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a) x3-0,25x=0
<=> x(x2-0,25)=0
<=> x(x-0,5)(x+0,5)=0
<=>\(\hept{\begin{cases}x=0\\x-0,5=0\\x+0,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\pm0,5\end{cases}}}\)
b) x2-10x=-25
<=> x2-10x+25=0
<=> (x-5)2=0
<=> x-5=0
<=> x=5
c) x2-4x=0
<=> x(x-4)=0
<=>\(\hept{\begin{cases}x=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=4\end{cases}}}\)
d1) x2-8x+16=0
<=> (x-4)2=0
<=> x-4=0
<=> x=4
d2) 2x2-4x=0
<=> 2x(x-2)=0
<=>\(\hept{\begin{cases}2x=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}}\)
e) x2-16=0
<=> (x-4)(x+4)=0
<=>\(\hept{\begin{cases}x-4=0\\x+4=0\end{cases}\Leftrightarrow x=\pm4}\)
a) x3-0.25x=0
(=) x(x2-0.25)=0
- x=0
- (x-0.5)(x+0.5)=0 (=)x=0.5 hoac -0.5
Vay x=0 hoac x=+-0.5
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)
\(\Leftrightarrow x=1\)
b, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)
d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)
e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)
f, \(4x^2-25=\left(2x-5\right)\left(2x+7\right)\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)
a) x( x + 1 ) - x( x - 5 ) = 6
⇔ x2 + x - x2 + 5x = 6
⇔ 6x = 6
⇔ x = 1
b) 4x2 - 4x + 1 = 0
⇔ ( 2x - 1 )2 = 0
⇔ 2x - 1 = 0
⇔ x = 1/2
c) x2 - 1/4 = 0
⇔ ( x - 1/2 )( x + 1/2 ) = 0
⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)
d) 5x2 = 20x
⇔ 5x2 - 20x = 0
⇔ 5x( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) 4x2 - 9 - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 - x ) = 0
⇔ ( 2x - 3 )( x + 3 ) = 0
⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )
⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0
⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0
⇔ ( 2x - 5 )(-2) = 0
⇔ 2x - 5 = 0
⇔ x = 5/2
a)\(x\left(x^2-0,25\right)=0\)
TH1:\(x=0\) TH2:\(x^2-0,25=0\)
\(x^2=0,25=>x=0,5\)
Vậy x E \(\hept{0,5;0}\)
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
a)x+5x^2=0
<=>x(x+5)=0
<=>x=0 hoặc x=-5
Vậy tập nghiệm của pt là 0;-5
a, \(x+5x^2=0\Leftrightarrow x\left(1+5x\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)
b, \(x+1=\left(x+1\right)^2\Leftrightarrow x+1-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\Leftrightarrow x=-1;x=0\)
c, \(x^3+x=0\Leftrightarrow x\left(x^2+1>0\right)=0\Leftrightarrow x=0\)
a) x+5x2=0
<=> x(1+5x)=0
<=>\(\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)
b) x+1=(x+1)2
<=> x+1-(x+1)2=0
<=> (x+1)(1-x-1)=0
<=> -x(x+1)=0
<=> \(\hept{\begin{cases}-x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
c) x3+x=0
<=> x(x2+1)=0
<=> x=0 (vì x2+1>0)
c)
d)
đây nhé