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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 ...
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
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
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)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
FM
7 tháng 10 2018
Ta có:
\(0.25x^3+x^2+x=0\)
\(\Leftrightarrow x^3+4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
7 tháng 10 2018
\(0,25x^3+x^2+x=0\)
\(x\left(0,25x^2+x+1\right)=0\)
\(x\left[\left(0,5x\right)^2+2\cdot0,5x\cdot1+1^2\right]=0\)
\(x\left(0,5x+1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\0,5x+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy.....
HH
2
11 tháng 9 2021
a: \(x^2+12x+36=0\)
\(\Leftrightarrow\left(x+6\right)^2=0\)
\(\Leftrightarrow x+6=0\)
hay x=-6
b: Ta có: \(x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c: Ta có: \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
11 tháng 9 2021
Lời giải:
a. $x^2+12x+36=0$
$\Leftrightarrow (x+6)^2=0$
$\Leftrightarrow x+6=0$
$\Leftrightarrow x=-6$
b.
$x^2-1=0$
$\Leftrightarrow (x-1)(x+1)=0$
$\Leftrightarrow x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=1$ hoặc $x=-1$
c.
$25x^2-9=0$
$\Leftrightarrow (5x)^2-3^2=0$
$\Leftrightarrow (5x-3)(5x+3)=0$
$\Leftrightarrow 5x-3=0$ hoặc $5x+3=0$
$\Leftrightarrow x=\frac{3}{5}$ hoặc $x=-\frac{3}{5}$
LC
1
10 tháng 7 2018
\(a,25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
\(b,8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-5=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
DT
1
16 tháng 9 2018
a) \(x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)
b) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Leftrightarrow2x^2-3x+6x-9=0\)
\(\Leftrightarrow x\left(2x-3\right)+3\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)
HL
1
23 tháng 9 2016
a) \(25x^2-2=0\)
\(=>\left(5x\right)^2-\left(\sqrt{2}\right)^2=0\)
\(=>\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)=0\)
\(=>\hept{\begin{cases}5x-\sqrt{2}=0\\5x+\sqrt{2}=0\end{cases}}\)
\(=>\hept{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)
b) \(10x-x^2-25=0\)
\(=>-x^2-5x-5x-25=0\)
\(=>-x\left(x+5\right)-5\left(x+5\right)=0\)
\(=>\left(x+5\right)\left(-x-5\right)=0\)
\(=>\hept{\begin{cases}x+5=0\\-x-5=0\end{cases}}\)
\(=>\hept{\begin{cases}x=-5\\x=-5\end{cases}}\)
25 tháng 10 2021
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
\(25x^2-9=0\)
\(25x^2=9\)
\(x^2=\frac{25}{9}\)
\(x^2=\left(\frac{5}{3}\right)^2=\left(-\frac{5}{3}\right)^2\)
\(\Rightarrow x=\frac{5}{3};x=-\frac{5}{3}\)
\(25x^2-9=0\)
\(\Leftrightarrow\)\(25x^2=0+9\)
\(\Leftrightarrow\)\(25x^2=9\)
\(\Leftrightarrow\)\(x^2=\frac{9}{25}\)
\(\Leftrightarrow\)\(x=-\frac{3}{5};\frac{3}{5}\)