\(S=\left(2.1\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+....+\left(2.10\right)^2\)

\(\Rightarrow S=2^2.1^2+2^2.2^2+....+2^2.10^2\)

\(\Rightarrow S=2^2\left(1^2+2^3+3^2+.....+10^2\right)\)

Áp dụng giả thiết từ đề

\(\Rightarrow S=2^2.385\)

\(\Rightarrow S=4.384=1540\)

\(S=2^2+4^2+6^2+...+20^2\)

    \(=1^2.4+2^2.4+3^2.4+...+10^2.4\)

    \(=4.\left(1^2+2^2+3^2+...+10^2\right)\)

    \(=4.385=1540\)

Bài 1:
a)1/9 x 27ⁿ= 3ⁿ

1/9=3ⁿ:27ⁿ

3ⁿ:27ⁿ=1/9

1ⁿ/9ⁿ=1/9

=>n=1

\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2^n-1.9=288=>2^n-1=32 nha)

=>2^n-1=2⁵=>n-1=5=>n=5+1=6

vậy......

~~~~~~~~~~~~~~~

a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\left(\frac{3}{5}\right)^2\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\frac{3}{5}\right)^{10}=\left(\frac{3}{5}\right)^5\)

b) \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=4+\frac{1}{27}=4\frac{1}{27}\)

c) \(2^3+3.\left(\frac{1}{2}\right)^0+\left(-2\right)^2:\frac{1}{2}.8=8+3.1+4:\frac{1}{2}.8=8+3+64=75\)

d) \(\left(-1\right)^{-1}-\left(-\frac{3}{5}\right)^0+\left(\frac{1}{2}\right)^{2:2}=-1-1+\left(\frac{1}{2}\right)^1=-2+\frac{1}{2}=-\frac{3}{2}\)

bạn giải thích cho mình phần a nhé

a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)

=>4^x=4^8

=>x=8

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)

=>2^x=2^11

=>x=11

c: =>1/6*6^x+6^x*36=6^15(1+6^3)

=>6^x=6*6^15

=>x=16

d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>x=9

e; 2\(^{-2}\).2\(^{x}\) + 2.2\(^{x}\) = 9.2\(^6\)

2\(^{x}\).(\(\frac14\) + 2) = 9.64

2\(^{x}\).(\(\frac14+\frac84\)) = \(576\)

2\(^{x}\).\(\frac94\) = 576

2\(^{x}\) = 576 : \(\frac94\)

2\(^{x}\) = 256

2\(^{x}\) = 2\(^8\)

\(x=8\)

Vậy \(x=8\)

e: \(2^{-2}\cdot2^{x}+2\cdot2^{x}=9\cdot2^6\)

=>\(2^{x}\left(2+2^{-2}\right)=9\cdot2^6\)

=>\(2^{x}\left(2+\frac14\right)=9\cdot2^6\)

=>\(2^{x}\cdot\frac94=\frac94\cdot2^8\)

=>\(2^{x}=2^8\)

=>x=8

f: \(3^{-22}\cdot3^4\cdot3^{x}=3^7\)

=>\(3^{x+4-22}=3^7\)

=>x-18=7

=>x=18+7=25

1.

\((\frac{1}{3}xy)^2.x^3+\frac{3}{2}(2x)^3(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)

\(=(\frac{1}{9}x^2y^2)x^3+\frac{3}{2}(8x^3)(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)

\(=\frac{1}{9}(x^2.x^3)y^2+(\frac{3}{2}.8.\frac{-7}{4})(x^3.x^2).y^2-\frac{2}{3}x^5y^2\)

\(=\frac{1}{9}x^5y^2-21x^5y^2-\frac{2}{3}x^5y^2=\frac{-194}{9}x^5y^2\)

2.

\(\frac{-2}{5}x^2y(-y^6)+\frac{3}{2}xy(\frac{-1}{15}xy^6)+(-2xy)^2y^5\)

\(=\frac{2}{5}x^2(y.y^6)+(\frac{3}{2}.\frac{-1}{15})(x.x).(y.y^6)+4x^2(y^2.y^5)\)

\(=\frac{2}{5}x^2y^7-\frac{1}{10}x^2y^7+4x^2y^7=\frac{43}{10}x^2y^7\)

3.

\(\frac{3}{7}xy^2z+\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2-\frac{3}{7}xy^2z\)

\(=(\frac{3}{7}xy^2z-\frac{3}{7}xy^2z)+(\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2)\)

\(=\frac{5}{6}x^3y^2\)

4.

\(\frac{2}{3}xy^2-\frac{5}{2}yz+\frac{1}{2}xy^2-\frac{2}{3}yz\)

\(=(\frac{2}{3}xy^2+\frac{1}{2}xy^2)-(\frac{5}{2}yz+\frac{2}{3}yz)\)

\(=\frac{7}{6}xy^2+\frac{19}{6}yz\)

5.

\(\frac{3}{2}xy^2z^5-\frac{5}{4}xyz^2+\frac{4}{3}xy^2z^5+\frac{1}{2}xyz^2\)

\(=(\frac{3}{2}xy^2z^5+\frac{4}{3}xy^2z^5)+(\frac{-5}{4}xyz^2+\frac{1}{2}xyz^2)\)

\(=\frac{17}{6}xy^2z^5-\frac{3}{4}xyz^2\)

1.

\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)

\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)

\(=2x^5y^4-4x^2y^3\)

2.

\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)

\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)

\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)

3.

\(5x-7xy^2+3x-\frac{1}{2}xy^2\)

\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)

\(=8x-\frac{15}{2}xy^2\)

4.

\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)

\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)

\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)

5.

\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)

\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)

\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)

6.

\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)

\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)

Ta có:
\(A=2^0+2^1+2^2+...+2^{40}\)

\(\Rightarrow A=1+2+2^2+...+2^{40}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{41}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{41}\right)-\left(1+2+2^2+...+2^{40}\right)\)

\(\Rightarrow A=2^{41}-1\)

Vì \(2^{41}-1< 2^{41}\) nên A < B

Vậy A < B

1) Tính

a) 25³ : 5² = (5²)³ : 5² = 5⁶ : 5² = 5⁴ = 625

\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 3² . \(\dfrac{1}{81}\) . 3² = 3² . 3² . \(\dfrac{1}{3^4}\) . 3² = 9

2) Tìm x thuộc Q, biết:

a) 3^{x + 2} = 27

=> 3^{x + 2} = 3³

x + 2 = 3

x = 3 - 2

x = 1

b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)

\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)

\(\dfrac{1}{2}x-3=3^{ }\)

\(\dfrac{1}{2}x=3+3\)

\(\dfrac{1}{2}x=9\)

\(x=9:\dfrac{1}{2}\)

\(x=18\)

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

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

\(x-\dfrac{1}{2}=-3\)

\(x=-3+\dfrac{1}{2}\)

\(x=\dfrac{-5}{2}\)

d) 5 . 5^{x + 1} = 125

5^{x + 1} = 125 : 5

5^{x + 1} = 25

5^{x + 1} = 5²

x + 1 = 2

x = 2 - 1

x = 1.