12+(5+x)=20

5+x=8

    x=8/5

vãy=8/5

12 + ( 5 + x ) = 20              5.2² + ( x + 3 ) = 5²                         2³ + ( x + 3 ) = 5²                                 4³ - ( x - 2 ) = 5²

17 + x = 20                        5.4 + x + 3 = 25                               8 + x + 3 = 25                                  64 - x + 2 = 25

x = 20 - 17                         20 + 3 + x = 25                                11 + x = 25                                       66 - x = 25

x = 3                                  23 + x = 25                                       x = 25 - 11                                        x = 66 - 25

                                          x = 25 - 23                                        x = 14                                               x = 41

                                          x = 2

Đăng nhìu v bn :) Đáng quan ngại đây :)

):(

a, \(390-\left(x-7\right)=13^2:12\)

\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)

\(x-7=390-\dfrac{169}{12}\)

\(x-7=\dfrac{4511}{12}\)

\(x=\dfrac{4511}{12}+7\)

\(x=\dfrac{4595}{12}\)

Vậy ...

b, \(\left(x-35.2^2\right):7=3^3-24\)

\(\left(x-35.4\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(\Leftrightarrow\left(x-140\right)=3.7\)

\(\Leftrightarrow x-140=21\)

\(\Leftrightarrow x=161\)

Vậy .....

c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)

\(x-3-\left(16.3-24\right):2:6=3\)

\(x-3-\left(48-24\right):2:6=3\)

\(x-3-24:2:6=3\)

\(x-3-2=3\)

\(x=3+2+3\)

\(x=8\)

Vậy ......

d) \(4x-5=5+5^2+5^3+.....+5^{99}\)

Đặt :

\(A=5+5^2+.........+5^{99}\)

\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)

\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)

\(\Leftrightarrow4A=5^{100}-5\)

\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)

\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)

Đến đây thì sao nữa nhỉ ?

e) \(\left(2x-1\right)^4=625\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

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

Vậy ....

a ) 4⁵ . 4 . 4⁷ = 4⁵ . 4¹ . 4⁷ = 4¹³

b ) 4 . 4³ . 4⁵ . 4⁶ = 4¹ . 4³ . 4⁵ . 4⁶ = 4¹⁵

c ) x . x³ . x⁴ = x¹ . x³ . x⁴ = x⁸

d ) x⁵ . x⁴ . x⁷ . x⁶ = x²²

dài thế

dai the giai bao gio xong chac mua quyt nam sau moi giai xong

Bài 2

a) x=5

b)x=1

1. 2^x=16\(\Rightarrow\)X=4

2. 2^2x-1=2⁷

\(\Rightarrow\)2⁷=2^2.4-1

Vậy x =4

x=4 nha chị

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

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