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\(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy x = 0 hoặc x = -1
\(\left(2x+1\right)^2=25\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy x = 2 hoặc x = -3
\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4,5\\x=-1,5\end{cases}}\)
Vậy x = 4,5 hoặc x = -1,5
a, (2x-3)3 = -64
=> (2x-3)3 = -43
=> 2x-3=-4
=> 2x = -1
=> x = -1 : 2
=> x = -1/2
b, (2x-3)2 =25
=> (2x-3)2 =5^2
=> 2x-3 = 5
=> 2x = 8
=> x = 4
c, (3x-4)2 =36
=> (3x-4)2 =62
=> 3x-4 = 6
=> 3x = 10
=> x = 3.(3)
d, 2x+1 = 64
=> 2x+1 = 26
=> x+1 = 6
=> x = 5
a. (2x-3)2 = 36
(2x-3)2 = 62
=> TH1: 2x - 3 = 6
2x = 9
x = 9/2
TH2: 2x - 3 = -6
2x = -6 + 3
2x = -3
x = -3/2
Vậy x \(\in\){ -3/2 ; 9/2)
Câu b tương tự
a.(2x-3)^2=36
\(\Rightarrow\left(2x-3\right)^2=6^2\)
\(\Rightarrow2x-3=6\)
\(\Rightarrow2x=9\)\(\Rightarrow x=9:2=\frac{9}{2}\)
a) \(x-1=27\)
\(\Rightarrow x=27+1\)
\(\Rightarrow x=28\)
Vậy \(x=28.\)
b) \(x^2+x=0\)
\(\Rightarrow x.\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;-1\right\}.\)
c) \(\left(2x+1\right)^2=25\)
\(\Rightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)
\(\Rightarrow2x+1=\pm5.\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:2\\x=\left(-6\right):2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{2;-3\right\}.\)
d) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow2x-3=\pm6.\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9:2\\x=\left(-3\right):2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{9}{2};-\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
Bài làm:
Ta có: \(\left(2x-\frac{1}{3}\right)^2=\frac{1}{36}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{1}{6}\\2x-\frac{1}{3}=-\frac{1}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{2}\\2x=\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{1}{12}\end{cases}}\)
\(\left(2x-\frac{1}{3}\right)^2=\frac{1}{36}=\left(\frac{1}{6}\right)^2=\left(-\frac{1}{6}\right)^2\)
\(< =>\orbr{\begin{cases}2x-\frac{1}{3}=\frac{1}{6}\\2x-\frac{1}{3}=-\frac{1}{6}\end{cases}< =>\orbr{\begin{cases}2x=\frac{1}{6}+\frac{1}{3}\\2x=-\frac{1}{6}+\frac{1}{3}\end{cases}}}\)
\(< =>\orbr{\begin{cases}2x=\frac{1}{6}+\frac{2}{6}\\2x=\frac{2}{6}-\frac{1}{6}\end{cases}< =>\orbr{\begin{cases}2x=\frac{3}{6}\\2x=\frac{1}{6}\end{cases}}}\)
\(< =>\orbr{\begin{cases}x=\frac{3}{6}:\frac{2}{1}=\frac{3}{6}.\frac{1}{2}=\frac{3}{12}=\frac{1}{4}\\x=\frac{1}{6}:\frac{2}{1}=\frac{1}{6}.\frac{1}{2}=\frac{1}{12}\end{cases}}\)
a)(2x+1)2=25=52=(-5)2
=>2x+1=5 hoặc 2x+1=-5
2x=5-1 2x=-5-1
2x=4 2x=-6
x=4/2 x=-6/2
x=2 x=-3
Vậy x=2 hoặc x=-3
b)(2x-3)2=36=62=(-6)2
=>2x-3=6 hoặc 2x-3=-6
2x=6+3 2x=-6+3
2x=9 2x=-3
x=9/2 x=-3/2
Vậy x=9/2 hoặc x=-3/2
(2x+1)2 = 25
=> 2x2 + 12 = 25
=> 2x2 + 1 = 25
=> 2x2 = 24
=> x2 = 12
=> x = \(\sqrt{12}\)
(2x - 3)2 = 36
=> 2x2 - (32) = 36
=> 2x2 - 9 = 36
=> 2x2 = 45
=> x2 = 22,5
=> x = \(\sqrt{22,5}\)
t i c k nha!! 65768769789890