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( 2x + 2x + 2x + ... + 2x + 2x ) + ( 1 + 2 + ... + 2015 ) = 0
4030x + 2031120 = 0
4030x = - 2031120
x = - 2031120 : 4030
x = - 504
vậy x = -504
(2x+1)+(2x+2)+.......+(2x+2015)=0
<=>2x+1+2x+2+2x+3+...+2x+2015=0
<=>(2x+2x+2x+2x+2x+2x+....+2x)+(1+2+3+4+5+....+2015)=0
<=>2x.2015+2031120=0
<=>4030x+2031120=0
<=>4030x= -2031120
<=>x=-2031120:4030=-504
Vậy x=-504
a: \(=\left(6+\dfrac{1}{3}+2+\dfrac{2}{3}\right)+\left(-3-\dfrac{2}{5}-1-\dfrac{3}{5}\right)+4=9-5+4=8\)
b: \(=\dfrac{5}{6}-\dfrac{1}{9}+\dfrac{4}{5}-\dfrac{4}{6}-\dfrac{7}{9}+\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{1}{9}-\dfrac{1}{6}\)
=-1+2=1
Vì AB+BC>AC; AC+BC>AB; AB+AC>BC
nên ba điểm A,B,C không thẳng hàng
a: \(\Leftrightarrow-\dfrac{4}{9}-\dfrac{2}{9}+\dfrac{2}{3}< =x< =\dfrac{11}{7}+\dfrac{3}{7}+\dfrac{2}{5}-\dfrac{7}{5}\)
=>0<=x<=2-1=1
hay \(x\in\left\{0;1\right\}\)
b: \(\Leftrightarrow-\dfrac{8}{13}+\dfrac{21}{13}+\dfrac{7}{17}< =x< =\dfrac{-9}{14}-\dfrac{5}{14}+3\)
=>24/17<=x<=2
hay x=2
\(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{100^2}>\dfrac{1}{100\cdot101}=\dfrac{1}{100}-\dfrac{1}{101}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Suy ra: \(\dfrac{9}{202}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{99}{100}\)
\(d,\dfrac{-25}{12}-\left(\dfrac{23}{12}+1\dfrac{1}{2}\right)=d,\left(\dfrac{-25}{12}-\dfrac{23}{12}\right)-\dfrac{3}{2}=\dfrac{-48}{12}-\dfrac{3}{2}=-4-\dfrac{3}{2}=\dfrac{-8}{2}-\dfrac{3}{2}=-\dfrac{11}{2}\\ e,\dfrac{-1}{9}.\dfrac{-3}{5}-\dfrac{5}{6}.\dfrac{3}{-5}+\dfrac{-5}{2}.\dfrac{3}{5}=\dfrac{1}{9}.\dfrac{3}{5}+\dfrac{5}{6}.\dfrac{3}{5}-\dfrac{5}{2}.\dfrac{3}{5}=\dfrac{3}{5}\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{5}{2}\right)=\dfrac{3}{5}.\dfrac{-14}{9}=\dfrac{-14}{15}\)
Bài 4:
Số thứ hai là 150x3/5=90
Số thứ ba là 90x2/3=60
Số thứ tư là 60x7/10=42
Trung bình của bốn số là:
(150+90+60+42):4=85,5
a: \(=\dfrac{5}{6}+\dfrac{7}{6}-\dfrac{1}{3}-\dfrac{2}{3}=2-1=1\)
b: \(=1+\dfrac{4}{5}-2-\dfrac{3}{4}-1-\dfrac{1}{4}+2+\dfrac{1}{5}=1-1=0\)
60x - 30x + 60 - 30 = 90
x.(60 - 30) + (60 - 30) = 90
x.30 + 30 = 90
30.(x+1) = 90
x+1 = 90 : 30
x+1 = 3
Vậy x = 3 - 1 = 2
Nhớ k cho mình nhé'! Thank you!!!
Giải
60x-30x+60-30=90
x.(60-30)+60-30=90
x.30+60-30=90
x.30+60=90+30
x.30+60=120
x.30=120-60
x.30=60
x=60:30
x=2