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Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)+\left(x+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow5x+\frac{31}{32}=1\frac{31}{32}\)
\(\Leftrightarrow5x=1\frac{31}{32}-\frac{31}{32}\Leftrightarrow5x=1\Rightarrow x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
a) \(\frac{-28}{19}\). \(\frac{\left(-38\right)}{14}\)= 4
b) \(\frac{-21}{16}\). \(\frac{\left(-24\right)}{7}\)= \(\frac{9}{2}\)
c) \(\frac{12}{17}\). \(\frac{\left(-34\right)}{9}\)= \(\frac{-8}{3}\)
d) \(\frac{\left(-15\right)}{4}\): \(\frac{21}{10}\)= \(\frac{-25}{14}\)
e) \(-2\frac{1}{7}\): \(-1\frac{1}{14}\)= 2
nhớ ủng hộ mik nha mn
a.=4
b.=119/24
c.= -8/3
d.= -25/14
Mình chỉ làm đc từng này thôi