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Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
6/7+5/8÷5-3/16×(-2)²
=6/7+1/8-3/4
=55/56-3/4
=13/56
b.2/3 + 1/3.( -4/9 + 5/6 ) : 7/12
=2/3 + 1/3. ( -8/18 + 15/18 ) : 7/12
=2/3 + 1/3 . 7/18 : 7/12
=2/3 + 7/54 : 7/12
= 2/3 + 2/9
=6/9 + 2/9
= 8/9
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)
\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)
\(=0\)
A=[(1+9)+(3+7)+(5+15)+(11+13)].2
A=[10+10+20+24].2
A=64.2
A=128
B=(17+23)+(19+21)+(23+27)+(25+29)
B=40+40+50+54
B=tự tính
A=(1+9)+(3+7)+(5+15)+(7+13)+(11.2)
A=10+10+10+10+22
A=40+22
A=62
mình làm câu A thôi nhé
A=\(\dfrac{9^5.30^3}{3^{15}.5^3.6^2}=\dfrac{3^{10}.3^3.10^3}{3^{15}.5^3.2^2.3^2}=\dfrac{3^{13}.2^3.5^3}{3^{15}.5^3.2^2.3^2}=\dfrac{2}{3^4}=\dfrac{2}{81}\)
(1) 5/2 x (3/4 - 1/3) -11/12+1/4 = 5/2 x (9/12-4/12) -11/12 +3/12=5/2 x 5/12 -11/12 +3/12 = 25/24 - 11/12 +3/12=25/24 -22/24 + 6/24=9/12=3/4
(2) 11/12 : (4/10 + 3/5) + (5/6 - 1/2) x2/3
= 11/12 : (4/10 + 6/10) + (5/6-3/6) x 2/3
= 11/12:1 +1/2 x 2/3
= 11/12 + 2/3 =11/12+8/12 = 19/12
Bài giải
Sử dụng công thức tính tổng dãy sô có quy luật của lớp 6, ta có:
12 + 22 + 32 + 42 +...+ 20082
= \(\frac{2008\left(2008^2+1\right)}{2}\)
= 4048193260