1: =>15-2x=-11

=>2x=26

hay x=13

\(a.\left[{}\begin{matrix}2x+9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(b.\left[{}\begin{matrix}x-3=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

a: \(\left(x+5\right)^2>=0\forall x\)

\(\left(2y-8\right)^2>=0\forall y\)

Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)

b: \(\left(x+3\right)\left(2y-1\right)=5\)

=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)

a) \(x^2\left(x-5\right)+x^2-4x-5=0\)

⇔\(x^2\left(x-5\right)+\left(x-5\right)\left(x+1\right)=0\)

⇔\(\left(x-5\right)\left(x^2+x+1\right)=0\)

Vì \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)∀x

⇒\(x=5\)

Vậy ...

b) \(x^8-1=0\)

⇔\(x^8=1\)

⇒ \(x=+-1\)

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

a) \(\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

b) \(\left(x^2+5\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\\x^2-25=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\\x^2=25\end{matrix}\right.\) \(\Leftrightarrow x^2=25\) \(\Leftrightarrow x=\pm5\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b.\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12+\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x^2-4\right)\)

\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm

\(a,x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\dfrac{x+1}{x-2}-\dfrac{5}{x-2}=\dfrac{12}{x^2-4}+1\) (ĐKXĐ : x ≠ 2 ; x ≠ -2)

\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow x^2+3x+2-5x-10=12+x^2+2x-2x+4\)

\(\Leftrightarrow2x=24\)

\(\Leftrightarrow x=12\left(N\right)\)

câu c chưa học :vv

a)

<=> x (x-2 ) = 0

<=> x =0 

x = 2

b)

đkxđ : x khác 2 , x khác -2

<=> \(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{12}{x^2-4}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\dfrac{x^2+3x+2}{....}-\dfrac{5x-10}{....}-\dfrac{12}{...}+\dfrac{x^2-4}{....}=0\)

<=> \(x^2+3x+2-5x+10-12+x^2-4=0\)

<=> \(2x^2-2x-4=0\)

<=> x =2 (ktm)

Vậy..

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)