tính
a; (1\16)3 : (1\8)2
b) (1\9)25 : (1\3)30
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Bài 1:
a: =25+75=100
b: =60-17-43+12=12
c: =-2-18=-20
d: =-3+36-17=36-20=16
Bài 2:
a: =-102
b: =-1000
c: =12x15=180
d: =21x(-10)=-210
a) (2 × 3 × 8)/(4 × 5 × 6 × 7)
= (6 × 2 × 4)/(4 × 5 × 6 × 7)
= 2/(5 × 7)
= 2/35
b) (36 × 22 × 51)/(11 × 17 × 72)
= (36 × 2 × 11 × 3 × 17)/(11 × 17 × 2 × 36)
= 3
Lời giải:
a.
\(\frac{2\times 3\times 8}{4\times 5\times 6\times 7}=\frac{2\times 24}{24\times 5\times 7}=\frac{2}{5\times 7}=\frac{2}{35}\)
b.
\(\frac{36\times 22\times 51}{11\times 17\times 72}=\frac{36\times 2\times 11\times 17\times 3}{11\times 17\times 72}=\frac{72\times 11\times 17\times 3}{11\times 17\times 72}=3\)
a) Ta có: \(1+\left(-2\right)+3+\left(-4\right)+...+2021\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)
\(=-1010+2021=1011\)
1+(-2)+3+(-4)+....+2021
=(1+3+5+.....+2021)-(2+4+6+....+2020)
=\(\dfrac{\left(1+2021\right)\cdot2021}{2}-\dfrac{\left(2+2020\right)\cdot2020}{2}=1010\cdot2021-1010\cdot2020\\ =1010\cdot\left(2021-2020\right)=1010\cdot1\\ =1010\)
\(a,\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\frac{1}{16^3}.8^2=\frac{\left(2^3\right)^2}{\left(2^4\right)^3}=\frac{2^{3.2}}{2^{4.3}}=\frac{2^6}{2^{12}}=\frac{1}{2^6}\)
\(b,\left(\frac{1}{9}\right)^{25}:\left(\frac{1}{3}\right)^{30}=\frac{1}{9^{25}}.3^{30}=\frac{3^{30}}{\left(3^2\right)^{25}}=\frac{3^{30}}{3^{50}}=\frac{1}{3^{20}}\)