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S
22 tháng 7 2016
Ta có: x3 - 0,25.x = 0
=> x.(x2 - 0,25) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-0,25=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=0,25=0,5^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=0,5\end{cases}}\)
22 tháng 7 2016
a) x3 - 0,25x = 0
x.x.x - 0,25x = 0
x. ( x2 - 0,25 ) = 0
TH1 : x = 0
TH2 : x2 - 0,25 = 0
x2 = 0 + 0,25
x2 = 0,25
=> x = 0,5
Vậy x = 0 ; 0,5
TT
4
14 tháng 8 2017
a, \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy...
b, \(4x^2-9=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy...
c, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
Vậy...
N
14 tháng 8 2017
a) \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x_1=0;x_2=0,5;x_3=-0,5\).
b) \(4x^2-9=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{3}{2};x_2=-\dfrac{3}{2}\).
c) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy x = 5.
24 tháng 6 2017
Lần sau ghi tách ra tí bạn ơi ;v
--------------------------------
1. a) \(5x-20y=5\left(x-4y\right)\)
b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)
c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
d) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=4xy\)
e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
24 tháng 6 2017
2. a) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)
Vậy...
b) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x+1-x^2-2x-1=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
c) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
Vì \(x^2+1>0\Rightarrow x=0\)
Vậy...
d) \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
Vậy..
e) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vậy...
NK
2
T
10 tháng 11 2023
a) \(x^2-36=0\)
\(\Rightarrow x^2-6^2=0\)
\(\Rightarrow\left(x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
b) \(x^3-0,25x=0\)
\(\Rightarrow x\left(x^2-0,25\right)=0\)
\(\Rightarrow x\left[x^2-\left(0,5\right)^2\right]=0\)
\(\Rightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
c) \(x^2-10x=-25\)
\(\Rightarrow x^2-10x+25=0\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2=0\)
\(\Rightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
d) \(x^2-2x=-1\) (sửa đề)
\(\Rightarrow x^2-2x+1=0\)
\(\Rightarrow\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
e) \(x^3+3x^2=-3x-1\)
\(\Rightarrow x^3+3x^2+3x+1=0\)
\(\Rightarrow\left(x+1\right)^3=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(\text{#}Toru\)
10 tháng 11 2023
a: \(x^2-36=0\)
=>\(x^2=36=6^2\)
=>\(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
b: \(x^3-0,25x=0\)
=>\(x\left(x^2-0,25\right)=0\)
=>\(x\left(x-0,5\right)\left(x+0,5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
c: \(x^2-10x=-25\)
=>\(x^2-10x+25=0\)
=>\(\left(x-5\right)^2=0\)
=>x-5=0
=>x=5
e: \(x^3+3x^2=-3x-1\)
=>\(x^3+3x^2+3x+1=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
BD
1
21 tháng 2 2017
\(\Leftrightarrow x^4+y^4+4x-4y+6\ge0\)
\(\Leftrightarrow\left(x^4-2x^2+1\right)+\left(y^4-2y^2+1\right)+2\left(x^2+2x+1\right)+2\left(y^2-2y+1\right)\ge0\)
\(\Leftrightarrow\left(x^2-1\right)^2+\left(y^2-1\right)^2+2\left(x+1\right)^2+2\left(y-1\right)^2\ge0\)
VPlaf tổng 4 số không ầm .=> dpcm
đẳng thức khi y=1; x=-1
HM
1
26 tháng 8 2021
a: Ta có: \(x\left(x-3\right)-x^2+5=0\)
\(\Leftrightarrow-3x+5=0\)
hay \(x=\dfrac{5}{3}\)
b: Ta có: \(x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
22 tháng 8 2021
\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)\left(x+3-x+3\right)=0\Leftrightarrow6\left(x-3\right)=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
22 tháng 8 2021
Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2=0\)
\(\Leftrightarrow x^2-9-x^2+6x-9=0\)
\(\Leftrightarrow6x=18\)
hay x=3
Lời giải:
\(x^3-0,25x=0\)
\(\Leftrightarrow x(x^2-0,25)=0\)
\(\Leftrightarrow x(x^2-0,5^2)=0\)
\(\Leftrightarrow x(x-0,5)(x+0,5)=0\Rightarrow \left[\begin{matrix} x=0\\ x-0,5=0\\ x+0,5=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x=0,5\\ x=-0,5\end{matrix}\right.\)