\(x-\dfrac{x}{12}+\dfrac{x}{\dfrac{18}{7}}-\dfrac{7}{12}+\dfrac{7}{18}=-\dfrac{4}{7}\)

\(\Rightarrow x\left(1-\dfrac{1}{12}+\dfrac{1}{\dfrac{18}{7}}\right)=-\dfrac{4}{7}+\dfrac{7}{12}-\dfrac{7}{18}\)

\(\Rightarrow x\left(\dfrac{216-18+7}{12.18}\right)=\dfrac{-864+882-588}{7.12.18}\)

\(\Rightarrow x\left(\dfrac{205}{12.18}\right)=\dfrac{-570}{7.12.18}\)

\(\Rightarrow x=\dfrac{-570}{7.12.18}.\dfrac{12.18}{205}\)

\(\Rightarrow x=\dfrac{-114}{287}\)

=>x/7=-4/7

=>x=-4

\(-\dfrac{7}{12}\cdot\dfrac{3}{8}+-\dfrac{7}{12}\cdot\dfrac{5}{8}+1\dfrac{7}{12}\)

\(=-\dfrac{7}{12}\cdot\left(\dfrac{3}{8}+\dfrac{5}{8}-1\right)\)

\(=-\dfrac{7}{12}\cdot\left(\dfrac{8}{8}-1\right)\)

\(=-\dfrac{7}{12}\cdot\left(1-1\right)\)

\(=-\dfrac{7}{12}\cdot0\)

\(=0\)

\(=\dfrac{7}{12}\times\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\dfrac{7}{12}\times\dfrac{7}{7}=\dfrac{7}{12}\)

`7/12xx3/7+7/12xx4/7=7/12xx(3/7+4/7)=7/12xx7/7=7/12`

\(\dfrac{7}{12}.\dfrac{3}{7}+\dfrac{7}{12}.\dfrac{4}{7}=\dfrac{7}{12}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)=\dfrac{7}{12}\)

\(=\dfrac{11}{12}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{11}{12}\cdot1=\dfrac{11}{12}\)

=11/12

11/12 nha

\(\dfrac{x-\dfrac{x}{12}+\dfrac{x}{18}}{7-\dfrac{7}{12}+\dfrac{7}{18}}=-\dfrac{4}{7}\\ \dfrac{x\left(1-\dfrac{1}{12}+\dfrac{1}{18}\right)}{7\left(1-\dfrac{1}{12}+\dfrac{1}{18}\right)}=-\dfrac{4}{7}\\ \dfrac{x}{7}=-\dfrac{4}{7}\\ x=-4\)

x :  7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

(\(x\) - 12) : 4 = 7

\(x\) - 12 = 7 \(\times\) 4

\(x\) - 12 = 28

\(x\)       = 28 + 12

\(x\)       = 40

\(x\) - 12 : 4 = 7

\(x\) - 3 = 7

\(x\) = 7 + 3

\(x\) = 10

\(\left(x-12\right)\) : 4 = 7

<=> \(x-12\) = 7 x 4

<=> \(x-12\) = 28

<=> \(x\) = 40