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tớ cũng học bồi dưỡng ,,k minh nha
giải
ta có:B=1/1x2x3+1/2x3x4+... +1/18x19x220
=>2B=2/1x2x3+2/2x3x4+...2/18x19x20
=(1/1x2-1/2x3)+(1/2x3-1/3x4)+..+(1/18x19-1/19x20)
=1/1x2-1/19x20=189/380
=>B=189/760<1/4
kb đi có gì tớ giải cho
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-2}{3}=\frac{27}{x-2}ĐKx\ne2\)
\(\Leftrightarrow\left(x-2\right)^2=81\Leftrightarrow x-2=\pm9\)
TH1 : \(x-2=9\Leftrightarrow x=11\)
TH2 : \(x-2=-9\Leftrightarrow x=-7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. |x-2|+2-x=0
=> |x-2|=x-2
+) x-2=x-2
=> x \(\in\)N
+) x-2=-(x-2)
=> x-2=-x+2
=> x+x=2+2
=> x=2
Vậy x > 2.
b. |x-3|-3=-x
=> |x-3|=-x+3
+) x-3=-x+3
=> x+x=3+3
=> x=3
+) x-3=-(-x+3)
=> x-3=x-3
=> x \(\in\)N
Vậy x < 3.
a) | x - 2 | + 2 - x = 0
=> x - 2 = 0
x = 0 + 2
x = 2
=> 2 - x = 0
x = 2 - 0
x = 2
Vậy x = 2 nhé
Bài kia tương tự
Tick ủng hộ nhé ^_^
a,\(27-\left\{-x-\left[-x-\left(-x\right)\right]\right\}=-18\)
\(=>\left\{-x-\left[-x+x\right]\right\}=27-\left(-18\right)\)
\(=>\left\{-x+x-x\right\}=27+18\)
\(=>-x=45\)
\(=>x=-45\)
\(b,x+\left\{\left(x+3\right)-\left[\left(x+3\right)-\left(-x-2\right)\right]\right\}=x\)
\(=>\left\{\left(x+3\right)-\left[x+3+x+2\right]\right\}=x-x\)
\(=>\left\{x+3-x-3-x-2\right\}=0\)
\(=>-x-2=0\)
\(=>-x=2=>x=-2\)
cảm ơn