120-[7.20-(134-110).5]
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a.120-[7.20-(134-110).5]
=120-(7.20-24.5)
=120-(140-120)
=120-20
=100
b.1987-200.8+1005:5
=1987-1600+201
=588
C, 387.2694+7305.387+387
=387.(2694+7305+1)
=387.10000
3870 000
d,
d.248:{[(368+232):120-3]+122}+2016
=248:{(600:120-3)+122}+2016
=248:(2+122)+2016
=248:144+2016
=2+2016
=2018
a)120-[7.20-(134-110).5]
=120-20
=100
b)1987-200.8+1005:5
=1987-1600+201
=588
c)387.2694+7305.387+387
=1042578+2827035+387
=3870000
d)248:{[368+232):120-3]+122}+2016
=248:124+2016
=2+2016
=2018
a)120-[7.20-(134-110).5]
=120-20
=100
b)1987-200.8+1005:5
=1987-1600+201
=588
c)387.2694+7305.387+387
=1042578+2827035+387
=3870000
d)248:{[368+232):120-3]+122}+2016
=248:124+2016
=2+2016
=2018
\(120+2\left(3x-17\right)=214\)
\(120+6x-34=214\)
\(86+6x=214\)
\(6x=214-86\)
\(6x=128\)
\(x=128:6\)
\(x=\frac{64}{6}\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=110\)
\(\left(x+x+...+x\right)+\left(1+2+3+...+10\right)=110\)
\(10x+55=110\)
\(10x=110-55\)
\(10x=55\)
\(x=55:10\)
\(x=5,5\)
\(10xX+\left(1+2+3+...+10\right)=110\)
\(1+2+3+...+10=\dfrac{10\left(1+10\right)}{2}=55\)
\(\Rightarrow10xX+55=110\Rightarrow10xX=55\)
\(\Rightarrow X=55:10=5,5\)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
(x/-5+2/3);11/5-1,4.15/49=|-5|/21
(x/-5+2/3);11/5-1,4.15/49=5/21
(x/-5+2/3);11/5-21/49=5/21
(x/-5+2/3);11/5=5/21+21/49
(x/-5+2/3);11/5=35/147+63/147
(x/-5+2/3);11/5=35+63/147
(x/-5+2/3);11/5=98/147
x/-5+2/3=98/147.11/5
x/-5+2/3=98.11/147.5
x/-5+2/3=1078/735
x/-5+2/3=154/105
x/-5=154/105-2/3
x/-5=154/105-70/105
x/-5=154-70/105
x/-5=84/105
x/-5=-4/-5
suy ra x=-4
120 - [7 . 20 - (134 - 110) . 5]
= 120 - [140 - 24 . 5]
= 120 - (140 - 120)
= 120 - 20
= 100
\(120-\left[7.20-\left(134-110\right).5\right]\)
\(=120-\left[140-24.5\right]\)
\(=120-\left[140-120\right]\)
\(=120-20\)
\(=100\)
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