a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>x=0(nhận)
b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
\(\Leftrightarrow x^2+2x-15=x^2-1-8\)
=>2x-15=-9
=>2x=-6
hay x=-3(nhận)
a, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne5\\x\ne-6\end{matrix}\right.\)
\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\\ \Leftrightarrow\dfrac{\left(x+6\right)^2}{\left(x+6\right)\left(x-5\right)}+\dfrac{\left(x-5\right)^2}{\left(x+6\right)\left(x-5\right)}-\dfrac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}=0\\ \Leftrightarrow\dfrac{x^2+12x+36+x^2-10x+25-2x^2-23x-61}{\left(x+6\right)\left(x-5\right)}=0\\ \Rightarrow-21x=0\\ \Leftrightarrow x=0\left(tm\right)\)
b, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\dfrac{8}{\left(x-3\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2+2x-15-x^2+1+8}{\left(x-3\right)\left(x-1\right)}=0\\ \Rightarrow2x-6=0\\ \Leftrightarrow x=3\left(ktm\right)\)
b. \(ĐK:x\ne1;3\)
\(\Rightarrow\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)-8}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)
\(\Leftrightarrow x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(ktm\right)\)
a. \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\\ ĐKXĐ:x\ne5;x\ne6\\ \Leftrightarrow\dfrac{\left(x+6\right)\left(x+6\right)}{x^2+x-30}+\dfrac{\left(x-5\right)\left(x-5\right)}{x^2+x-30}=\dfrac{2x^2+23x+61}{x^2+x-30}\\ \Leftrightarrow x^2+6x+6x+36+x^2-5x-5x+25=2x^2+23x+61\\ \Leftrightarrow x^2+6x+6x+x^2-5x-5x-2x^2-23x=61-36-25\\ \Leftrightarrow-21x=0\\ \Leftrightarrow x=0\left(n.h.ận\right)\)
b. \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(ĐKXĐ:x\ne1;x\ne3\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{x^2-4x+3}=\dfrac{\left(x+1\right)\left(x-1\right)}{x-3}-\dfrac{8}{x^2-4x+3}\\ \Leftrightarrow x^2-3x+5x-15=x^2-1-8\\ \Leftrightarrow x^2-3x+5x-x^2=-1-8+15\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\left(l.o.ại\right)\)