/x+1/=6+3+2x=9+2x

=> \(\left[{}\begin{matrix}x+1=9+2x\\x+1=-\left(9+2x\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+8=0\\3x=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-8\\x=-\dfrac{10}{3}\end{matrix}\right.\)

/x-1/=6+3+2x=9+2x

=> \(\left[{}\begin{matrix}x-1=9+2x\\x-1=-\left(9+2x\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+10=0\\3x=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-10\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(\left(3x-6\right)-\left(2x+3\right)=15\)

\(\Leftrightarrow3x-6-2x-3=15\)

\(\Leftrightarrow x-9=15\)

\(\Leftrightarrow x=15+9\)

\(\Leftrightarrow x=24\)

\(\left(3x-6\right)-\left(2x+3\right)=15\)

\(3x-6-2x-3=15\)

\(x-3=15\)

\(x=18\)

(3x - 6) - (2x + 3) = 15

<=> 3x - 6 - 2x + 3 = 15

<=> 3x - 2x + 3 = 15 + 6

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

<=> 3x - (2x + 3) = 21

<=> x - 3 = 21

<=> x = 21 + 3

<=> x = 24

=> x = 24

K hiểu phần nào ib mình 

a,b: x= 1 hoặc 0

x=8

x¹⁰=1^x                        x¹⁰=x                     (2x-15)⁵=(2x-15)³

=>x=1(suy đoán)     =>x=1(suy đoán)       Hình như đề sai (suy đoán)

a)x¹⁰=1^x=>x¹⁰=1=>x=1

b)x¹⁰=x=>x¹⁰-x=0=>x(x⁹-1)=0

TH1.x=0

TH2.x⁹-1=0

=>x⁹=1=>x=1

c)(2x-15)⁵=(2x-15)³

như câu b

ủa câu hỏi đâu

Các bạn ơi giúp mik với

bạn nhấn vô đọc tiếp á

giúp mik với

a) \(\left|x-3\right|+\left|2x-6\right|=8\)

\(x-3+2x-6=8\)

\(3x-9=8\)

\(3x=17\)

\(\Rightarrow x=\frac{17}{3}\)

b) Tương tự câu a .

c) \(\left|2x-3\right|=6-\left|3-2x\right|\)

\(2x-3=6-3-2x\)

\(2x-3=x\)

\(-2x=3\)

\(x=\frac{-3}{2}\)

d)  \(\left|3x-2\right|-\left|6-9x\right|=-\left|-16\right|\)

\(3x-2-6-9x=-16\)

\(3x-8-9x=-16\)

\(-6x-8=-16\)

\(-6x=-8\)

\(\Rightarrow x=\frac{8}{6}\)

\(\)

hình như sai rồi đó bạn ạ

\(3x\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)

\(=\frac{6}{7}\)

Tìm x

\(a,3x(2x+1)=0\)

\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)

\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)

\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)

\(\frac{4}{3}x=\frac{2}{3}-5\)

\(\frac{4}{3}x=\frac{-13}{3}\)

\(x=\frac{-13}{3}\div\frac{4}{3}\)

\(x=\frac{-13}{4}\)

Chúc ban học tốt

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}{7.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}=\frac{6}{7}\)

Tìm x

a)bạn tự làm vì nó rất dễ nha

b)\(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}.\left(2x+1\right)=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-\frac{1}{2}.2x-\frac{1}{2}=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-\frac{2}{2}x-\frac{1}{2}=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-x-\frac{1}{2}=5\)

\(\frac{1}{3}x-x=5+\frac{1}{2}+\frac{1}{2}+\frac{2}{3}\)

\(\left(\frac{1}{3}-1\right)x=5+1+\frac{2}{3}\)

\(\frac{-2}{3}x=6+\frac{2}{3}\)

\(\frac{-2}{3}x=\frac{20}{3}\)

\(x=\frac{20}{3}:\left(\frac{-2}{3}\right)\)

\(x=\frac{20}{3}.\left(\frac{3}{-2}\right)\)

x=-10

Vậy x=-10

Chúc bạn học tốt

2^x+84=116

2^x=116-84

2^x=32

2^x=2^5

=>x=5

Bn cũg z nè,chúc bn học giỏi và lun là con ngoan trò giỏi

Cho mk xin cái

Mk nha

Kb đy nha!