a) \(\left(2x-15\right)^3=\left(2x-15\right)^5\)

\(\Rightarrow\left(2x-15\right)^3-\left(2x-15\right)^5=0\)

\(\Rightarrow\left(2x-15\right)^3\left[1-\left(2x-15\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\1-\left(2x-15\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)

TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)

TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

Vậy  \(x\in\left\{\frac{15}{2};8;7\right\}\)

b) \(x^{10}=x\)

\(\Rightarrow x^{10}-x=0\)

\(\Rightarrow x\left(x^9-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy x = 0 hoặc x = 1

c) \(x^{10}=1^0\)

\(\Rightarrow x^{10}=1\)

\(\Rightarrow\orbr{\begin{cases}x^{10}=1^{10}\\x^{10}=\left(-1\right)^{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy x = 1 hoặc x = -1

_Chúc bạn học tốt_

Mơn pn

vghgdhjg

a)\(x^{10}=1^x\Leftrightarrow x=1\)

b) \(x^{10}=x\Leftrightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3.4\)

\(\Leftrightarrow2x-15=0\Leftrightarrow2x=15\Leftrightarrow x=\frac{15}{2}\)

nếu thắc mắc vì sao \(2x-15=1\)vì nếu như vậy ta có vế trái =1 nhưng vế phải lại phải  nhân cho 4 nên vế trái  ko = vế phải 

d) \(a^x-15=17\)

\(\Leftrightarrow a^x=32\)

\(\Leftrightarrow a^x=2^5\)

g) \(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=2^5.5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=800+200\)

\(\Leftrightarrow\left(7x-11\right)^3=1000\)

\(\Leftrightarrow\left(7x-11\right)^3=10^3\)

\(\Leftrightarrow7x-11=10\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=21:7=3\)

2^x=64

x=6

x¹⁵=x³¹<=>x={0;1} kkk tui nha

a) 

\(4^{10}\cdot8^{15}=2^{20}\cdot2^{45}=2^{65}\)

b)

\(27^{16}\cdot9^{10}=3^{48}\cdot3^{20}=3^{68}\)

a) 4¹⁰ . 8¹⁵ = (2²)¹⁰ . (2³)¹⁵ = 2²⁰ . 2⁴⁵ = 2⁶⁵
b) 27¹⁶ . 9¹⁰ = (3³)¹⁶ .(3²)¹⁰=3⁴⁸ . 3²⁰ = 3⁶⁸
c) \(\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}=\frac{2^{10}\cdot13\cdot\left(1+5\right)}{2^5\cdot7\cdot13}=\frac{2^5\cdot6}{7}=\frac{192}{7}\)

\(2x+4⋮x-1\Rightarrow2\left(x-1\right)+6⋮x-1\)

\(\Rightarrow6⋮x-1\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\Rightarrow x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

Vậy...........................................

\(2x^2+\left(-3\right)^2=41\)

\(\Rightarrow2x^2=41-9=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\pm4\)

\(2\left(x-5\right)-3\left(x+7\right)=14\)

\(\Rightarrow2x-10-3x-21=14\)

\(\Rightarrow2x-3x=14+21+10\)

\(\Rightarrow-x=45\Rightarrow x=-45\)

\(-7\left(5-x\right)-2\left(x-10\right)=15\)

\(\Rightarrow-35+x-2x+20=15\)

\(\Rightarrow x-2x=15-20+35\)

\(\Rightarrow-x=30\Rightarrow x=-30\)

Bài 1 tự làm!

Bài 2:

a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)

b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)

\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)

c, Đề chưa rõ

d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)

e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)

\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)

f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x⁰ = 1)

\(\Rightarrow x+1=1\Rightarrow x=0\)

Làm phần c nè!

c, \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

\(\Rightarrow\left(x-5\right)^4.\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\Rightarrow x=5\)

giúp với

ai nhanh nhất mk sẽ k mà

Bài 1 :

a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)

= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)

b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)

= \(10+45-455+750=350\)

c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)

= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)