Rút gọn:
a) \(\frac{x^2+2x+1}{5x^3+5x^2}\)
b) \(\frac{10xy^2.\left(x+y\right)}{15xy.\left(x+y\right)^3}\)
c) \(\frac{2x^2+2x}{x^2+2x+1}\)
Giải từng bước giúp mình nhé. Thanks nhiều ạ!
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a)\(\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}\)=\(\frac{x+1}{5x^2}\)
b)\(\frac{10y}{15\left(x+y\right)^2}\)
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
b/ ĐKXĐ: ...
\(2x^3-2y^3+5x-5y=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)
\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)
Thế vào pt dưới:
\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)
Đặt \(x+\frac{1}{x}+1=t\)
\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)
\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}+1=-1\\x+\frac{1}{x}+1=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-2x+1=0\end{matrix}\right.\) \(\Leftrightarrow...\)
a/ ĐKXĐ: ...
\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)
TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)
TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)
\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)
\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)
Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)
\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)
\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)
Vế trái luôn dương nên pt vô nghiệm
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)