Tìm x biết :
a) [x-$\frac{2}{3}$]$^{2}$ = 0
b) (x-4)$^{2}$ = 1
c) (2x-1)$^{3}$ = -27
d) ( x+ $\frac{1}{3}$)$^{2}$= $\frac{1}{25}$
Tìm x biết :
a) [x-$\frac{2}{3}$]$^{2}$ = 0
b) (x-4)$^{2}$ = 1
c) (2x-1)$^{3}$ = -27
d) ( x+ $\frac{1}{3}$)$^{2}$= $\frac{1}{25}$
Đáp án:
Giải thích các bước giải:
mk gửi
xin bn ctlhn
Đáp án:
$\\$
`a,`
`(x – 2/3)^2 = 0`
`↔ x – 2/3 = 0`
`↔ x = 0+ 2/3`
`↔ x = 2/3`
Vậy `x=2/3`
$\\$
`b,`
`(x-4)^2=1`
`↔` \(\left[ \begin{array}{l}(x-4)^2=1^2\\(x-4)^2=(-1)^2\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x-4=1\\x-4=-1\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=1+4\\x=-1+4\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy `x=5` hoặc `x=3`
$\\$
`c,`
`(2x – 1)^3 =-27`
`↔ (2x-1)^3=-3^3`
`↔2x-1=-3`
`↔2x=-3+1`
`↔2x=-2`
`↔x=-2÷2`
`↔x=-1`
Vậy `x=-1`
$\\$
`d,`
`(x+1/3)^2 = 1/25`
`↔` \(\left[ \begin{array}{l}(x+\dfrac{1}{3})^2=(\dfrac{1}{5})^2\\(x+\dfrac{1}{3})^2=(\dfrac{-1}{5})^2\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=\dfrac{-1}{5}\end{array}\right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{1}{5}-\dfrac{1}{3}\\x=\dfrac{-1}{5}-\dfrac{1}{3}\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{-2}{15}\\x=\dfrac{-8}{15}\end{array} \right.\)
Vậy `x=(-2)/15` hoặc `x=(-8)/15`