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2.
\(\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot15\cdot37\right)\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left(65\cdot111-13\cdot5\cdot3\cdot37\right)\\=\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-\left(13\cdot5\right)\cdot\left(3\cdot37\right)\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot\left[65\cdot111-65\cdot111\right]\\ =\left(1+2+3+...+100\right)\cdot\left(1^2+2^2+3^2+...+10^2\right)\cdot0\\ =0\)

Vậy xét là \(\frac{1}{2}+1\)nhé.
a,\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{1000}{999}\)
=3x4x5x...x1000/2x3x4x...x999
=1000/2=500
b, c tương tự câu a
)(1/2+1)x(1/3+1)x(1/4+1)x...x(1/999+1)
b)(1/2-1)x(1/3-1)x(1/4-1)x...x(1/1000-1)
c)3/22 x 8/32 x 15/42 x .... x 99/102
mình ko biết làm chép lại de thui

a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

1: =>-2x+6x2-4x+8=3/2+18x2
=>\(18x^2+\dfrac{3}{2}=6x^2-6x+8\)
\(\Leftrightarrow12x^2+6x-\dfrac{13}{2}=0\)
hay \(x\in\left\{\dfrac{-3+\sqrt{87}}{12};\dfrac{-3-\sqrt{87}}{12}\right\}\)
2: Đề thiếu vế phải rồi bạn
3: \(\Leftrightarrow x^2\cdot\dfrac{2}{3}-2x-\dfrac{4}{3}x+4=\dfrac{2}{3}x^2-\dfrac{2}{3}x\)
=>-10/3x+2/3x=-4
=>-8/3x=-4
=>x=4:8/3=4x3/8=12/8=3/2

\(3^x-15=12\\ \Rightarrow3^x=27\\ \Rightarrow x=3\)
\(5^{x-1}-12=13\\ \Rightarrow5^{x-1}=25\\ \Rightarrow x=3\)
\(3^{x-2}.9=81\\ \Rightarrow3^{x-2}=9\\ \Rightarrow x=4\)
\(3^{x+1}:9=3\\ \Rightarrow3^{x+1}=27\\ \Rightarrow x=2\)
\(2.x^3=16\\ \Rightarrow x^3=8\\ \Rightarrow x=2\)
\(75:x^2=3\\ \Rightarrow x^2=25\\ \Rightarrow x=\pm2\)
Câu cuối cộng trừ 5

a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
\(\left[1-\frac{1}{2^2}\right]\left[1-\frac{1}{3^2}\right]...\left[1-\frac{1}{15^2}\right]\)
\(=\left[1-\frac{1}{4}\right]\left[1-\frac{1}{9}\right]...\left[1-\frac{1}{225}\right]\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{224}{225}\)
Đến đây tự tính :v
Tính tiếp cho Forever Alone
\(=\frac{3}{4}.\frac{8}{9}.....\frac{224}{225}\)
\(=\frac{3}{225}\)