1.Tính giá trị biểu thức
a.7 5/9-(3 5/9-2 3/4)
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a) 7/5 x 3/4 : 4/5 = 21/20 : 4/5
= 21/16
Tìm x:
b) x - 3/9 = 8/7
x = 8/7 + 3/9
X = 31/21
1.a,=(54+45+1).113
=100.113
=11300
b,=(3/7+8/14)+(4/9+10/18)
=1+1
=2
2.a,=13/10+1/3
=49/30
b,=12/9.(1/12+1/6)
=12/9.1/4
=1/3
c,=3/4.3/2
=9/8
d,=3/2-1/3
=7/6
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113
= 54 x 113 + 45 x 113 + 113x1
=113 x(54+45+1)
= 113x100
=1300
b)3/7 + 4/9 + 8/14 + 10/18
=(3/7+8/14)+(4/9+10/18)
= 1 + 1
=2
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: \(x=-\dfrac{134}{81}\)
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
a) \(\dfrac{1}{3}+\dfrac{4}{3}\times\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{4}{6}=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)
b) \(\dfrac{3}{5}\times\dfrac{4}{7}:\dfrac{16}{21}=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{21}{16}=\dfrac{12}{35}\times\dfrac{21}{16}=\dfrac{252}{560}=\dfrac{9}{20}\)
`a, 3/4 + 1/2 xx 7/2`
`= 3/4 + 7/4`
`=10/4`
`=5/2`
`b, 6/15 - 1/3 : 5/3`
`= 6/15 - 1/3 xx 3/5`
`= 6/15 - 3/15`
`= 3/15`
`=1/5`
`c, x-4/9 = 3/7 : 9/4`
`=> x-4/9= 3/7 xx 4/9`
`=> x-4/9= 12/63`
`=> x-4/9=4/21`
`=> x= 4/21 +4/9`
`=>x= 40/63`
`d, 7/9 xx 3/5 -1/2=1/5`
`->` sao lại bằng có `x` ko vậy ạ?
`a,`
`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`
`b,`
`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`
`c,` Tìm x?
`x-4/9=3/7 \div 9/4`
`x-4/9=4/21`
`x=4/21+4/9`
`x=40/63`
`d, 7/9x \times 3/5-1/2=1/5`
`7/9x \times 3/5=1/5+1/2`
`7/9x \times 3/5=7/10`
`7/9x=7/10 \div 3/5`
`7/9x=7/6`
`x=7/6 \div 7/9=3/2`
\(a,\) Số số hạng là \(\left(40-2\right):2+1=20\left(số\right)\)
Tổng là \(\left(40+2\right)\times20:2=420\)
\(b,\) Số số hạng là \(\left(39-1\right):2+1=20\left(số\right)\)
Tổng là \(\left(39+1\right)\times20:2=400\)
a) 2 + 4 + 6 + 8 + ... + 34 + 36 + 38 + 40
= ( 2 + 42 ) + ( 4 + 38 ) + .... + ( 20 + 22 )
= 42 \(\times\) 10
= 420
b) 1 + 3 + 5 + 7 + ... + 35 + 37 + 39
= ( 1 + 39 ) + ( 3 + 37 ) + ...+ ( 19 + 21 )
= 40 \(\times\) 10
= 400
\(log_575+log_53=log_5\left(75.3\right)=log_5225\)
\(4log_{12}2+2log_{12}3=log_{12}16+log_{12}9=log_{12}\left(16.9\right)=log_{12}144=log_{12}12^2=2\)
\(\dfrac{1}{3}log_3\dfrac{9}{7}+log_37^{\dfrac{1}{3}}=\dfrac{1}{3}\left(log_3\dfrac{9}{7}+log_37\right)=\dfrac{1}{3}log_3\left(\dfrac{9}{7}.7\right)=\dfrac{1}{3}log_39=\dfrac{2}{3}\)
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)