\(\left|x\right|=-10\left(vl\right)\) vì \(\left|x\right|\ge9\forall x\)

\(\Leftrightarrow x\in\varnothing\)

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KB với mình nha

Để \(13⋮\left(x-2\right)\Rightarrow\left(x-2\right)\in U\left(13\right)=\left\{-1;1;-13;13\right\}\)

\(\Rightarrow x\in\left\{1;3;-11;15\right\}\)

a. \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

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

b. \(x^{2019}=x^{2018}\Leftrightarrow x^{2019}-x^{2018}=0\)

\(\Leftrightarrow x^{2018}\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^{2018}=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

c. \(3^{x+1}+3^x=108\)

\(\Leftrightarrow3^x\left(3+1\right)=108\Leftrightarrow3^x.4=108\)

\(\Leftrightarrow3^x=27\Leftrightarrow x=3\)

d. \(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Leftrightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

\(\Leftrightarrow\left(x-5\right)^4.\left[1-\left(x-5\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

\(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

\(\Leftrightarrow9x^2+12x+4-9x^2+12x-4=5x+38\)

\(\Leftrightarrow24x-5x=38\Leftrightarrow19x=38\Leftrightarrow x=2\)

\(\Rightarrow S=\left\{2\right\}\)

\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (ĐKXĐ: \(x\ne0;x\ne\dfrac{1}{2}\))

\(\)\(\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow2x^2+6x-x-3=2x^2+2x\)

\(\Leftrightarrow2x^2-2x^2+6x-x-2x=3\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)\)

\(\Rightarrow S=\left\{1\right\}\)

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

Gọi chiều dài ban đầu là x, chiều rộng là y (x,y > 0)

Theo bài ra, ta có hpt:

\(\left\{{}\begin{matrix}x+y=17\\\left(x+3\right)\left(y-2\right)=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=17\\xy-2x+3y-6=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=17\\-2x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=34\\-2x+3y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=40\\x+y=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=8\end{matrix}\right.\left(tm\right)\)

Vậy, ...

a. \(\left(x-3\right)\in U\left(2\right)=\left\{-1;-2;1;2\right\}\)

\(\Rightarrow x\in\left\{2;1;4;5\right\}\)

b. \(\left(x+2\right)\in U\left(3\right)=\left\{-1;-3;1;3\right\}\)

\(\Rightarrow x\in\left\{-3;-5;-1;1\right\}\)

\(\left|-x+7\right|=8\Leftrightarrow\left[{}\begin{matrix}-x+7=8\\-x-7=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)