Ta có : \(\frac{4^{x+2}+4^{x+1}+4^x}{21}=\frac{3^{2x}+3^{2x+1}+3^{2x+3}}{31}\)

\(\Rightarrow\frac{4^x\left(4^2+4+1\right)}{21}=\frac{3^{2x}\left(1+3+3^3\right)}{31}\)

\(\Rightarrow\frac{4^x.21}{21}=\frac{3^{2x}.31}{31}\)

=> 4^x = 3^2x

=> 4^x = (3²)^x

=> 4^x = 9^x

=> \(\frac{4^x}{9^x}=1\)(vì lũy thừa của một số khác 0 luôn luôn là 1 số khác 0)

=> \(\left(\frac{4}{9}\right)^x=1\)

=> x = 0 

Vậy x = 0

\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)

Vậy `x=41/35`

__

\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)

Vậy `x=-2`

a)

-4/7 - x = 3/5 - 2x

2x - x = 3/5 + 4/7

x = 41/35

Vậy x = 41/35

b)

3/7.x - 2/3.x = 10/21

x(3/7 - 2/3) = 10/21

x.(-5/21) = 10/21

x = 10/21 : (-5/21) = -2

Vậy x = -2

Lời giải:

\(|2x-5|+x=21\)

\(\Leftrightarrow |2x-5|=21-x\)

\(\Rightarrow \left[\begin{matrix} 2x-5=21-x\\ 2x-5=x-21\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{26}{3}\\ x=-16\end{matrix}\right.\) (đều thỏa mãn)

Vậy \(x=\frac{26}{3}; x=-16\)

a) 9            d) -18

b) 1

c) -3

a, \(\left|9+x\right|=2x\)

\(\Leftrightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}\Leftrightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)

b, \(\left|5x\right|-3x=2\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3x=2\\5x-3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

c, \(\left|x+6\right|-9=2x\)

\(\Leftrightarrow\orbr{\begin{cases}x+6-9=2x\\x+6-9=-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x+9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=7x\end{cases}\Leftrightarrow}\orbr{\begin{cases}x+6=3\\x=7x-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)

d. Tương tự 

❤                                              ✔

a) Tìm số nguyên x, biết:

Với \(x\in Z\), ta có:

9 - x = 8 - (2x + 16)

<=> 9 - x = 8 - 2x - 16 

<=> 9 - x = -2x - 8

<=> x  = -17 (TM)

Vậy x = -17

b) Với \(x\in Z\), ta có:

18 - 2x = 21 - (3x - 5)

<=> 18 - 2x = 21 - 3x + 5

<=> 18 - 2x = 26 - 3x

<=> x = 8 (TM)

Vậy x = 8

a) Ta có: 9-x=8-(2x+16)

\(\Leftrightarrow9-x=8-2x-16\)

\(\Leftrightarrow9-x=-2x-8\)

\(\Leftrightarrow9-x+2x+8=0\)

\(\Leftrightarrow x+17=0\)

hay x=-17

Vậy: x=-17

b) Ta có: 18-2x=21-(3x-5)

\(\Leftrightarrow18-2x=21-3x+5\)

\(\Leftrightarrow18-2x+3x-26=0\)

\(\Leftrightarrow x-8=0\)

hay x=8

Vậy: x=8

a) \(\left|9+x\right|=2x\)

\(\Rightarrow\left[{}\begin{matrix}9+x=2x\\9+x=-2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-x=9\\-2x-x=9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\-3x=9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

b) \(\left|5x\right|-3x=2\)

\(\Leftrightarrow\left|5x\right|=2+3x\)

\(\Rightarrow\left[{}\begin{matrix}5x=2+3x\\5x=-2-3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x-3x=2\\5x+3x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=2\\8x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

c) \(\left|x+6\right|-9=2x\)

\(\Leftrightarrow\left|x+6\right|=2x+9\)

\(\Rightarrow\left[{}\begin{matrix}x+6=2x+9\\x+6=-2x-9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-2x=9-6\\x+2x=-9-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-3x=3\\3x=-15\end{matrix}\right.\)

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

d) \(\left|2x-3\right|+x=21\)

\(\Leftrightarrow\left|2x-3\right|=21-x\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=21-x\\2x-3=-21+x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x+x=21+3\\2x-x=-21+3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=24\\x=-18\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-18\end{matrix}\right.\)

dễ thế cũng hỏi . chịu

a,\(\left|9+x\right|=2x\)

\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

Vậy...

Trường hợp 2 chưa chắc chắn lắm!!!

a) \(\left|9+x\right|=2x\)

Xét trường hợp 1:

\(9+x=2x\)

\(\Leftrightarrow9+x-2x=0\)

\(\Leftrightarrow9-x=0\)

\(\Leftrightarrow x=9\)

Xét trường hợp 2:

\(9+x=-2x\)

\(\Leftrightarrow9+x-\left(-2x\right)=0\)

\(\Leftrightarrow9+x+2x=0\)

\(\Leftrightarrow9+3x=0\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-9:3\)

\(\Leftrightarrow x=-3\)

Vậy x=9 hoặc x=-3

b) \(\left|x+6\right|-9=2x\)

\(\Leftrightarrow\left|x+6\right|=2x+9\)

Xét trường hợp 1:

\(x+6=2x+9\)

\(\Leftrightarrow x+6-\left(2x+9\right)=0\)

\(\Leftrightarrow x+6-2x-9=0\)

\(\Leftrightarrow-3-x=0\)

\(\Leftrightarrow x=-3\)

Xét trường hợp 2:

\(x+6=-\left(2x+9\right)\)

\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)

\(\Leftrightarrow x+6+\left(2x+9\right)=0\)

\(\Leftrightarrow x+6+2x+9=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-15:3\)

\(\Leftrightarrow x=-5\)

Vậy x=-3 hoặc x=-5

ở câu b,bạn Nguyễn Nam phải thêm điều kiện cho mỗi trường hợp

TH1: phải đi với ĐK x phải lớn hơn hoặc bằng -6

TH2:phải đi với Đk x phải bé hơn -6 nên kết quả x=-5(KTM)

Vậy x=-3