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HH
1 tháng 7 2018
a/ \(25x^2-9=0\)
<=> \(\left(5x-3\right)\left(5x+3\right)=0\)
<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
1 tháng 7 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)
Vậy S = {3/5 ; -3/5}
b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)
\(\Leftrightarrow9=0\left(vl\right)\)
Vậy S = \(\varnothing\)
30 tháng 12 2016
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
30 tháng 12 2016
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
6 tháng 6 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=0\Leftrightarrow x=0,6\\5x+3=0\Leftrightarrow x=-0,6\end{matrix}\right.\)
b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-0,125\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)
\(\Leftrightarrow2x=-255\)
\(\Leftrightarrow x=-127,5\)
LH
4
9 tháng 11 2018
a) \(x^2-2x=0\)
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2-4^2=0\)
\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\left(3x-5\right)\left(3x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
c) \(x^2-25x=0\)
\(x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2-3^2=0\)
\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)
\(\left(4x-4\right)\left(4x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}}\)
N
9 tháng 11 2018
a) \(x^2-2x=0\)
\(x.\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
vậy..
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2=16\)
\(\left(3x-1\right)^2=4^2=\left(-4\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
vậy ...
c) \(x^2-25x=0\)
\(x.\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy ....
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}4x-1=3\\4x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
vậy ...
L
24 tháng 8 2024
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
\(x^2\left(x+1\right)+2x\left(x+1\right)=0\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\left\{{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
a, \(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)
=> ( x + 1 ) ( \(x^2\) + 2x ) = 0
=> ( x + 1 ) x (x + 2 ) = 0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\)
=> \(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
=> \(\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=\dfrac{-2}{15}\end{matrix}\right.\)