\(4^2x-1=7.3^2\)

<=>\(16x-1=7.9\)

<=>\(16x-1=63\)

<=>\(16x=64\)

<=>\(x=4\)

4² . x - 1 = 7 . 3²

4² . x - 1 = 7 . 9

4² . x - 1 = 63

16 . x      = 63 + 1

16 . x      = 64

x             = 64 : 16

x             = 4

Hoc tot!!!

lớp 7 mà mình ấn nhằm

Đáp án:

13. -12(x-5)+7(3-x)=5

      -12x-(-12).5+7.3-7x=5

       -12x-(-60)+21-7x=5

       -12x+60+21-7x=5

       -12x+(60+21)-7x=5

        -12x-7x+81=5

        -19x+81     =5

        -19x            =5-81

        -19x            =-76

              x            =(-76):(-19)

              x            =4

Vậy x=4.

Lời giải:

1.

$3^{x+2}+4.3^{x+1}=7.3^6$

$3^{x+1}.3+4.3^{x+1}=7.3^6$

$3^{x+1}(3+4)=7.3^6$

$3^{x+1}.7=7.3^6$

$\Rightarrow 3^{x+1}=3^6$

$\Rightarrow x+1=6$

$\Rightarrow x=5$

2.

$5^{x+4}-3.5^{x+3}=2.5^{11}$

$5^{x+3}.5-3.5^{x+3}=2.5^{11}$

$5^{x+3}(5-3)=2.5^{11}$

$2.5^{x+3}=2.5^{11}$

$\Rightarrow 5^{x+3}=5^{11}$

$\Rightarrow x+3=11$

$\Rightarrow x=8$

3.

$4^{x+3}-3.4^{x+1}=13.4^{11}$

$4^{x+1}.4^2-3.4^{x+1}=13.4^{11}$

$4^{x+1}.16-3.4^{x+1}=13.4^{11}$

$13.4^{x+1}=13.4^{11}$

$\Rightarrow 4^{x+1}=4^{11}$

$\Rightarrow x+1=11$
$\Rightarrow x=10$

5.3^x = 5.3⁴

=> x = 4

5.3⁴ = 7.3⁵ - 2.3⁵

       = 3⁵ . (7 - 2)

       = 5. 3⁵

  => 3^x = 3⁵

=> x = 5

a) Ta có: \(5\cdot3^x=5\cdot3^4\)

\(\Leftrightarrow3^x=3^4\)

hay x=4

\(7.3^x-5^{x+2}=3^{x+4}-5^{x+3}\)

\(7.3^x-5^25^x=3^43^x-5^35^x\)

\(7.3^x-25.5^x=81.3^x-125.5^x\)

\(7.3^x-25.5^x=81.3^x-125.5^x\)

\(74.3^x=100.5^x\)

\(\left(\frac{3}{5}\right)^x=\frac{100}{74}=\frac{50}{37}\)

\(x=log_{\frac{3}{5}}\frac{50}{37}\)

Bài 2:

a) \(\frac{8^{14}}{4^{12}}\)

\(=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}\)

\(=\frac{2^{42}}{2^{24}}\)

\(=2^{18}\)

\(=262144.\)

b) \(\left(-\frac{1}{3}\right)^7.3^7\)

\(=\left[\left(-\frac{1}{3}\right).3\right]^7\)

\(=\left(-1\right)^7\)

\(=-1.\)

c) \(\frac{90^2}{15^2}\)

\(=\left(\frac{90}{15}\right)^2\)

\(=6^2\)

\(=36.\)

d) \(\frac{790^4}{79^4}\)

\(=\left(\frac{790}{79}\right)^4\)

\(=10^4\)

\(=10000.\)

Chúc bạn học tốt!

Mk làm tiếp cho bạn Vũ Minh Tuấn nhé!

Bài 1:

\(-\frac{64}{343}=x^3\)

\(\Rightarrow x^3=\left(-\frac{4}{7}\right)^3\)

\(\Rightarrow x=-\frac{4}{7}\)

Vậy \(x=-\frac{4}{7}\)

\(\left(x+20\right)^{100}+\left|y+4\right|=0\)

Ta có: \(\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)

Vậy \(x=-20;y=-4\)

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

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

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

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

Vậy \(x=\frac{5}{6}\)

\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)

Đề có phải như thế này không vậy bạn?

\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\)

\(3\cdot3^{x+1}+4\cdot3^{x+1}=7\cdot3^6\)

\(\left(3+4\right)\cdot3^{x+1}=7\cdot3^6\)

\(7\cdot3^{x+1}=7\cdot3^6\)

x + 1 = 6

x = 6 - 1 = 5

Vậy x = 5