tìm,x
x+3+5=16
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a)\(\Leftrightarrow\)\(x(27+30+43)=210500\)
\(\Leftrightarrow\)\(100x=210500\)
\(\Leftrightarrow\)\(x=2105\)
b)\(\Leftrightarrow\)\(x(128-12-16)=5208000\)
\(\Leftrightarrow\)\(100x=5208000\)
\(\Leftrightarrow\)\(x=52080\)
\(x\times27+x\times30+x\times43=210500\)
\(\Leftrightarrow x\times\left(27+30+43\right)=210500\)
\(\Leftrightarrow x\times100=210500\)
\(\Leftrightarrow x=2105\)
Giải:
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{16}{99}\)
\(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x.\left(x+2\right)}\right)=\dfrac{16}{99}\)
\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{16}{99}\)
\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{x+2}\right)=\dfrac{16}{99}\)
\(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{16}{99}:\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{32}{99}\)
\(\dfrac{1}{x+2}=\dfrac{1}{3}-\dfrac{32}{99}\)
\(\dfrac{1}{x+2}=\dfrac{1}{99}\)
\(\Rightarrow x+2=99\)
\(x=99-2\)
\(x=97\)
Chúc em học tốt!
\(\dfrac{1}{3x5}+\dfrac{1}{5x7}+\dfrac{1}{7x9}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{16}{99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{16}{99}\)
\(=\dfrac{2}{3x5}\)\(+\dfrac{2}{5x7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{32}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}+.....+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{32}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{32}{99}=>x=97\)
\(\dfrac{x}{12}=x+\dfrac{15}{16}\)
\(\Leftrightarrow\dfrac{x}{12}=\dfrac{16x+15}{16}\)
\(\Rightarrow12\left(16x+15\right)=16x\)
\(\Leftrightarrow192x+180=16x\)
\(\Leftrightarrow176x=-180\)
\(\Leftrightarrow x=-\dfrac{45}{44}\)
Vậy \(x=-\dfrac{45}{44}\)
\(\dfrac{x}{12}=\dfrac{x+15}{16}\)
\(\Rightarrow12\left(x+15\right)=16x\)
\(\Leftrightarrow12x+180=16x\)
\(\Leftrightarrow4x=180\)
\(\Leftrightarrow x=45\)
Vậy x = 45
\(x=-\dfrac{3}{16}+\dfrac{1}{4}=\dfrac{1}{16}\)
bạn viết rõ đề nhé
\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)
\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)
\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)
\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)
\(\dfrac{3}{4}-x=\dfrac{91}{66}\)
\(x=\dfrac{-83}{132}\)
a. (80x - 801) . 12 = 0
<=> 80x - 801 = 0
<=> 80x = 801
<=> x = \(\dfrac{801}{80}\)
(Mấy câu tiếp mik ko hiểu đề, bn viết lại để dễ hiểu hơn nhé)
c: Ta có: \(\overline{xxx}=16\)
\(\Leftrightarrow100x+10x+1=16\)
\(\Leftrightarrow101x=16\)
hay \(x=\dfrac{16}{101}\)
\(x-\dfrac{3}{4}=2\)
\(x=2+\dfrac{3}{4}\)
\(x=\dfrac{11}{4}\)
\(\dfrac{x}{10}=\dfrac{-3}{5}\\ \Rightarrow x=\dfrac{10.\left(-3\right)}{5}=-6\\ ------------\\ x+\dfrac{5}{12}=\dfrac{-2}{3}\\ x=\dfrac{-2}{3}-\dfrac{5}{12}\\ x=\dfrac{-8}{12}-\dfrac{5}{12}\\ x=\dfrac{-13}{12}\)
x=16-3-5
x=8
k minh nha
mình đang âm
\(x=16-3-5\)
\(x=8\)