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5) 32.2+ ( x+52) =102
=> 18 + ( x + 52 )= 102
=> x + 52 = 102 -18
=> x + 52 = 84
=> x = 84 -52
=> x = 32
Ta có:
1) 1 + 3 + 5 + ... + 99 = ( x - 2 ) 2
=> 100 x 50 / 2 = ( x - 2 ) 2
=> 2500 = ( x - 2 ) 2
=> x - 2 = \(\sqrt{2500}=50\)
=> x = 52
2) 3x + 25 = 26.2 + 2.30
3x + 25 = 2.( 26 + 30 )
3x + 25 = 2.56 = 112
3x = 112 - 25 = 87
không tìm được x.Có thể bạn đã sai đề.
3) 49. 7x = 204
=> 7x = 204 : 49 = 204/49
không tìm được x nếu x là số tự nhiên.
4) 22 (x + 32 ) - 5 = 55
22 (x + 32 ) = 60
x + 32 = 15
x = 15 - 9 = 6
5) 32.2 + ( x + 52 ) = 102
x + 52 = 102 - 9.2
x + 52 = 102 - 18 = 84
x = 84 - 52 = 32
\(2x^3-50=0\)
\(\Rightarrow2\left(x^3-25\right)=0\)
\(\Rightarrow x^3-25=0\Rightarrow x^3=25\)
\(\Rightarrow x=\sqrt[3]{25}\)
\(x^2-5x=-6\)
\(\Rightarrow x\left(x-5\right)=-6\)
Xét ước
\(\left(2x-1\right)^2-\left(3x+5\right)=0\)
\(\Rightarrow4x^2-4x+1-3x-5=0\)
\(\Rightarrow4x^2-4-7x=0\)
\(\Rightarrow4x^2-7x=4\)
\(\Rightarrow x\left(4x-7\right)=4\)
Xét ước
\(4x^2-20x+25=0\)
\(\Rightarrow\left(2x-5\right)^2=0\)
\(\Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\)
\(\left(3x-1\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(3x-1\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left|3x-1\right|=\left|x-2\right|\)
Xét dấu:v
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 ...
\(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
\(\left(x-1\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(x^2-9=0\Leftrightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(\left(2x+3\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
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
nick thằng trong truy kích đấy là TrumStream thôi ko phải Trùm