\(a,\Leftrightarrow\left(x^2-8x+16\right)-10=0\\ \Leftrightarrow\left(x-4\right)^2-10=0\\ \Leftrightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\\ b,\Leftrightarrow10\left(2x-1\right)+6x=9x\\ \Leftrightarrow20x-10-3x=0\\ \Leftrightarrow17x=10\Leftrightarrow x=\dfrac{10}{17}\)

a:Ta có: \(16-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-5\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)

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

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

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

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

b) \(x\left(3x-1\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)

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

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

a) \(\dfrac{x+5}{4}-\dfrac{x}{2}+1=0\)

<=> \(\dfrac{x+5-2x+4}{4}=0\)

<=> -x + 9 = 0 <=> x = 9

b) \(3\left(x+2\right)=\dfrac{x-4}{3}\)

<=> 9x + 18 = x-4

<=> 8x = -22

<=> x = \(\dfrac{-11}{4}\)

a, 0,16 : x = 2 – 1,6.         

0,16 : x = 0,4

x = 0,16 : 0,4

x = 0,4

câu b thiếu nha

a: =>0,16:x=0,4

hay x=0,4

b: =>x-2,5=5.18

hay x=7,68

\(x^2-x^2-2x-1-2=0\)

\(-2x-3=0\Leftrightarrow x=\dfrac{-2}{3}\)

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

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

a)\(x^2-x\left(x+2\right)-1=2\\ \Rightarrow x^2-x^2-2x-1=2\\ \Rightarrow-2x=3\\ \Rightarrow x=-\dfrac{3}{2}\)

b) \(x^2=\left(2x-1\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=2x-1\\x=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

a. x² - x(x + 2) - 1 = 2

<=> x² - x² - 2x = 3

<=> -2x = 3

<=> \(x=-\dfrac{3}{2}\)

b. x² = (2x - 1)²

<=> x² - (2x - 1)² = 0

<=> (x - 2x + 1)(x + 2x - 1) = 0

<=> (1 - x)(3x - 1) = 0

<=> \(\left[{}\begin{matrix}1-x=0\\3x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)