\(\dfrac{3x}{2\cdot5}+\dfrac{3x}{5\cdot8}+\dfrac{3x}{8\cdot11}+\dfrac{3x}{11\cdot14}=\dfrac{1}{21}\\ x\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\\ x\cdot\dfrac{3}{7}=\dfrac{1}{21}\\ x=\dfrac{1}{21}:\dfrac{3}{7}\\ x=\dfrac{1}{9}\)

Lời giải:

\(3x(3x-8)-9x^2+8=0\)

\(\Leftrightarrow 9x^2-24x-9x^2+8=0\)

\(\Leftrightarrow 8-24x=0\Leftrightarrow x=\frac{1}{3}\)

3x ( 3x - 8 ) - 9x² + 8 = 0  

9x² - 24x - 9x² + 8 = 0

( 9x² - 9x² ) - 24x + 8 = 0

- 24x + 8 = 0

-24x = -8

x = \(\dfrac{1}{3}\)

Vậy x = \(\dfrac{1}{3}\)

Lời giải:
ĐK: $x\geq \frac{1}{3}$

PT $\Leftrightarrow \sqrt{(3x-1)+6\sqrt{3x-1}+9}+\sqrt{(3x-1)-6\sqrt{3x-1}+9}=3x+4$
$\Leftrightarrow \sqrt{(\sqrt{3x-1}+3)^2}+\sqrt{(\sqrt{3x-1}-3)^2}=3x+4$

$\Leftrightarrow |\sqrt{3x-1}+3|+|\sqrt{3x-1}-3|=3x+4$

Nếu $x\geq \frac{10}{3}$ thì:

$\sqrt{3x-1}+3+\sqrt{3x-1}-3=3x+4$

$\Leftrightarrow 2\sqrt{3x-1}=3x+4$

$\Leftrightarrow 2\sqrt{3x-1}=(3x-1)+5$

$\Leftrightarrow (\sqrt{3x-1}-1)^2=-4< 0$ (vô lý)

Nếu $\frac{1}{3}\leq x< \frac{10}{3}$ thì:

$\sqrt{3x-1}+3+3-\sqrt{3x-1}=3x+4$

$\Leftrightarrow 2=3x\Leftrightarrow x=\frac{2}{3}$ (thỏa mãn)

Vậy.......

\(\left(3x-8\right)^{10}=\left(3x-8\right)^2\\ \Rightarrow\left(3x-8\right)^{10}-\left(3x-8\right)^2=0\\ \Rightarrow\left(3x-8\right)^2\left[\left(3x-8\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(3x-8\right)^2=0\\\left(3x-8\right)^8=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-8=0\\3x-8=1\\3x-8=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

\(243\le3^{2x-1}\le3^8\\ \Rightarrow3^5\le3^{2x-1}\le3^8\\ \Rightarrow2x-1\in\left\{5;6;7;8\right\}\\ \Rightarrow x\in\left\{3;\dfrac{7}{2};4;\dfrac{9}{2}\right\}\)

\(\left(3x+4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)

\(27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)

\(27x^3-27x^3+108x^2+24x^2+144x+72x=64-64=0\)

\(132x^2+216x=0\)

\(x\left(132x+216\right)=0\)

\(\Rightarrow x=\hept{\begin{cases}0\\\frac{216}{132}=\frac{18}{11}\end{cases}}\)

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(3x.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)

\(3x.\frac{1}{7}=\frac{1}{21}\)

\(\frac{3}{7}x=\frac{1}{21}\)

\(x=\frac{1}{21}:\frac{3}{7}\)

\(x=\frac{7}{81}\)

5(8+3x)+2(3x-8)=0

=>40+15x+6x-16=0

=>21x=16-40

=>21x=-24

=>x=-24/21

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

=> \(x\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)=\frac{1}{21}\)

=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

=> \(x.\frac{3}{7}=21\)

=> x = 49 

Vậy x = 49

Xin lỗi bạn nhé x = 1/9 bước cuối mình ghi sót 

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