Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, (-45)+(-2018)+/-145/+/-2018/
=[(-45)+/-45/] +[(-2018)+/-2018/)
=(-45+45)+(-2018+2018)
=0
b, 5^3.4^3+85.5^3-5^7:5^4
=5^3.4^3+85.5^3-5^3
=5^3.(4^3+85-1)
=5^3.148
=18500
c, (7^108.50-7^108):7^110
=[7^108.(50-1)];7^110
=(49.7^108):7^110
=49:7^2
=49:49=1
a) (-45)+(-2018)+/-145/+/-2018/
=(-45)+(-2018)+145+2018
=(-45+145)+(-2018+2018)
=100+0
=100
b)53.42+85.53-57:54
=53.42+85.53-53
=53.(42+85-1)
=53.100
=12500
c) (7108.50-7108):7110
=7108.(50-1):7110
=7108.49:7110
=7108.72:7110
=7110:7110
=1
\(\Leftrightarrow2\cdot2\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{x+1}=\dfrac{252}{1009}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{4036}\)
=>x+1=4036
hay x=4035
b: \(\Leftrightarrow1-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{34}+...+\dfrac{1}{89}-\dfrac{1}{100}+x=\dfrac{5}{3}\)
=>x+99/100=5/3
=>x=5/3-99/100=203/300
c: \(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
\(\Leftrightarrow\dfrac{10}{231}-x+4+\dfrac{221}{231}=\dfrac{7}{3}\)
=>5-x=7/3
hay x=8/3
\(B=\frac{1010+1007+\frac{2017}{113}+\frac{2017}{117}-\frac{1010}{119}-\frac{1007}{119}}{1010+1008+\frac{2018}{113}+\frac{2018}{117}-\frac{1010}{119}-\frac{1008}{119}}\)
\(B=\frac{2017+\frac{2017}{113}+\frac{2017}{117}-\frac{2017}{119}}{2018+\frac{2018}{113}+\frac{2018}{117}-\frac{2018}{119}}\)
\(B=\frac{2017.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}{2018.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)
\(B=\frac{2017}{2018}\)
Vậy \(B=\frac{2017}{2018}\)
Chúc bạn học tốt !!!
a) \(15.3-x=2\left(x-4\right)\)
\(\Leftrightarrow15-x=2x-8\)
\(\Leftrightarrow15+8=2x+x\)
\(\Leftrightarrow23=3x\)
\(\Leftrightarrow x=\dfrac{23}{3}\)
b) \(10-\left|3-x\right|=6.\left(-7\right)\)
\(\Leftrightarrow10-\left|3-x\right|=-42\)
\(\Leftrightarrow\left|3-x\right|=10+42\)
\(\Leftrightarrow\left|3-x\right|=52\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=52\\3-x=-52\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-49\\x=55\end{matrix}\right.\)
c) \(\left(x-2\right)\left(2y-1\right)=-6\)
\(\Leftrightarrow\left(x-2\right)\left(2y-1\right)=\left(-1\right).6=6.\left(-1\right)=\left(-2\right).3=3.\left(-2\right)=\left(-6\right).1=1.\left(-6\right)=\left(-3\right).2=2.\left(-3\right)\)
Ta có bảng sau:
\(x-2\) | \(-1\) | \(6\) | \(-6\) | \(1\) | \(-2\) | \(3\) | \(-3\) | \(2\) |
\(2y+1\) | \(6\) | \(-1\) | \(1\) | \(-6\) | \(3\) | \(-2\) | \(2\) | \(-3\) |
\(x\) | \(1\) | \(8\) | \(-4\) | \(3\) | \(0\) | \(5\) | \(-1\) | \(4\) |
\(y\) | \(\dfrac{5}{2}\) | \(-1\) | \(0\) | \(\dfrac{-7}{2}\) | \(1\) | \(\dfrac{-3}{2}\) | \(\dfrac{1}{2}\) | \(-2\) |
KL: Vậy...
D = 2018 + 2018 x 113 - 2018 x 14
= 2018 x ( 1 + 113 - 14 )
= 2018 x 100
= 201 800
D = 2018 + 2018 x 113 - 2018 x 14
D = 2018 x ( 1+113-14)
D = 2018 x 100 = 201800 .
k nha!
a) = \(\frac{-2}{5}\)
b) = \(\frac{-5}{4}\)
c)= \(\frac{-11}{15}\)
d) = \(\frac{3}{2}\)
(15.3-21).x2+108-12018=113
=> (45-21).x2+108-1=113
=>24x2+107=113
=>24x2=6
=> x2=\(\dfrac{6}{24}=\dfrac{1}{4}\)
=> x = \(\pm\dfrac{1}{2}\)