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a/ \(x^2=5\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
vậy .....
b/ \(x^2-9=0\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=3^2\\x^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy .......( nhầm cái ngoặc)
c/ \(x^2+1=0\)
\(\Leftrightarrow x^2=-1\)
Mà \(x^2\ge0\Leftrightarrow x\in\varnothing\)
Vậy ....
d/ \(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3^2\\\left(x-1\right)^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy ...
e/ \(\left(2x+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+3\right)^2=5^2\\\left(2x+3\right)^2=\left(-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy .....
f/ Ta có :
\(x^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1^2\\x^2=\left(-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
\(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
\(\left(x-1\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(x^2-9=0\Leftrightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(\left(2x+3\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x= 1/8 hoặc -1/8
b) x= 4
c) x= 2
d) Hương ơi Nhi thấy câu này wrong hay sao ấy.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(5^2.7^3.11^1.x-5^2.7^4.11^2=0\)
\(\left(5^2.7^3.11\right)\left(x-7.11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(5^2.7^3.11\right)=0\left(l\right)\\\left(x-7.11\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}\\x-77=0\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=77\end{cases}}}\)
Vậy x = 77
b) \(\left(2x+1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}\left(2x+1\right)^2=5^2\\\left(2x+1\right)^2=\left(-5\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
(2x+1)^2=5^2
2x+1=5
2x=5-1
2x=4
x=4÷2
x=2
câu1khó mik làm câu2
(2x+3/5)^2-9/25=0
(2x+3/5)62=0+9/25=(3/5)^2
=>2x+3/5=3/5
2x=3/5-3/5
2x=0
x=0/2
x=0