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a: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
d: \(\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}\)
\(\dfrac{1}{3-x}=\dfrac{-1}{x-3}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x^2-9}=\dfrac{1}{\left(x+3\right)\left(x-3\right)}\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề của bạn hơn nhé.
a: \(\dfrac{1}{2a}+\dfrac{2}{3b}\)(ĐKXĐ: a<>0 và b<>0)
\(=\dfrac{1\cdot3b+2\cdot2a}{2a\cdot3b}\)
\(=\dfrac{3b+4a}{6ab}\)
b: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}\)(ĐKXĐ: \(x\notin\left\{1;-1\right\}\))
\(=\dfrac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{-4x}{x^2-1}\)
c: \(\dfrac{x+y}{xy-y}+\dfrac{z}{yz}\)(ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >0\\z< >0\end{matrix}\right.\))
\(=\dfrac{x+y}{y\left(x-1\right)}+\dfrac{1}{y}\)
\(=\dfrac{x+y+x-1}{y\left(x-1\right)}\)
\(=\dfrac{2x+y-1}{y\left(x-1\right)}\)
d: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{2}{x-3}-\dfrac{12}{x^2-9}\)
\(=\dfrac{2}{x-3}-\dfrac{12}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: ĐKXĐ: x<>2
\(\dfrac{1}{x-2}+\dfrac{2}{x^2-4x+4}\)
\(=\dfrac{1}{x-2}+\dfrac{2}{\left(x-2\right)^2}\)
\(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)