Tính 2 : 1 2 - 2 3 3
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Bài 1:
a, 3\(\dfrac{2}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{17}{5}\) - \(\dfrac{1}{2}\)
= \(\dfrac{34}{10}\) - \(\dfrac{5}{10}\)
= \(\dfrac{29}{10}\)
b, \(\dfrac{4}{5}\) + \(\dfrac{1}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{4\times4}{5\times4}\) + \(\dfrac{1\times3}{5\times4}\)
= \(\dfrac{16}{20}\) + \(\dfrac{3}{20}\)
= \(\dfrac{19}{20}\)
c, 4\(\dfrac{4}{9}\) : 2\(\dfrac{2}{3}\) + 3\(\dfrac{1}{6}\)
= \(\dfrac{40}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{5}{3}\) + \(\dfrac{19}{6}\)
= \(\dfrac{10}{6}\) + \(\dfrac{19}{6}\)
= \(\dfrac{29}{6}\)
Bài 2:
3\(\dfrac{2}{5}\) + 2\(\dfrac{1}{5}\)
= \(\dfrac{17}{5}\) + \(\dfrac{11}{5}\)
= \(\dfrac{28}{5}\)
b, 7\(\dfrac{1}{6}\) : 5\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
2 - 1 = 1 3 - 1 = 2 1 + 1 = 2 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 2 - 1 = 1 3 - 2 = 1
3 - 2 = 1 2 - 1 = 1 3 - 1 = 2 3 - 1 = 2
ok nhá
Lời giải chi tiết:
1 + 2 = 3 | 3 – 1 = 2 | 1 + 1 = 2 | 2 – 1 = 1 |
3 – 2 = 1 | 3 – 2 = 1 | 2 – 1 = 1 | 3 – 1 = 2 |
3 – 1 = 2 | 2 – 1 = 1 | 3 – 1 = 2 | 3 – 2 = 1 |
1+2=3 | 3-1=2 | 1+1=2 | 2-1=1 |
3-2=1 | 3-2=1 | 2-1=1 | 3-1=2 |
3-1=2 | 2-1=1 | 3-1=2 | 3-2=1 |
#HT#
a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
5:
a: \(3^{2n}=\left(3^2\right)^n=9^n\)
\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)
=>\(3^{2n}>2^{3n}\)
b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
mà \(1568239201< 8036054027\)
nên \(199^{20}< 2003^{15}\)
4: \(100< 5^{2x-1}< 5^6\)
mà \(25< 100< 125\)
nên \(125< 5^{2x-1}< 5^6\)
=>3<2x-1<6
=>4<2x<7
=>2<x<7/2
mà x nguyên
nên x=3