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A ) Đặt
\(A=0,1+0,2+...+1,9\\ \Rightarrow10A=1+2+3+..+19\\ =\left(1+19\right)\cdot\dfrac{19}{2}\\ =20\cdot\dfrac{19}{2}\\ =10\cdot19=190\\ \Rightarrow A=19\)
b) \(\left(1999\cdot1998+1998\cdot1997\right)\cdot\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(=1998\cdot\left(1999+1997\right)\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(=1998\cdot3996\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(=1998\cdot3996\cdot0=0\)
75% x X + 3/4 x X +X=30
3/4 x X +3/4 x X + X =30
X x (3/4 +3/4 +1)=30
X x (3/2 +1)=30
X x 5/2 = 30
X = 30 : 5/2
X = 30x2/5
X = 12
Ta có: \(75\%\cdot x+\dfrac{3}{4}\cdot x+x=30\)
\(\Leftrightarrow\dfrac{3}{4}\cdot x+\dfrac{3}{4}\cdot x+x=30\)
\(\Leftrightarrow\dfrac{5}{2}\cdot x=30\)
\(\Leftrightarrow x=30:\dfrac{5}{2}=30\cdot\dfrac{2}{5}=12\)
Vậy: x=12
\(\Rightarrow x+\dfrac{1}{6}=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{7}{12}\)
\(C=\dfrac{2}{5}+\dfrac{2}{45}+...+\dfrac{2}{6885}\)
\(=\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{85}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{84}{85}=\dfrac{42}{85}\)
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
⇔\(x\) x 1=\(\dfrac{8}{3}\)
⇔\(x\) =\(\dfrac{8}{3}\)
Vậy \(x\)=\(\dfrac{8}{3}\)
\(\dfrac{6}{x}+\dfrac{1}{2}=2\\ \dfrac{6}{x}=2-\dfrac{1}{2}\\ \dfrac{6}{x}=\dfrac{4}{2}-\dfrac{1}{2}\\ \dfrac{6}{x}=\dfrac{3}{2}\\ x=6:\dfrac{3}{2}\\ x=\dfrac{6x2}{3}\\ x=4\)
Đủ chi tiết chưa nhỉ ??
x=4