\(5^{20}:\left(5^{15}\cdot6+5^{15}\cdot19\right)\\ =5^{20}:\left[5^{15}\cdot\left(6+19\right)\right]\\ =5^{20}:\left(5^{15}\cdot25\right)\\ =5^{20}:5^{17}\\ =5^3\\ =125\)

____________________

\(7^{18}:7^{16}+2^2\cdot3^3\\ =7^{18-16}+4\cdot27\\ =7^2+108\\ =49+108\\ =157\)

__________________

\(59\cdot73-30^2+27\cdot59\\ =59\cdot\left(73+27\right)-30^2\\ =59\cdot100-30^2\\ =5900-900\\ =5000\)

125
157
5000

\(151-2^{91}:2^{88}+1^2\cdot3\\ =151-2^{91-88}+3\\ =154-2^3\\ =154-8\\ =146\)

_______________

\(2^{38}:2^{36}+5^1\cdot3^2-7^2\\ =2^{38-36}+5\cdot9-49\\ =2^2+45-49\\ =4-4\\ =0\)

_____________

\(7^{91}:7^{89}+5\cdot5^2-124\\ =7^{91-89}+5^3-124\\ =7^2+125-124\\ =49+1\\ =50\)

______________

\(4\cdot15+28:7-6^{20}:6^{18}\\ =60+4-6^{20-18}\\ =64-6^2\\ =64-36\\ =28\)

______________

\(\left(3^2+2^3\cdot5\right):7\\ =\left(9+8\cdot5\right):7\\ =\left(9+40\right):7\\ =49:7\\ =7\)

_______________

\(11^{25}:11^{23}-3^5:\left(1^{10}+2^3\right)-60\\ =11^{25-23}-3^5:\left(1+8\right)-60\\ =11^2-3^5:9-60\\ =121-3^5:3^2-60\\ =61-3^3\\ 61-27\\ =34\)

146
131
50
28
7
34

\(a)71-\left(33+x\right)=26\\ 33+x=71-26=45\\ x=45-33\\ x=12\\ b)97-\left(64-x\right)=44\\ 64-x=97-44\\ 64-x=53\\ x=64-53\\ x=11\\ c)x-36:18=12\\ x-2=12\\ x=2+12\\ x=14\\ d)3636:\left(12\cdot x-91\right)=36\\ 12\cdot x-91=3636:36\\ 12\cdot x-91=101\\ 12\cdot x=101+91\\ 12\cdot x=192\\ x=\dfrac{192}{12}\\ x=16\\ e)\left(x:23+45\right)\cdot67=8911\\ x:23+45=8911:67\\ x:23+45=133\\ x:23=133-45=88\\ x=88\cdot23\\ x=2024\)

Cứu voiii

\(x^2\) + 6\(x\) + 9 = 25

\(x^2\) + 6\(x\) + 9 - 25 = 0

\(x^2\) + 6\(x\) + (9 - 25) = 0

\(x^2\) + 6\(x\) - 16 = 0

\(x^2\) - 2\(x\) + 8\(x\) - 16 = 0

(\(x^2\) - 2\(x\)) + (8\(x\) - 16) = 0

 \(x\)(\(x\) - 2) + 8(\(x-2\)) = 0

   (\(x\) - 2)(\(x\) + 8) = 0

  \(\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.\)

   \(\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

Vậy \(x\) \(\in\) {2; - 8} 

cứu mình với

tích mà bạn=)

e

a) 

\(3^x=81\\ 3^x=3^4\\ x=4\)

b) 

\(\left(3x-5\right)^2=49\\ \left(3x-5\right)^2=7^2\)

TH1: 3x - 5 = 7 

3x = 7 + 5

3x = 12

x = 12 : 3 

x = 4 

TH2: 3x - 5 = -7

3x = -7 + 5

3x = -2

x = -2/3 

c) 68 - ? = 36

d) 

\(\left(7-2x\right)^3=27\\ \left(7-2x\right)^3=3^3\\ 7-2x=3\\ 2x=7-3\\ 2x=4\\ x=\dfrac{4}{2}\\ x=2\)

ê

15583 là số nguyên tố bạn nhé

ê

=> E = {12; 14; 16; ... ; 100} 

657 x 8 -1423

=5256 - 1423

=3833

Đặt: 

\(A=\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}+...+\dfrac{1}{25^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{\left(5^2\right)^{10}}\\ A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\\ 5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{19}}\\ 5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{19}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{20}}\right)\\4A =1-\dfrac{1}{5^{20}}\\ 4A=\dfrac{5^{20}-1}{5^{20}}\\ A=\dfrac{5^{20}-1}{4\cdot5^{20}}\)

ê

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)

\(50x+\left(1+2+...+50\right)=1175\)

\(50x+\dfrac{50\cdot\left(50+1\right)}{2}=1175\)

\(50x+1275=1175\)

\(50x=1175-1275\)

\(50x=-100\)

\(x=-100:50\)

\(x=-2\)

Vậy...

sửa \(\left(x+1\right)+\left(x+2\right)+...+\left(x+50\right)=1175\)

\(\Leftrightarrow50x+1275=1175\Leftrightarrow x=-2\)