
\(=5x\left(x-2\right)\)
thank's pạn nha pạn có biết 3 bài còn lại ko giúp mk với