Giải phương trình
18+1/11(x-8)=36+1/11{x-[18+1/11(x-8)]-36
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\(18+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-36\right]\)
\(\Leftrightarrow\dfrac{198}{11}+\dfrac{1}{11}\times\left(x-18\right)=36+\dfrac{1}{11}\times\left[\dfrac{10}{11}\times\left(x-18\right)-\dfrac{396}{11}\right]\)
\(\Leftrightarrow\dfrac{198+x-18}{11}=36+\dfrac{1}{11}\times\dfrac{10x-180-396}{11}\)
\(\Leftrightarrow\dfrac{180+x}{11}=36+\dfrac{10x-576}{121}\)
\(\Leftrightarrow\dfrac{1980+11x}{121}=\dfrac{4356}{121}+\dfrac{10x-576}{121}\)
\(\Leftrightarrow1980+11x=4356+10x-576\)
\(\Leftrightarrow11x-10x=4356-1980-576\)
\(\Leftrightarrow x=1800\)
`(5/6 -x+7/12) : ( 11/24 -1/8)=11/36`
`=>(5/6 -x+7/12) : (11/24 - 3/24)=11/36`
`=>(5/6 -x+7/12) : 8=11/36`
`=>5/6 -x+7/12=11/36 xx 8`
`=>5/6 -x+7/12=22/9`
`=> x+7/12=5/6-22/9`
`=> x+7/12=-29/18`
`=>x=-29/18 -7/12`
`=>x=-79/36`
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a) 26 x 11 = 26 x (10 + 1)
= 26 x 10 + 26 x 1 = 260 + 26 = 286
35 x 101 = 35 x (100 + 1)
= 35 x 100 + 35 x 1 = 3500 + 35 = 3535
b) 213 x 11 = 213 x (10 +1)
= 213 x 100 + 213 x 1 = 2130 + 213 = 2343
123 x 101 = 123 x (100 + 1)
= 123 x 100 + 123 x 1
= 12300 + 123 = 12423
Nói thêm: Muốn nhân một số với 11 (hoặc 101) ta nhân số đó với 10 (hoặc 100) rồi cộng thêm chính số đó.
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{18}\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{18}:\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}\)
<=>x+3=30
<=>x=27
Vậy x=27