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a)\(\left(x-2\right)^2=1\)
\(\Rightarrow\left(x-2\right)^2=\orbr{\begin{cases}1^2\\\left(-1\right)^2\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}3\\1\end{cases}}\)
b)\(\left(2x-1\right)^3=-27\)
\(\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow x=-1\)
c)\(\frac{3}{2}:\left(x-\frac{5}{3}-\frac{17}{3}\right)=\frac{11}{13}\)
\(\Rightarrow\frac{3}{2}:\left(x-\frac{22}{3}\right)=\frac{11}{13}\)
\(\Rightarrow\frac{3}{2x}-11=\frac{11}{13}\)
\(\frac{3}{2x}=\frac{11}{13}+\frac{39}{13}=\frac{50}{13}\)
\(\Rightarrow39=100x\)
\(\Rightarrow x=\frac{39}{100}\)
a. (x-2)2 = 1
(x-2)2 = 12
x-2 = 1
x = 1+2
x = 3
b. (2x-1)3 = -27
(2x-1)3 = (-3)3
2x-1 = -3
2x = -3 + 1
x = -2:2
x = -1
c. \(\frac{3}{2}:\left(x-\frac{5}{3}-\frac{17}{3}\right)=\frac{11}{3}\)
\(x-\frac{5}{3}-\frac{17}{3}=\frac{3}{2}:\frac{11}{3}\)
\(x-\frac{5}{3}=\frac{9}{22}+\frac{17}{3}\)
\(x=\frac{406}{66}+\frac{5}{3}\)
\(x=\frac{511}{66}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 : bạn cứ đóng ngoặc bài lại rồi cho thêm mũ nào đó vào là xong
bài 2:
a,(2x-3)^2=4
(2x-3)^2=(+-2)^2
=> 2x-3=(+-2)
(bn cứ phân ra 2 trường hợp rồi từ từ làm
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{3}=\frac{2x-5}{7}\)
\(\Rightarrow\)7(x-1)=3(2x-5)
7x-7=6x-15
7x-6x=-15+7
x=-8
Vậy x=-8
\(\frac{3}{x}=\frac{4x}{27}\)
\(\Rightarrow3.27=4x.x\)
81=4x2
x2=81:4
bạn tự tính
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)x+\left(-5\right)=-14\)
\(\Leftrightarrow x=-14-\left(-5\right)\)
\(\Leftrightarrow x=-14+5\)
\(\Leftrightarrow x=-9\)
\(b)-x+7=-23\)
\(\Leftrightarrow-x=-23+ \left(-7\right)\)
\(\Leftrightarrow-x=-30\)
\(\Leftrightarrow x=30\)
\(c)112-x=\left(-3\right).\left(-15\right)\)
\(\Leftrightarrow112-x=45\)
\(\Leftrightarrow x=112-45\)
\(\Leftrightarrow x=67\)
\(d)\left(x-15\right)-27=5^5:5^3\)
\(\Leftrightarrow\left(x-15\right)-27=5^2\)
\(\Leftrightarrow\left(x-15\right)-27=25\)
\(\Leftrightarrow x-15=52\)
\(\Leftrightarrow x=67\)
\(e)\left(2x+1\right)^2=81\)
\(\Leftrightarrow\left(2x+1\right)^2=9^2\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
\(f)(x-5^3)=-27\)
\(f)(x-5^3)=-9^3\)
\(\Leftrightarrow x-5=-9\)
\(\Leftrightarrow x=-4\)
P/s: Bạn tự kết luận.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{27}+\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}\\ \Leftrightarrow\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}-\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\\ \Leftrightarrow2x-\dfrac{1}{3}=\dfrac{2}{3}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
\(\dfrac{1}{27}\)+\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)
⇔\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)−\(\dfrac{1}{27}\)=\(\dfrac{8}{27}\)=\(\left(\dfrac{2}{3}\right)^3\)
⇔\(2x-\dfrac{1}{3}\)=\(\dfrac{2}{3}\)
⇔2\(x\)=1⇔\(x\)=\(\dfrac{1}{2}\)
Vậy \(x\)=\(\dfrac{1}{2}\)