(-7+3x)\(^4\)= 81
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a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)
<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)
<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)
<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)
<=> \(2x^2-1=\pm\frac{7}{9}\)
<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)

Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)

x - ( 81 + 5x ) = 19
=> x - 81 - 5x = 19
=> ( x - 5x ) - 81 = 19
=> -4x = 19 + 81
=> -4x = 100
=> x = 100 : ( -4 )
=> x = -25
2x - ( 8 + 3x ) = 92
=> 2x - 8 - 3x = 92
=> ( 2x - 3x ) = 8 + 92
=> -x = 100
=> x = -100

b: 1/2x-4=0
=>1/2x=4
hay x=8
a: x+7=0
=>x=-7
e: 4x2-81=0
=>(2x-9)(2x+9)=0
=>x=9/2 hoặc x=-9/2
g: x2-9x=0
=>x(x-9)=0
=>x=0 hoặc x=9

a: x+7=0
nên x=-7
b: x-4=0
nên x=4
c: -8x+20=0
=>-8x=-20
hay x=5/2
d: x2-100=0
=>(x-10)(x+10)=0
=>x=10 hoặc x=-10

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=-18\\3x-2=18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{16}{3}\\x=\dfrac{20}{3}\end{matrix}\right.\)

\(\left(12x-5\right)\left(4x-1\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(48x^2-12x-20x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(48x^2-32x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(144x^3-336x^2-96x^2+224x+15x-35\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(144x^3-432x^2+239x-35\right)\left(1-16x\right)=81\)
\(\Rightarrow144x^3-2304x^4-432x^2+6912x^3+239x-3824x^2-35+560x=81\)
\(\Rightarrow-2304x^4+7056x^3-4256x^2+799x-116=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1\approx2,3\\x_2\approx0,5\end{matrix}\right.\)
`(-7 + 3x)^4 = 81`
`=> (-7 + 3x)^4 = 3^3`
`=> 3x - 7 = -3` hoặc `3x - 7 = 3`
`=> 3x = 4` hoặc `3x = 10`
`=> x = 4/3` hoặc `x = 10/3`
Vậy ...
\(\left(-7+3x\right)^4=81\)
\(\Rightarrow\left(-7+3x\right)^4=\left(\pm3\right)^4\)
\(\Rightarrow\left[{}\begin{matrix}-7+3x=3\\-7+3x=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=3--7\\3x=-3--7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=3+7\\3x=-3+7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=10\\3x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy ...