K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 10 2022

\(x:\dfrac{-1}{5}=\left(\dfrac{-1}{5}\right)^{12}\)

\(\Leftrightarrow x=\left(\dfrac{-1}{5}\right)^{12}.\left(\dfrac{-1}{5}\right)\)

\(\Leftrightarrow x=\left(-\dfrac{1}{5}\right)^{13}\)

4 tháng 10 2022

x : (-1/5) = (-1/5)12

⇔ x = (-1/5) . (-1/5)12

⇔ x = (-1/5)13

Vậy x = (-1/5)13

Tích đúng 5 sao cho mình nhé.

OK bạn.

31 tháng 7 2016

8x = 1612

=> (23)x = (24)12

=> 23x = 248

=> 3x = 48

=> x = 48 : 3

=> x = 16

Vậy x = 16

31 tháng 7 2016

x=16 nhé

ủng hộ đi

 

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

a: x>-3/5 nên x+3/5>0

x<1/7 nên x-1/7<0

A=1/7-x-(x+3/5)+4/5

=1/7-2x-3/5+4/5

=-2x+12/35

b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)

\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)

-3/5<x<1/7

nên x-1/7<0; x+3/5>0

\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)

c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)

Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0

\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)

1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)

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

\(\Leftrightarrow-2x\left(x+1\right)=0\)

Vì -2≠0

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

Vậy: x∈{0;-1}

2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)

\(\Leftrightarrow11x-2=0\)

\(\Leftrightarrow11x=2\)

hay \(x=\frac{2}{11}\)

Vậy: \(x=\frac{2}{11}\)

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

\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)

\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)

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

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

Vậy: \(x=\frac{-13}{4}\)

4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)

2 tháng 4 2020

1. \(2x+x\left(x-5\right)=3x^2-x\)

\(\Leftrightarrow2x+x^2-5x=3x^2-x\)

\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)

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

\(\Leftrightarrow-2x\left(1+x\right)=0\)

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

2. \(15-5\left(1-2x\right)=12-x\)

\(\Leftrightarrow15-5+10x=12-x\)

\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)

\(\Leftrightarrow-2+11x=0\)

\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)

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

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

\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)

\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)

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

4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)

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

\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)

\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)

2 tháng 4 2020

1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)

\(x+\frac{4}{5}=\pm\frac{4}{5}\)

\(TH1:x+\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)

\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)

\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)

Vậy x ∈ {0; \(\frac{-8}{5}\)}

2 tháng 4 2020

Hỏi đáp ToánHỏi đáp ToánHai câu cuối khó nhìn nên ko giải