Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
150 + 1,03 : [ 10,3 . ( x - 1 ) ] = 160
1,03 : [ 10,3 . ( x - 1 ) ] = 160 - 150
1,03 : [ 10,3 . ( x - 1 ) ] = 10
[ 10,3 . ( x - 1 ) ] = 1,03 : 10
[ 10,3 . ( x - 1 ) ] = 0,103
( x - 1 ) = 0,103 : 10,3
( x - 1 ) = 0,01
x = 0,01 + 1
x = 1,01
b) Từ đề bài ,ta có: \(\hept{\begin{cases}\frac{-\left(x+3\right)}{27}=\frac{-121}{33}\left(1\right)\\\frac{11}{1-2y}=\frac{-121}{33}\left(2\right)\end{cases}}\)
Giải (1):
\(\frac{-\left(x+3\right)}{27}=\frac{-121}{33}=>-\left(x+3\right)=-121.27:33=-99=>-x-3=-99=>-x=-96=>x=96\)
Giải (2) :
\(\frac{11}{1-2y}=\frac{-121}{33}=>1-2y=11.33:\left(-121\right)=-3=>2y=4=>y=2\)
Vậy x=96;y=2
\(\frac{\left(x-1\right)^3}{-3}\)=\(\frac{-27}{x-1}\)
<=>(x-1)\(^3\).(x-1)=(-3).(-27)
<=>(x-1)\(^4\)=81
<=>(x-1)\(^4\)=3\(^4\)
<=>x-1=3
<=>x=4(Thỏa mãn )
Vậy x=4
\(\left(1-2x\right)3=27\)
\(3-6x=27\)
\(6x=3-27\)
\(6x=-24\)
\(x=-24:6\)
\(x=-4\)
( 1 - 2x ) x 3 = 27
( 1 - 2x ) = 27 : 3
( 1 - 2x ) = 9
<=> 1 - 2x = 9
<=> 2x = ( 9 + 1 )
<=> 2x = 10
<=> x = 10 : 2 = 5
=> x = 5
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
<=> \(x-\frac{1}{2}=\frac{1}{3}\)
<=> x = \(\frac{5}{6}\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
<=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{array}\right.}\)
Vậy...
a)\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{5}{6}\)
b)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\pm\left(\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}\)
các bạn giúp mình với
Bó tay