Question

tìm x

a,\(\dfrac{6}{7}-\dfrac{3}{4}x=-1\dfrac{1}{2}x-3\)

b\(\left(4x-1\right)^2=\left(4x-1\right)^4\)

c\(\left(\dfrac{4}{5}-\dfrac{3}{4}:x\right)^3=\dfrac{-1}{27}\)

d\(\left|\dfrac{4}{3}-\dfrac{1}{4}x\right|:2\dfrac{1}{3}=0,5\)

giúp mình nhé

Answer 1

a) \(\dfrac{6}{7}-\dfrac{3}{4}x=-1\dfrac{1}{2}x-3\)

<=> \(\dfrac{6}{7}-\dfrac{3}{4}x=-\dfrac{3}{2}x-3\)

<=> \(-\dfrac{3}{4}x+\dfrac{3}{2}x=-3-\dfrac{6}{7}\)

<=> \(x\left(-\dfrac{3}{4}+\dfrac{3}{2}\right)=-\dfrac{27}{7}\)

<=> \(x=-\dfrac{36}{7}\)

b) (4x-1)² = (4x-1)⁴

<=> (4x-1)² - (4x-1)⁴ = 0

<=> (4x-1)²[1-(4x-1)²] = 0

<=> \(\left\{{}\begin{matrix}4x-1=0\\\left(4x-1\right)^2=1\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\\left\{{}\begin{matrix}4x=2\\4x=0\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

c) \(\left(\dfrac{4}{5}-\dfrac{3}{4}:x\right)^3=-\dfrac{1}{27}\)

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

<=> \(\dfrac{3}{4}:x=\dfrac{17}{15}\)

<=> \(x=\dfrac{45}{68}\)

d) \(\left|\dfrac{4}{3}-\dfrac{1}{4}x\right|:2\dfrac{1}{3}=0,5\)

<=> \(\left|\dfrac{4}{3}-\dfrac{1}{4}x\right|=\dfrac{7}{6}\)

<=> \(\left\{{}\begin{matrix}\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{7}{6}\\\dfrac{4}{3}-\dfrac{1}{4}x=-\dfrac{7}{6}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\dfrac{1}{4}x=\dfrac{1}{6}\\\dfrac{1}{4}x=\dfrac{5}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)