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Câu 2: Ta có \(S=6^2+18^2+30^2+...+126^2\)
\(S=6^2\left(1^2+3^2+5^2+...+21^2\right)\)
\(=6^2.1771=36.1771=63756\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(3A=n\left(n+1\right)\left(n+2\right)\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Bài 2:
\(B=1^2+2^2+...+n^2\)
\(B=1\left(2-1\right)+2\left(3-1\right)+...+n\left[\left(n+1\right)-1\right]\)
\(B=\left[1\cdot2+2\cdot3+...+n\left(n+1\right)\right]-\left(1+2+...+n\right)\)
\(B=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)
\(B=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
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Bài 2:
a, 1/3 + 1/2 : x = -4
=> 1/2 : x = -4 - 1/3
=> 1/2 : x = -13/3
=> x = 1/2 ; -13/3
=> x = -3/26
Vậy x = -3 / 26
Bài 2:
b, x2 - 4x = 0
=> x.(x - 4) =0
=> x=0 hoặc x - 4 = 0
x - 4= 0 => x=4
Vậy x=0 và x=4
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a/ Công thức tính là \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Áp dụng vào tính nha
b/\(\frac{\left(n\left(n+1\right)\right)^2}{4}\)
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mk chỉ biết câu a thôi
\(-3+\frac{1}{3}\)=\(\frac{-3}{1}\)+\(\frac{1}{3}\)= \(\frac{-9}{3}\)+\(\frac{1}{3}\)=\(\frac{-8}{3}\)
Bài 1:
= \(\frac{-8}{3}\)
= \(\frac{-9}{4}\)
= \(\frac{-42}{19}\)
Bài 2:
A = \(\frac{3}{7}\)
Ai thấy mình đúng tk nha !
![](https://rs.olm.vn/images/avt/0.png?1311)
BÀI 1
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)
bài 2
a) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
b) \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)
Study well
Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
Bài 2:
a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)
b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)
\(A=1^2+2^3+3^1=1+8+3=12\)
A=12+23+31
=1+8+3
=12