Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)

\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)

Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)

      \(\left|x-\frac{3}{7}\right|\ge0\forall x\)

Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)

\(9x^2-1+\left(3x-1\right).\left(x+2\right)=0\)

\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)

\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)

\(\Leftrightarrow12x^2+5x=3\)

\(\Leftrightarrow12x^2+5x-3=0\)

\(\Leftrightarrow12x^2-4x+9x-3=0\)

\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{4}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x^3-8x\left(x+2\right)=6\\ \Leftrightarrow\left(x^2+3x+2\right).\left(x+3\right)-x^3-8x^2-16x=6\\ \Leftrightarrow x^3+6x^2+11x+6-x^3-8x^2-16x-6=0\\ \Leftrightarrow-2x^2-5x=0\\ \Leftrightarrow x.\left(-2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

10 - { [ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 } = 5

[ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 = 10 - 5 = 5

( x : 3 + 17 ) : 10 + 3 : 2⁴ = 5 x 10

( x : 3 + 17 ) : 10 + 48 = 50

( x : 3 + 17 ) : 10 = 50 - 48

( x : 3 + 17 ) : 10 = 2

x : 3 + 17 = 2 x 10

x : 3 + 17 = 20

x : 3 = 20 - 17 = 3

x = 3 x 3 = 9

a) [(2x+14) : 4 - 3] : 2 = 1

(2x+14) : 4 - 3 = 1/2

(2x+14) : 4  = 1/2 + 3

(2x+14) : 4  = 7/2

2x+14 = 7/2 . 1/4

2x = 7/8 - 1/4

2x = 5/8

x= 5/8.1/2

x= 5/16

Còn phần b nữa ak

a. Ta có: ( x-2)² \(\ge\) 0 , \(\forall\) x

=> ( x-2)² +2023 \(\ge\) 2023

Vậy ...

Dấu bằng xảy ra khi x-2 = 0

b. (x-3)²+(y-2)²-2018

Ta có: \((x-3)^2 \ge0,\forall x\)

           \((y-2) ^2 \ge0,\forall y\) 

=> ( x-3)² + ( y-2)² \(\ge\) 0

=>  ( x-3)² + ( y-2)²-2018 \(\ge\) -2018, \(\forall\) x,y 

Vậy ...

Dấu bằng xảy ra khi x-3=0

                                 y-2=0

c. ( x+1)² +100

Ta có : ( x+1)² \(\ge0,\forall x\) 

=> ( x+1)²+100 \(\ge\) 100

Vậy ...

Dấu bằng xảy ra khi x+1=0

\(\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}=0\)

Ta thấy: \(\begin{cases}\left(x-\frac{1}{5}\right)^{2014}\ge0\\\left(y+0,4\right)^{2016}\ge0\\\left(z-3\right)^{2018}\ge0\end{cases}\)

\(\Rightarrow\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}\ge0\)

\(\Rightarrow\begin{cases}\left(x-\frac{1}{5}\right)^{2014}=0\\\left(y+0,4\right)^{2016}=0\\\left(z-3\right)^{2018}=0\end{cases}\)\(\Rightarrow\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}\)

lần sau viết đề cẩn thận hơn nhé

c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

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