7 . 2 ^x = 2 ⁹ + 5 . 2 ⁸

7 . 2 ^x =  2 ⁸ . ( 2 + 5 )

7 . 2 ^x = 2 ⁸ . 7

=> 2 ^x  = 2 ⁸

=> x = 8

Tìm số tự nhiên x,biết: 7.2^x=2⁹+5.2⁸

\(7.2^x=1792\)

\(2^x=1792:7=256\)

\(2^x=2^8\)

x=8

bấm **** nha

Ta có: \(C=\dfrac{5.2^9.6^{19}-7.2^{29}.27^6}{5.2^{25}.9^9-4.3^{20}.8^9}\)

\(=\dfrac{5.2^9.2^{19}.3^{19}-7.2^{29}.3^6.9^6}{5.2^{25}.3^9.3^9-4.3^{18}.3^2.2^9.2^9.2^9}\)

\(=\dfrac{5.2^{28}.3^{19}-7.2^{28}.2.3^{18}}{5.2^{25}.3^{18}-4.3^{18}.3^2.2^{25}.2^2}\)

\(=\dfrac{5.2^{28}.3^{18}.3-7.2^{28}.2.3^{18}}{3^{18}.2^{25}\left(5-4.3^2.2^2\right)}\)

\(=\dfrac{2^{28}.3^{18}\left(15-14\right)}{3^{18}.2^{25}\left(-139\right)}\)

\(=\dfrac{2^{25}.2^3}{2^{25}\left(-139\right)}=\dfrac{-8}{139}\)

Vậy \(C=\dfrac{-8}{139}.\)

Các bạn ơi mình cần gấp lắm ai giải được mình cho 3 like luông mình nói là giữ lời không thì mình cho các bạn nick này của mình

\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)  (sửa \(-9^9\rightarrow.9^9\))

\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\dfrac{2^{29}.3^{18}\left(15-9\right)}{2^{28}.3^{18}\left(15-14\right)}\)

\(=2.6=12\)

2^x+2-2^x=96

=>2^x.(2²-1)=96

=>2^x.3=96

=>2^x=32

=>2^x=2⁵

=>x=5

7.2^x=2⁹+5.2⁸

=>7.2^x=2⁸.(2+5)=

=>7.2^x=2⁸.7

=>2^x=2⁸

=>x=8

(x+7)²=7

=>\(x+7=\sqrt{7}\) hoặc \(x+7=-\sqrt{7}\)

=>\(x=\sqrt{7}-7\) hoặc \(x=-\sqrt{7}-7\)

a, Tự chép đề bài :v

=> 2^2x-3 = ( 8³. 16⁵ ) : 4¹⁰

2^2x-3 = ( 2⁹. 2²⁰ ) : 2²⁰

2^2x-3 = 2²⁹ : 2²⁰

2^2x-3 = 2⁹

=> 2x - 3 = 9

2x = 9 + 3

2x = 12

x = 6 

Vậy....

b, 7. 2^x = 2⁹ + 5. 2⁸

7. 2^x = 1792

2^x = 1792 : 7

2^x = 256

2^x = 2⁸

=> x = 8 

Vậy ....

a  \(\frac{2^{2x-3}}{4^{10}}=8^3.16^5\)

\(\Leftrightarrow\frac{2^{2x-3}}{4^{10}}=2^{29}\)

\(\Leftrightarrow2^{2x-3}=2^{29}.4^{10}\)

\(\Leftrightarrow2^{2x-3}=2^{49}\)

\(\Leftrightarrow2x-3=49\)

\(\Leftrightarrow x=26\)

b  \(7.2^x=2^9+5.2^8\)

\(\Leftrightarrow7.2^x=2^8.(2+5)\)

\(\Leftrightarrow7.2^x=7.2^8\)

\(\Leftrightarrow x=7\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)