Tính (√5-2)/(2+2√5)-1/(2+2√5)-2/(2+√5)
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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\(\dfrac{2}{3}\)
= \(\dfrac{43}{6}\) : \(\dfrac{17}{3}\)
= \(\dfrac{43}{34}\)
Lời giải chi tiết:
5 – 1 = 4 | 4 – 1 = 3 | 3 – 1 = 2 | 2 + 3 = 5 |
5 – 2 = 3 | 4 – 2 = 2 | 3 – 2 = 1 | 3 + 2 = 5 |
5 – 3 = 2 | 4 – 3 = 1 | 2 – 1 = 1 | 5 – 2 = 3 |
5 – 4 = 1 | 5 – 3 = 2 |
5-1=4 4-1=3 3-1=2 2+3=5
5-2=3 4-2=2 3-2=1 3+2=5
5-3=2 4-3=1 2-1=1 5-2=3
5-4=1 5-3=2
Lời giải chi tiết:
5 – 2 – 1 = 2 | 4 – 2 – 1 = 1 | 3 – 1 – 1 = 1 |
5 – 2 – 2 = 1 | 5 – 1 – 2 = 2 | 5 – 1 – 1 = 3 |
5 - 1 = 4 1 + 4 = 5 2 + 3 = 5
5 - 2 = 3 4 + 1 = 5 3 + 2 = 5
5 - 3 = 2 5 - 1 = 4 5 - 2 = 3
5 - 4 = 1 5 - 4 = 1 5 - 3 = 2
5-1=4 1+4=5 2+3=5
5-2=3 4+1=4 3+2=5
5-3=2 5-1=4 5-2=3
5-4=1 5-4=1 5-3=2
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
5 - 1 - 1 = 3 4 - 1 - 1 = 2 3 - 1 - 1 = 1
5 - 1 - 2 = 2 5 - 2 - 1 = 2 5 - 2 - 2 = 1