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3x - 55 + 2x = -10 => 3x + 2x - 55 = -10 => 5x - 55 = -10 => 5x = (-10) + 55 = 45 => x = 45 : 5 = 9
Ta có: \(5.\left|x\right|=\left|-2017\right|+\left|13\right|\)
\(5.\left|x\right|=2017+13\)
\(5.\left|x\right|=2030\)
\(\left|x\right|=406\)
\(\Rightarrow x\in\left\{-406;406\right\}\)
Sắp thi rùi, chúc bn hok "tuốt" nha^^
** Bổ sung thêm điều kiện $x,y$ là số nguyên.
Lời giải:
Với $x,y$ nguyên thì $x-2, x-y+5$ cũng là số nguyên. Mà $(x-2)(x-y+5)=11$ nên ta có các TH sau:
TH1:
$x-2=1, x-y+5=11\Rightarrow x=3; y=-3$ (thỏa mãn)
TH2:
$x-2=-1, x-y+5=-11\Rightarrow x=1; y=17$ (thỏa mãn)
TH3:
$x-2=11, x-y+5=1\Rightarrow x=13; y=17$ (thỏa mãn)
TH4:
$x-2=-11; x-y+5=-1\Rightarrow x=-9; y=-3$ (thỏa mãn)
A=2+2^2+2^3+....+2^10:3
A=(2+2^2)+(2^3+2^4)+....+(2^9+2^10):3
A=2.(1+2)+2^3.(1+2)+...+2^9.(1+2):3
A=2.3+2^3.3+...+2^9.3:3
A=3.(2+2^3+...+2^9):3
vậy A:3
= 28/15 . 3/4 - ( 11/20 + 1/4 ) : 7/3
= 28/15 . 3/4 - 4/5 : 7/3
= 7/5 - 12/35
= 37/35
= \(\dfrac{28}{15}\) . \(\dfrac{3}{4}\) - (\(\dfrac{11}{20}\) + \(\dfrac{1}{4}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - (\(\dfrac{11}{20}\) + \(\dfrac{5}{20}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) x \(\dfrac{3}{7}\)
= \(\dfrac{7}{5}\) - \(\dfrac{12}{35}\)
= \(\dfrac{49}{35}\) - \(\dfrac{12}{35}\)
= \(\dfrac{37}{35}\)
\(\frac{x}{2}-\frac{x}{3}=25\%\\ \frac{3x}{6}-\frac{2x}{6}=\frac{1}{4}\\ \frac{x}{6}=\frac{1}{4}\\ \Rightarrow4x=6\cdot1\\ 4x=6\\ x=6:4\\ x=\frac{3}{2}\)Vậy \(x=\frac{3}{2}\)
\(\frac{x}{2}-\frac{x}{3}=25\%\)
\(=\frac{3x}{6}-\frac{2x}{6}=\frac{25}{100}=\frac{1}{4}\)
\(=\frac{x}{6}=\frac{1}{4}\)
\(\Rightarrow6.1=x.4\)
\(6=x.4\)
\(6:4=x\)
\(\frac{3}{2}=x\\\)
KL: \(x=\frac{2}{3}\)
\(A=\dfrac{4}{1.2}+\dfrac{4}{2.3}+\dfrac{4}{3.4}+...+\dfrac{4}{2014.2015}\)
\(=4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)
\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(=4\left(1-\dfrac{1}{2015}\right)\)
\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
\(A=\dfrac{4}{1\cdot2}+\dfrac{4}{2\cdot3}+...+\dfrac{4}{2014\cdot2015}\)
\(=4\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2014\cdot2015}\right)\)
\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(=4\left(1-\dfrac{1}{2015}\right)=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)