1+1=
2+2=
5+5=
6+6=
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Lời giải chi tiết:
6 – 5 < 2 | 6 – 1 = 4 + 1 |
6 – 4 = 2 | 5 – 3 < 5 – 2 |
5 – 2 > 2 | 6 – 3 < 6 – 2 |
Lời giải chi tiết:
5 + 4 = 9 | 6 + 3 = 9 | 7 + 2 = 9 | 1 + 8 = 9 |
5 + 3 + 1 = 9 | 6 + 2 + 1 = 9 | 7 + 1 + 1 = 9 | 1 + 2 + 6 = 9 |
5 + 2 + 2 = 9 | 6 + 3 + 0 = 9 | 7 + 0 + 2 = 9 | 1 + 5 + 3 = 9 |
5+4=9 | 6+3=9 | 7+2=9 | 1+8=9 |
5+3+1=9 | 6+2+1=9 | 7+1+1=9 | 1+2+6=9 |
5+2+2=9 | 6+3+0=9 | 7+0+2=9 | 1+5+3=9 |
Bài 1:
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+1986}\right)\)
Nhận xét: \(1-\frac{1}{1+2+...+n}=1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Do đó: \(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+1986}\right)\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot...\cdot\frac{1985\cdot1988}{1986\cdot1987}=\frac{1\cdot4\cdot1988}{1986\cdot3}=\frac{3976}{2979}\)
Bài 2:
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
\(\Rightarrow\frac{4\cdot4^5}{3\cdot3^5}\cdot\frac{6\cdot6^5}{2\cdot2^5}=2^x\)\(\Rightarrow\frac{4^6}{3^6}\cdot\frac{6^6}{2^6}=2^x\)
\(\Rightarrow\frac{\left(2^2\right)^6}{3^6}\cdot\frac{\left(2\cdot3\right)^6}{2^6}=2^x\)\(\Rightarrow\frac{2^{12}}{3^6}\cdot\frac{2^6\cdot3^6}{2^6}=2^x\)
\(\Rightarrow\frac{2^6\cdot3^6\cdot2^{12}}{2^6\cdot3^6}=2^x\)\(\Rightarrow2^{12}=2^x\Rightarrow x=12\)
a: 14/5-7/5=7/5
b: 7/8-1/3+5/4
=21/24-8/24+30/24
=43/24
c; =7/6+5/6+2/15+13/15
=2+1
=3
d: =4*5/3*11=20/33
e: =2/9*1/6*1/4=2/9*1/24=1/108
2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)
b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)
c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)
1+1=2
2+2=4
3+3=6
1+1=2
2+2=4
5+5=10
6+6=12