`@` `\text {Ans}`

`\downarrow`

`3/(x-7) = 27/135`

`3/(x - 7) = 1/5`

`3 \times 5 = (x - 7) \times 1`

`15 = x - 7`

`x - 7 = 15`

`x = 15 + 7`

`x = 22`

Vậy, `x = 22.`

\(\dfrac{3}{x-7}=\dfrac{27}{135}\) (ĐK: \(x\ne7\))

\(27\times\left(x-7\right)=3\times135\)

\(27\times x-189=405\)

\(27\times x=405+189\)

\(27\times x=594\)

\(x=594:27\)

\(x=22\)

\(-\dfrac{x}{27}-1=\dfrac{2}{3}\)

\(\Rightarrow-\dfrac{x}{27}=\dfrac{2}{3}+1=\dfrac{5}{3}\)

\(\Rightarrow-3x=27.5\)

\(\Rightarrow x=-135:\left(-3\right)\)

\(\Rightarrow x=-45\)

`->B`

Nếu 22% số đo cần tìm là 1,32 tạ thì số đo cần tìm là

B.-45

a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

=> 6(2x-7) = 9(x-3)

=> 12x - 42 = 9x - 27

=> 12x - 9x = -27 + 42

=> 3x = 15 

=> x = 5

Vậy x = 5

b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

=> -7(x + 27) = 6(x + 1)

=> -7x - 189 = 6x + 6

=> -7x - 6x = 6 + 189

=> -13x = 195

=> x = -15

Vậy x = -15

a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)

\(\Leftrightarrow12x-42=9x-27\)

\(\Leftrightarrow12x-9x=-27+42\)

\(\Leftrightarrow3x=15\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)

\(\Leftrightarrow6x+6=-7x+189\)

\(\Leftrightarrow6x+7x=189-6\)

\(\Leftrightarrow13x=183\)

hay \(x=\dfrac{183}{13}\)

Vậy: \(x=\dfrac{183}{13}\)

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

\(a,\dfrac{8}{14}=\dfrac{x+1}{7}\\ \Leftrightarrow\dfrac{x+1}{7}=\dfrac{4}{7}\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\\ b,\dfrac{-4}{3}=\dfrac{5x+1}{-27}\\ \Leftrightarrow\dfrac{36}{-27}=\dfrac{5x+1}{-27}\\ \Leftrightarrow5x+1=36\\ \Leftrightarrow5x=35\\ \Leftrightarrow x=7\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

a)

\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)

c)

\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

d)

\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

e)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)

f)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)

help