`( x+ 1,2)+(x+1,6)+(x+ 2,0)+ ..... +(x+15,2)= 331,2`

`(x+x+x+...+x)+(1,2+1,6+2,0+...+15,2)=331,2`

Số hạng của dãy là :

`(15,2-1,2) : 0,4 +1=36`

Tổng dãy là :

`(15,2 +1,2) xx 36 :2=295,2`

`36x+295,2=331,2`

`36x=331,2-295,2`

`36x=36`

`x=36:36`

`x=1`

\(\sqrt{x^2+x+5}=x+1\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{\left(x^2+x+5\right)^2}=\left(x+1\right)^2\\ \Leftrightarrow x^2+x+5=x^2+2x+1\\ \Leftrightarrow x^2+x-x^2-2x=1-5\\ \Leftrightarrow-x=-4\\ \Leftrightarrow x=4\left(tm\right)\)

\(\dfrac{x-3}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{2}{3-\sqrt{x}}\\ =\dfrac{x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-3}\\ =\dfrac{x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3+\sqrt{x}-3+2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(a,2\left|3x-1\right|+1=5\\ \Rightarrow2\left|3x-1\right|=5-1\\ \Rightarrow2\left|3x-1\right|=4\\ \Rightarrow\left|3x-1\right|=4:2\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(b,\left|\dfrac{x}{2}-1\right|=3\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

\(c,\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3\\ \Rightarrow\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-x=\dfrac{13}{5}\\-x=-\dfrac{17}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=\dfrac{17}{5}\end{matrix}\right.\)

\(d,\left|x-\dfrac{1}{3}\right|=2\dfrac{3}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{13}{5}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{13}{5}\\x-\dfrac{1}{3}=-\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}+\dfrac{1}{3}\\x=-\dfrac{13}{5}+\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{44}{15}\\x=-\dfrac{34}{15}\end{matrix}\right.\)

`4*(3x-1)^3 -5^2=475`

`=>4*(3x-1)^3 = 475 +25`

`=>4*(3x-1)^3 = 500`

`=>(3x-1)^3=500:4`

`=>(3x-1)^3= 125`

`=> (3x-1)^3=5^3`

`=>3x-1=5`

`=>3x=5+1`

`=>3x=6`

`=>x=6/3=2`

`(x-1)(x+1)`

`=x^2 -1^2`

`=x^2-1`

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HHDT số `3` : `x^2-y^2=(x-y)(x+y)`

\(5\dfrac{3}{4}=2\dfrac{1}{2}+3\dfrac{1}{x}\\ \dfrac{23}{4}=\dfrac{5}{2}+3\dfrac{1}{x}\\ 3\dfrac{1}{x}=\dfrac{23}{4}-\dfrac{5}{2}\\ 3\dfrac{1}{x}=\dfrac{23}{4}-\dfrac{10}{4}\\ 3\dfrac{1}{x}=\dfrac{13}{4}\\ \Rightarrow x=4\)

\(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=\dfrac{7}{5}+\dfrac{7}{10}x\\ \Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}+\dfrac{13}{5}=\dfrac{7}{5}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}+\dfrac{26}{10}=\dfrac{14}{10}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}-\dfrac{7x}{10}=\dfrac{14}{10}-\dfrac{26}{10}\\ \Rightarrow\dfrac{-8x}{10}=-\dfrac{12}{10}\\ \Rightarrow\dfrac{-4x}{5}=-\dfrac{6}{5}\\ \Rightarrow-20x=-30\\ \Rightarrow x=\dfrac{3}{2}\)