Tìm x
x : 12/13 = 3/4
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\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)
\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)
\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)
\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)
\(\dfrac{3}{4}-x=\dfrac{91}{66}\)
\(x=\dfrac{-83}{132}\)
a. \(=\left(\dfrac{12}{15}+\dfrac{3}{15}\right)+\left(\dfrac{4}{3}+\dfrac{5}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =1+2+1=4\)
b. \(x\times\dfrac{3}{4}=\dfrac{1\times3\times6\times5}{6\times5\times5\times3}\\ x\times\dfrac{3}{4}=\dfrac{1}{5}\\ x=\dfrac{1}{5}:\dfrac{3}{4}\\ x=\dfrac{4}{15}\)
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(x-\dfrac{3}{11}=\dfrac{9}{12}\Rightarrow x=\dfrac{9}{12}+\dfrac{3}{11}=\dfrac{45}{44}\)
\(\dfrac{x}{10}=\dfrac{-3}{5}\\ \Rightarrow x=\dfrac{10.\left(-3\right)}{5}=-6\\ ------------\\ x+\dfrac{5}{12}=\dfrac{-2}{3}\\ x=\dfrac{-2}{3}-\dfrac{5}{12}\\ x=\dfrac{-8}{12}-\dfrac{5}{12}\\ x=\dfrac{-13}{12}\)
\(a,\Rightarrow x=\dfrac{7}{3}+\dfrac{6}{7}=\dfrac{49+18}{21}=\dfrac{67}{21}\\ b,\Rightarrow x=\dfrac{15}{7}-\dfrac{13}{14}=\dfrac{30-13}{14}=\dfrac{17}{14}\\ c,\Rightarrow3x=\dfrac{9}{4}-\dfrac{5}{6}=\dfrac{27-10}{12}=\dfrac{17}{12}\\ \Rightarrow x=\dfrac{17}{12}\cdot\dfrac{1}{3}=\dfrac{17}{36}\)
a, => |x+4| = 3-(-7) = 10
=> x+4=10 hoặc x+4=-10
=> x=6 hoặc x=-14
b, => |x+5| = 13 - 13 = 0
=> x+5=0
=> x=0-5 = -5
c, => |x-10| + 12 = 4
=> |x-10| = 4-12 = -8
=> ko tồn tại x tm vì |x-10| > = 0
Tk mk nha
a) -7 + | x - 4 | = 3 b) 13 - | x + 5 | = 13 c) | x - 10 | - (-12) = 4
=> | x - 4 | = 3 - (-7) => | x + 5 | = 13 - 13 =>| x - 10 | + 12 = 4
=> | x - 4 | = 10 => | x + 5 | = 0 =>| x - 10 | = 4 -12
=> x - 4 = 10 hoặc - 10 => x + 5 = 0 =>|x - 10 | = - 8
=> x = 14 hoặc x = -6 => x = 0 - 5 => Với mọi |x - 10| luôn lớn hơn hoặc bằng 0
=> x = - 5 Nên : Không có x thỏa mãn
\(x+\left(13-15\right)=5+\left(10-7\right)\)
\(< =>x-2=8=>x=10\)
Ta có: \(x+\left(13-15\right)=5+10-7\)
\(\Leftrightarrow x-2=15-7=8\)
hay x=10
`x : 12/13 = 3/4`
`=> x = 3/4 xx 12/13`
`=> x = 9/13`
Vậy `x = 9/13`
x : 12/13 = 3/4
x = 3/4 x 12/13
x = 9/13
HT