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a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) [ MTC: x(x+3) ]
\(=\dfrac{x.2}{x\left(x+3\right)}+\dfrac{1\left(x+3\right)}{x\left(x+3\right)}\)
\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)
\(=\dfrac{3x+3}{x\left(x+3\right)}\)
\(=\dfrac{3\left(x+1\right)}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)
\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) \(\left[MTC:2\left(x-1\right)\left(x+1\right)\right]\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2+2x+1\right)-4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x^2-2x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)}{2\left(x+1\right)}\)
a) Ta có :
\(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}\)
b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)-2x.2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{-3x+1}{2\left(x-1\right)\left(x+1\right)}\)
c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12+36}{6y\left(y-6\right)}=\dfrac{y^2-24}{6y\left(y-6\right)}\)
d) \(\dfrac{6+x}{x+3x}+\dfrac{3}{2x+6}=\dfrac{6+x}{4x}+\dfrac{3}{2\left(x+3\right)}\)
\(=\dfrac{\left(6+x\right)\left(2x+6\right)+12x}{8x\left(x+3\right)}\)(Đề câu này phải sửa thành\(\dfrac{6+x}{x^2+3x}chứ\)) ???
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
a/ \(\dfrac{4x+2}{3x^2-x}:\dfrac{x^2+3x}{1-3x}=-\dfrac{4x+2}{x\left(1-3x\right)}\cdot\dfrac{1-3x}{x^2+3x}=-\dfrac{4x^2+2}{x\left(x^2+3x\right)}\)
b/ \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2-12xy+9y^2}{1-x^2}=-\dfrac{2\left(2x+3y\right)}{1-x}\cdot\dfrac{\left(1-x\right)\left(1+x\right)}{\left(2x+3y\right)^2}=\dfrac{-2\left(x+1\right)}{2x+3y}=\dfrac{-2x-2}{2x+3y}\)
c/ \(\dfrac{x^4-xy^3}{2xy+y^2}:\dfrac{x^3+x^2y+xy^2}{2x+y}=\dfrac{x\left(x^3-y^3\right)}{y\left(2x+y\right)}\cdot\dfrac{2x+y}{x\left(x^2+xy+y^2\right)}=\dfrac{x\left(x-y\right)\left(x^2+xy+y^2\right)}{y}\cdot\dfrac{1}{x\left(x^2+xy+y^2\right)}=\dfrac{x-y}{y}\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
Lời giải
a)
\(\left(\frac{3}{2x-y}-\frac{2}{2x+y}-\frac{1}{2x-5y}\right).\frac{4x^2-y^2}{y^2}\)
\(=\frac{3(4x^2-y^2)}{(2x-y)y^2}-\frac{2(4x^2-y^2)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x-y)(2x+y)}{(2x-y)y^2}-\frac{2(2x-y)(2x+y)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x+y)-2(2x-y)}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{2x+5y}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{(2x+5y)(2x-5y)-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{4x^2-25y^2-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}=\frac{-25}{2x-5y}+\frac{1}{2x-5y}=\frac{-24}{2x-5y}\)
Ta có đpcm.
b)
\(\frac{x^2-x+1}{x^2+x}.\frac{x+1}{3x-2}.\frac{9x-6}{x^2-x+1}\)
\(=\frac{(x^2-x+1)(x+1).3(3x-2)}{x(x+1)(3x-2)(x^2-x+1)}\)
\(=\frac{3}{x}\) (đpcm)
\(a,\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2.\left(x-1\right)}+\dfrac{-2x}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)}{2.\left(x-1\right).\left(x+1\right)}+\dfrac{\left(-2x\right).x}{x.\left(x+1\right).\left(x-1\right)}=\dfrac{\left(x+1\right).\left(x+1\right)-2x^2}{x.\left(x+1\right)\left(x-1\right)}\)
b: \(=\dfrac{y^2-12y+24}{6y\left(y-6\right)}\)
c: \(=\dfrac{12-2x+3x}{2x\left(x+3\right)}=\dfrac{x+12}{2x\left(x+3\right)}\)