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Bài 1:
a) Ta có: \(M=\left(\dfrac{x+2}{x^2+2x+1}+\dfrac{x-2}{1-x^2}\right)\cdot\dfrac{x+1}{x}\)
\(=\left(\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}-\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\right)\cdot\dfrac{x+1}{x}\)
\(=\dfrac{x^2-x+2x-2-\left(x^2+x-2x-2\right)}{\left(x+1\right)^2\cdot\left(x-1\right)}\cdot\dfrac{x+1}{x}\)
\(=\dfrac{x^2+x-2-x^2+x+2}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x}\)
\(=\dfrac{2x}{\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x}\)
\(=\dfrac{2}{x^2-1}\)
Bài 2:
1: Ta có: \(\left(x-5\right)^2+\left(x+3\right)^2=2\left(x-4\right)\left(x+4\right)-5x+7\)
\(\Leftrightarrow x^2-10x+25+x^2+6x+9=2\left(x^2-16\right)-5x+7\)
\(\Leftrightarrow2x^2-4x+34=2x^2-32-5x+7\)
\(\Leftrightarrow2x^2-4x+34-2x^2+5x+25=0\)
\(\Leftrightarrow x+59=0\)
hay x=-59
Vậy: S={-59}
Bài 1:
b) Ta có: \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Leftrightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)
\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)
nên x-89=0
hay x=89
Vậy: S={89}
Bài 1:
a)ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}\)
Suy ra: \(x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhân\right)\\x=6\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;6}
a)|7x-5|=|2x-3|
=>7x-5=2x-3 hoặc 7x-5=3-2x
=>5x=2 hoặc 9x=8
=>x=\(\frac{2}{5}\) hoặc x=\(\frac{8}{9}\)
Vậy x=\(\frac{2}{5}\) hoặc x=\(\frac{8}{9}\)
b)|4x-5|=x-7
\(VT\ge0\Rightarrow VP\ge0\Rightarrow x-7\ge0\Rightarrow x\ge7\)
=>4x-5=x-7 hoặc 4x-5=-(x-7)
=>3x=-2 hoặc 5x=12
=>x=\(-\frac{2}{3}\)(loại do \(x\ge7\)) hoặc x=\(\frac{12}{5}\)(loại do \(x\ge7\))
Vậy pt vô nghiệm
c)Ta thấy: \(\hept{\begin{cases}\left(x+8\right)^4\ge0\\\left|y-7\right|\ge0\end{cases}}\)
\(\Rightarrow\left(x+8\right)^4+\left|y-7\right|\ge0\)
Dấu = khi \(\hept{\begin{cases}\left(x+8\right)^4=0\\\left|y-7\right|=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x+8=0\\y-7=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=-8\\y=7\end{cases}}\)
Vậy \(\hept{\begin{cases}x=-8\\y=7\end{cases}}\)
1: \(\Leftrightarrow x-2-7x+7=-1\)
=>-6x+5=-1
hay x=1(loại)
3: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)-\left(x+1\right)\left(x+3\right)=4\)
\(\Leftrightarrow x^2+x-2-x^2-4x-3=4\)
=>-3x=9
hay x=-3(loại)
4: \(\Leftrightarrow x^2+2x+1-x^2+2x-1=3x\cdot\dfrac{x+1-x+1}{x+1}\)
\(\Leftrightarrow4x=\dfrac{6x}{x+1}\)
\(\Leftrightarrow4x^2+4x-6x=0\)
\(\Leftrightarrow4x^2-2x=0\)
=>2x(2x-1)=0
hay \(x\in\left\{0;\dfrac{1}{2}\right\}\)
Bài 4:
a: Ta có: \(\widehat{OAB}=\widehat{ODC}\)
\(\widehat{OBA}=\widehat{OCD}\)
mà \(\widehat{ODC}=\widehat{OCD}\)
nên \(\widehat{OAB}=\widehat{OBA}\)
hay ΔOAB cân tại O