32 < 2^x < 2^2x-3 . 2^8-2x

=>  2⁵  <  2^x  <  2^{2x - 3} . 2^{8 - 2x}

=>  2⁵  <  2^x  <  2⁵

=> 2^x = 2⁵

=> x = 5

\(2^5\le2^x\le2^{2x-3+8-2x}=2^5\)

\(\Rightarrow2^x=2^5\Rightarrow x=5\)

a: (x-3)²=49

=>x-3=7 hoặc x-3=-7

=>x=10 hoặc x=-4

b: \(\left(x^4\right)^2=\dfrac{x^{12}}{x^5}\)

\(\Leftrightarrow x^8-x^7=0\)

\(\Leftrightarrow x^7\left(x-1\right)=0\)

=>x=0 hoặc x=1

c: \(\Leftrightarrow x^{10}-25x^8=0\)

\(\Leftrightarrow x^8\left(x^2-25\right)=0\)

\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)

hay \(x\in\left\{0;5;-5\right\}\)

Ta có \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{1}{2}x+5\right)^{20}+\left(y^2-\frac{1}{2}\right)^{10}\ge0\forall x;y\)

mà \(\left(\frac{1}{2}x+5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\le0\)

=> Đẳng thức xảy ra <=> \(\hept{\begin{cases}\frac{1}{2}x+5=0\\y^2-\frac{1}{4}=0\end{cases}}\Rightarrow\hept{\begin{cases}\frac{1}{2}x=-5\\y^2=\frac{1}{4}\end{cases}}\Rightarrow\hept{\begin{cases}x=-10\\y=\pm\frac{1}{2}\end{cases}}\)

Vậy các cặp (x;y) thỏa mãn là \(\left(-10;\frac{1}{2}\right);\left(-10;-\frac{1}{2}\right)\)

( 1/2x - 5 )²⁰ + ( y² - 1/4 )¹⁰ ≤ 0 (1)

Ta có : \(\hept{\begin{cases}\left(\frac{1}{2}x-5\right)^{20}\ge0\forall x\\\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall y\end{cases}\Rightarrow}\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\forall x,y\)(2)

Từ (1) và (2) => Chỉ xảy ra trường hợp ( 1/2x - 5 )²⁰ + ( y² - 1/4 )¹⁰ = 0

=> \(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

Vậy ( x ; y ) = { ( 10 ; 1/2 ) , ( 10 ; -1/2 ) }

d) 

\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)

\(=\left(\frac{6}{7^3}\right)^2\)

\(=\frac{6^2}{7^{3^2}}\)

\(=\frac{36}{7^6}\)

\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)

\(=\left(\frac{6}{7^3}\right)^2\)

\(=\left(\frac{6^2}{7^{3^2}}\right)\)

\(=\frac{36}{7^6}\)

Code : Breacker

Ta có: \(8< 2^x< 2^9.2^{-5}\)

\(\Leftrightarrow2^3< 2^x< 2^4\)

Mà x là số tự nhiên

=> Không tồn tại x thỏa mãn

Bài kia làm nhầm rồi,sửa lại từ dòng 4:

\(4+x=25\)

\(x=21\)

Câu 2:

\(\left(-13x^4y^m\right)\left(-3x^ny^6\right)=39x^{15}y^8\)

\(\left[\left(-13\right)\left(-3\right)\right]\left[x^4x^n\right]\left[y^my^6\right]=39x^{15}y^8\)

\(39x^{4+n}y^{m+6}=39x^{15}y^8\)

\(\Rightarrow\left\{{}\begin{matrix}4+n=15\\m+6=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=11\\m=2\end{matrix}\right.\)

Vậy \(\left(m,n\right)=\left(2;11\right)\)

#hoctotnhe

a, 3 : ( 1 - 3/2x ) = 4 : ( 2 - x )

<=> \(\frac{3}{1-\frac{3}{2}x}=\frac{4}{2-x}\)

<=> 3 ( 2 - x ) = 4 ( 1 - 3/2x )

<=> 6 - 3x = 4 - 6x

<=> -3x + 6x = 4 - 6

<=> 3x = -2

<=> x = -2/3

b, 2.3^x + 3^x-1 = 7( 3² + 2.6² )

b, 2.3^x + 3^x-1 = 7( 3² + 2.6² )

<=> 2.3^x + 3^x-1 = 7.81

<=> 3^x-1(2.3 + 1) = 7.81

<=> 3^x-1.7 = 7.81

<=> 3^x-1=81

<=> 3^x-1 = 3⁴

=> x - 1 = 4 => x = 5