Tìm x:
\(6\dfrac{2}{9}.x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
giải giúp mình chi tiết nhé!!!
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Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
`5/9+4/9:x=1/3`
`=>4/9:x=1/3-5/9`
`=>4/9:x=3/9-5/9`
`=>4/9:x=-2/9`
`=>x=4/9:(-2/9)`
`=>x=4/9.(-9/2)`
`=>x=-4/2`
`=>x=-2`
`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`
`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`
`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`
`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`
`<=> x * 88/315 = -352/45`
`<=> x = -28`
Vậy `S={-28}`
\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)
\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)
\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)
\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)
\(=-\dfrac{69}{45}\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)
Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)
Lấy (4) trừ (3) ta có:
\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)
`6/7 . 8/13 +6/7 . 9/13+3/13 . 6/7`
`=6/7 . (8/13+9/13+3/13)`
`=6/7 . 20/13`
`=120/91`
\(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}+\dfrac{3}{13}.\dfrac{6}{7}\)
\(=\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}+\dfrac{3}{13}\right)\)
\(=\dfrac{6}{7}.\left(\dfrac{8+9+3}{13}\right)\)
\(=\dfrac{6}{7}.\dfrac{20}{13}\)
\(=\dfrac{6.20}{7.13}\)
\(=\dfrac{120}{91}\)
chỉ cho biết mẫu số nên ta quy đồng mẫu số
4/7 < x/10 < 5/7 vậy
4/7 = 40/70
x/10 = x/70
5/7 = 50/70
vậy x là
41;42;43;44;45;46;47;48;49
\(\dfrac{4}{7}< \dfrac{x}{10}< \dfrac{5}{7}\)
\(\Rightarrow\dfrac{4.10}{7.10}< \dfrac{7.x}{10.7}< \dfrac{5.10}{7.10}\)
\(\Rightarrow\dfrac{40}{70}< \dfrac{7.x}{70}< \dfrac{50}{70}\)
\(\Rightarrow7.x\in\left\{41;42;43;...49\right\}\)
\(\Rightarrow7.x\in\left\{42;49\right\}\) \(\left(x\in N\Rightarrow7.x⋮7\right)\)
\(7.x=42\Leftrightarrow x=6\)
\(7.x=49\Leftrightarrow x=7\)
\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)
a) x=24/35 -2/7
x=14/35
b) x=7/8+5/6
x=41/24
c) x-11/5=3/5
x=11/5+3/5
x=14/5
tick cho mình nhé
\(6\dfrac{2}{9}.x+3\dfrac{10}{27}=22\dfrac{1}{7}\)
\(\dfrac{56}{9}.x+\dfrac{91}{27}=\dfrac{155}{7}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{155}{7}-\dfrac{91}{27}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{4185}{189}-\dfrac{637}{189}\)
\(\left(\dfrac{56}{9}.x\right)\) \(=\dfrac{3548}{189}\)
\(x\) \(=\dfrac{3548}{189}:\dfrac{56}{9}\)
\(x\) \(=\dfrac{3548}{189}.\dfrac{9}{56}\)
\(x\) \(=\dfrac{887}{294}\)
Vậy \(x\\\) \(=\dfrac{887}{294}\)
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