\(4^{x-5}=16\)

\(4^{x-5}=4^2\)

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

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

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

tu hoc moi gioi

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

Vũ Hồng Linh bạn check lại bài đầu dùm =_=" 

\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

(3\(x\) + 1)³ =  - 27

(3\(x\) + 1)³ = (-3)³

3\(x\) + 1 = -3

 3\(x\)       = - 3 - 1

 3\(x\) = -4

    \(x=-\dfrac{4}{3}\)

Vậy \(x=-\dfrac{4}{3}\)

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

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

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

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

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

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

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

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

\(\Leftrightarrow x=\frac{1}{6}\)

a, \(x=\dfrac{3}{4}-1=\dfrac{3}{4}-\dfrac{4}{4}=-\dfrac{1}{4}\)

b, \(x=\dfrac{1}{5}-4=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}\)

c, \(x=2+\dfrac{1}{5}=\dfrac{10}{5}+\dfrac{1}{5}=\dfrac{11}{5}\)

d, \(x=\dfrac{1}{81}-\dfrac{5}{3}=\dfrac{1}{81}-\dfrac{135}{81}=-\dfrac{134}{81}\)

\(a,\dfrac{3}{4}-x=1\\ \Rightarrow x=1,75\\ b,x+4=\dfrac{1}{5}\\ \Rightarrow x=-3,8\\ c,x-\dfrac{1}{5}=2\\ \Rightarrow x=2,2\\ d,x+\dfrac{5}{3}=\dfrac{1}{81}\\x=\dfrac{1}{81}-\dfrac{5}{3}\\ \Rightarrow x=-\dfrac{134}{81}\)

a) \(\dfrac{3}{4}-x=1\)

\(x=\dfrac{3}{4}-1\)

\(x=-\dfrac{1}{4}\)\(=0,25\)

b) \(x+4=\dfrac{1}{5}\)

\(x=\dfrac{1}{5}-4\)

\(x=-\dfrac{19}{5}\)\(=-3,8\)

c) \(x-\dfrac{1}{5}=2\)

\(x=2+\dfrac{1}{5}\)

\(x=\dfrac{11}{5}=2,2\)

d) \(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(x=-\dfrac{134}{81}\)

2*(4x+1)+3*(2x+3)=81

8x+2+6x+9=81

14x+11=81

14x=81-11

14x=70

x=70:14

x=5

\(\left(2x+1\right)^2=49\)

\(\Rightarrow2x+1=7\left(h\right)2x+1=-7\)

\(\Rightarrow2x=6\left(h\right)2x=-8\)

\(\Rightarrow x=3\left(h\right)x=-4\)

\(\left(2x-3\right)^4=81\)

\(\Rightarrow2x-3=3\left(h\right)2x-3=-3\)

\(\Rightarrow2x=6\left(h\right)2x=0\)

\(\Rightarrow x=3\left(h\right)x=0\)

Giải:

(2.x + 1)²=49

(2.x + 1)²=7²

=>2.x + 1=7

2.x=7-1

2.x=6

x=6:2

x=3

(2.x-3)⁴=81

(2.x-3)⁴=3⁴

=>2.x-3=3

2.x=3+3

2.x=6

x=6:2

x=3

Chúc bạn học tốt

\(125\)