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Áp dụng công thức: Nhân 2 lũy thừa cùng cơ số.
Ta có:
\(x\times x^2\times x^3\times...\times x^{100}\)
\(=x^{1+2+3+...+100}\)
\(x=5050\)
\(x.x^2.x^3...x^{100}=x^{1+2+3+...+100}\)
Đặt \(3^{1+2+3+...+100}=3^A\)
Ta có:
\(A=1+2+3+...+100\)
\(\Rightarrow A=100+99+98+...+1\)
\(\Rightarrow A=\left(1+100\right)+\left(2+99\right)+\left(3+98\right)+...+\left(100+1\right)\) ( 50 cặp số )
\(\Rightarrow A=101+101+101+...+101\) ( 50 số 101 )
\(\Rightarrow A=101.50\)
\(\Rightarrow A=5050\)
\(\Rightarrow3^A=3^{5050}\)
Vậy \(x.x^2.x^3...x^{100}=x^{5050}\)
Dấu chấm thay cho dấu nhân nhé!
a) \(\left(\dfrac{1}{2}x-3\right)\left(-\dfrac{1}{3}+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=0\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=0+3\\-\dfrac{1}{3}+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{1}{2}\\x=0-\left(-\dfrac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{1}{3}\end{matrix}\right.\)
d) \(9x^2=1\)
\(\Leftrightarrow x^2=1:9\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
\(\Leftrightarrow x^2=\left(\dfrac{1}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
a) \(\left(x-3\right)^2\) = \(\left(-3\right)^2\) + \(\left(-4\right)^2\)
\(\left(x-3\right)^2=9+16\)
\(\left(x-3\right)^2=25\)
\(\left(x-3\right)^2=5^2\)
\(\Rightarrow x-3=5\)
\(\Rightarrow x=5+3=8\)
Vậy x = 8
b) -12 . (x - 5) +7 . (3 - x) = 5
-12x + 60 + 21 - 7x = 5
-12x - 7x = 5 - 60 - 21
-19x = -76
x = -76 : (-19) =4
Vậy x = 4
a) \(\dfrac{2}{3}.x-\dfrac{1}{2}.x=\dfrac{5}{12}\)
=> \(\left(\dfrac{2}{3}-\dfrac{1}{2}\right).x=\dfrac{5}{12}\)
=> \(\left(\dfrac{4}{6}-\dfrac{3}{6}\right).x=\dfrac{5}{12}\)
=> \(\dfrac{1}{6}\) . x = \(\dfrac{5}{12}\)
=> \(x=\dfrac{5}{12}:\dfrac{1}{6}\)
=> x =\(\dfrac{5}{12}.\dfrac{6}{1}\)
=> x = \(\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
a) -12.(x-5)+7.(3.x)=5
<=> -12x+60+21+7x=5
<=>-5x+81=5
<=>-5x=5-81=-76
<=>x=-76/-5=76/5=15,2
b) 30.(x+2)-6.(x-5)-24.x=100
<=> 30x+60-6x+30-24x=100
<=> 0x=100-60-30=10
=> không có giá trị nào của x để 0x=10
c) \(|5.x-2|< 13\)
Khi 5x-2 < 13
<=> 5x<15 <=> x<3
Khi 5x-2 <-13
<=> 5x<-11 <=> x<-11/5 <=> x<-2,2
= x ^1 +100 = x^101
\(x\times x^2\times x^3\times x^4\times...\times x^{100}=x^{1+2+3+4+...+100}=x^{101\times500}=x^{5050}\)