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1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$
1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)
\(=x^2-2x+5-x^2-2x+7x-14\)
\(=3x-9\)
2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)
\(=-5x^2+25x+x^3-7x-3x^2+21\)
\(=x^3-8x^2+18x+21\)
3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)
\(=x^3-x^2-2x-x^2-4x+5\)
\(=x^3-2x^2-6x+5\)
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
1.
\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
2.
\(\left(x^2-9\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
3.
\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right).3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
4.
\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
\(x-5=\frac{1}{3}\left(x+2\right)\)
\(3x-15=x+2\)
\(2x=17\)
\(x=\frac{17}{2}\)
vậy x = 17/2
\(x-5=\frac{1}{3}\left(x+2\right)\)
\(x-5=\frac{1}{3}x+\frac{2}{3}\)
\(x-\frac{1}{3}x=\frac{2}{3}+5\)
\(\frac{2}{3}x=\frac{17}{3}\)
\(x=\frac{17}{2}\)