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a) 100 - 7.(x-5) = 68
7.(x - 5) = 100 - 68 = 32
x - 5 = \(\frac{32}{7}\)
x = \(\frac{32}{7}+5=\frac{67}{7}\)
b) 2(x-6)+2227:17=151
2( x -6) + 131 = 151
2 (x - 6) = 20
x - 6 = 20 : 2 = 10
x = 10 + 6 = 16
a) 2x-3=-x+6
2x - 3 + x - 6 =0
3x -9 = 0
3x = 9
x = 9 : 3
x= 3
c/ \(\left(-12x-4^3\right).8^3=4.8^4\)
\(\left(-12x-64\right).512=16384\)
\(-12x-64=\dfrac{16384}{512}=32\)
\(-12x=32+64=96\)
\(x=\dfrac{96}{\left(-12\right)}=-8\)
\(a,x+\frac{1}{2}=\frac{3}{4}.\frac{8}{6}\)
\(x+\frac{1}{2}=1\)
\(x=\frac{1}{2}\)
\(b,x-\left(\frac{1}{6}+\frac{1}{3}\right)=\frac{5}{2}.\frac{7}{5}\)
\(x-\frac{1}{6}-\frac{1}{3}=\frac{7}{2}\)
\(x-\frac{1}{6}=\frac{23}{6}\)
\(x=4\)
\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
Gọi biểu thức cần tình giá trị là A.
Ta có: \(A=\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\)
\(\Leftrightarrow4A=4\left(\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\right)\)
\(\Leftrightarrow4A=\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+...+\dfrac{1}{2008}-\dfrac{1}{2012}\)
\(\Leftrightarrow4A=\dfrac{1}{4}-\dfrac{1}{2012}=\dfrac{251}{1006}\)
\(\Leftrightarrow A=\dfrac{\dfrac{251}{1006}}{4}=\dfrac{251}{4024}\)
Ta gọi Biểu thức trên là A ta có:
A= \(\dfrac{1}{4.8}+\dfrac{1}{8.12}+\dfrac{1}{12.16}+...+\dfrac{1}{2008.2012}\)
4.A=\(\dfrac{4}{4.8}+\dfrac{4}{8.12}+\dfrac{4}{12.16}+...+\dfrac{4}{2008.201}\)
4.A=\(\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{16}+...+\dfrac{1}{2008}-\dfrac{1}{2012}\)
4.A=\(\dfrac{1}{4}-\dfrac{1}{2012}\)
4.A=\(\dfrac{502}{2012}\)
A=\(\dfrac{502}{2012}:4\)
A=\(\dfrac{502}{8048}\)
A=\(\dfrac{251}{4024}\)
Vậy biểu thức trên có giá trị là \(\dfrac{251}{4024}\)
\(4^2x+4^2.10=160\)
\(4^2\left(x+10\right)=160\)
\(16\left(x+10\right)=160\)
\(x+10=10\)
\(x=0\)
vay \(x=0\)
\(\left(12x-4^3\right).8^3=4.8^4\)
\(\left(12x-4^3\right).8^3-4.8^4=0\)
\(\left(12x-64\right).8^3-4.8^4=0\)
\(8^3\left(12x-64-32\right)=0\)
\(12x-96=0\)
\(12x=96\)
\(x=8\)
vay \(x=8\)
\(5x-2x+2=-32\)
\(3x+2=-32\\ 3x=-32-2=-34\\ x=\dfrac{-34}{3}\)
5x-2x+2=-4.8
<=>3x=-4.8-2
<=>3x=-6.8
<=>x=\(\dfrac{-34}{15}\)