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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+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)
bài dưới chỗ quy đồng thôi nha
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^3-1\right)\left(x^3+1\right)}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)}\)
<=>\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
<=>\(\frac{\left(x^2-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\)=\(\frac{2\left(x+2\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
<=>\(\left(x^2-1\right)\left(x^3+1\right)-\left(x^2-1\right)\left(x^3-1\right)=2\left(x+2\right)^2\)
Bài 1:
a,\(3x\left(5x^2-2x-1\right)\)
\(=3x.5x^2-3x.2x-3x=15x^3-6x^2-3x\)
b,\(\left(x^2+2xy-3\right)\left(-xy\right)\)
\(=x^2.\left(-xy\right)+2xy.\left(-xy\right)-3.\left(-xy\right)\)
\(=-x^3y-2x^2y^2+3xy\)
c,\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=\dfrac{1}{2}x^2y.\left(2x^3\right)-\dfrac{1}{2}x^2y.\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
Chúc bạn học tốt!!!
Bài 1:
a) \(3x\left(5x^2-2x-1\right)\\ =15x^3-6x^2-3x\)
b) \(\left(x^2+2xy-3\right)\left(-xy\right)\\ =-x^3y-2x^2y+3xy\)
c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\\ =x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
a) x+5x2 = 0
x(1+5x) = 0
Ta có các trường hợp sau
TH1: x = 0
TH2: 1+5x=0
5x = -1
x = \(\frac{-1}{5}\)
Vậy x = {\(\frac{-1}{5}\) ; 0}
b) x+1 = (x+1)2
(x + 1) - (x+1)2 = 0
(x+1)[1 - (x + 1)] = 0
Ta có các trường hợp sau
TH1: x + 1 = 0
x = -1
TH2: 1 - (x +1) = 0
1- x - 1 = 0
-x = 0
=> x = 0
Vậy x = {-1 ; 0}
HT