\(a,\Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{49+18}{21}=\dfrac{67}{21}\\ b,\Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{30-13}{14}=\dfrac{17}{14}\\ c,\Rightarrow3x=\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{27-10}{12}=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}\cdot\dfrac{1}{3}=\dfrac{17}{36}\)

\(x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{67}{21}\)

\(x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{17}{14}\)

\(x=\left(\dfrac{9}{4}-\dfrac{5}{6}\right):3=\dfrac{17}{36}\)

\(x-\dfrac{6}{7}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}\\ \Rightarrow x=\dfrac{67}{21}\)

\(x+\dfrac{13}{14}=\dfrac{15}{7}\\ \Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}\\ \Rightarrow x=\dfrac{17}{14}\)

\(\dfrac{9}{4}-3\times x=\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{9}{4}-\dfrac{5}{6}\\ \Rightarrow3\times x=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}:3\\ \Rightarrow x=\dfrac{17}{36}\)

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

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

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

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

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

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

Lời giải:

PT $\Leftrightarrow 4x^2+4x+1=3(x^2-4)+18$

$\Leftrightarrow 4x^2+4x+1=3x^2+6$

$\Leftrightarrow x^2+4x-5=0$

$\Leftrightarrow (x-1)(x+5)=0$

$\Leftrightarrow x-1=0$ hoặc $x+5=0$

$\Leftrightarrow x=1$ hoặc $x=-5$

\(\left(2x+1\right)^2=3\left(x-2\right)\left(x+2\right)+18\)

\(\Leftrightarrow4x^2+4x+1=3\left(x^2-4\right)+18\)

\(\Leftrightarrow4x^2+4x+1=3x^2-12+18\)

\(\Leftrightarrow4x^2+4x+1=3x^2+6\)

\(\Leftrightarrow4x^2-3x^2+4x=6-1\)

\(\Leftrightarrow x^2+4x=5\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow x^2+5x-x-5=0\)

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

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

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

Vậy: \(S=\left\{-5;1\right\}\)

[3x-1]⁵=[3x-1]⁸

=>[3x-1]⁸-[3x-1]⁵⁼⁰

   [3x-1]⁵.[[3x-1]³-1]=0

=>[3x-1]⁵ =0​ hoặc [3x-1]³-1=0

=>3x-1=0 hoặc3x-1 =1

=>x=1/3 hoac x=2/3

Vay....

=>

x=3,507135583 nha bạn

Ai tk mình đi mình bị âm nè!

Ai tk mình mình tk lại!!!

Xy-3x=-19  

=> x(y - 3) = -19

Xy+3x-2y=11

=> x(y + 3) - 2y - 6 = 5

=> x(y + 3) - 2(y + 3) = 5

=> (x - 2)(y + 3) = 5

xét bảng như câu a nha

3x+4y-xy=16 

=> x(3 - y) - 12 + 4y = 4

=> x(3 - y) -4(3 - y) = 4

Xy+3x+2y=-3 

=> x(y + 3) + 2y + 6 = 3

=> x(y + 3) + 2(y + 3) = 3

=> (x + 2)(y + 3) = 3

\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left(3x-1+1\right)\left(3x-1-1\right)=0\)

\(\Rightarrow\left(3x-1\right)^{10}.3x.\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x=0\\3x-1=0\\3x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=1\\3x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

x = 0 hoặc 1/3 hoặc 2/3

*THAM KHẢO