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

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

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

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3\\ =\left[\left(\dfrac{2}{3}\right)^2\right]^4:\left(\dfrac{4}{9}\right)^3\\ =\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3\\ =\dfrac{4}{9}\)

\(27^3:3^2\\ =\left(3^3\right)^3:3^2\\ =3^9:3^2\\ =3^7\\ =2187\)

\(\left(-\dfrac{3}{5}\right)^4:\left(\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{3}{5}:\dfrac{5}{2}\right)^4\\ =\left(-\dfrac{6}{25}\right)^4\)

\(\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{9}{25}\right)^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left[\left(\dfrac{3}{5}\right)^2\right]^5\\ =\left(\dfrac{3}{5}\right)^{12}:\left(\dfrac{3}{5}\right)^{10}\\ =\left(\dfrac{3}{5}\right)^2\\ =\dfrac{9}{25}\)

\(\left(\dfrac{2}{3}\right)^8:\left(\dfrac{4}{9}\right)^3=\left(\dfrac{4}{9}\right)^4:\left(\dfrac{4}{9}\right)^3=\dfrac{4}{9}\)

\(27^3:3^2=3^9:3^2=3^7=2187\)

a) Ta có: 

\(64^8=\left(2^6\right)^8=2^{6\cdot8}=2^{48}\)

\(16^{12}=\left(2^4\right)^{12}=2^{4\cdot12}=2^{48}\)

\(\Rightarrow64^8=16^{12}\)

b) Ta có:

\(\left(\dfrac{1}{16}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{4\cdot10}=\left(\dfrac{1}{2}\right)^{40}\)

Mà: 50 > 40 => `(1/2)^50<(1/2)^40` 

c) Ta có: 

\(\left(\dfrac{9}{16}\right)^{100}=\left[\left(\dfrac{3}{4}\right)^2\right]^{100}=\left(\dfrac{3}{4}\right)^{200}\)

Mà: `3/4>2/3=>(3/4)^200>(2/3)^200`

\(^{^{ }}\)a,64^8=16^12

b,(1/16)^10<(1/2)^50

c,(2/3)^200>(9/16)^100

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