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a) x3-x2-21x+45=0
<=> x3+5x2-6x2-30x+9x+45=0
<=> (x+5)(x2-6x+9)=0
<=> (x+5)(x2-3x-3x+9)=0
<=> (x+5)(x-3)2=0
Vậy S={-5;3}
b) X3+3X2+4X+2=0
<=> X3+X2+2X2+2X+2X+2=0
<=> (X+1)(X2+2X+2)=0
VÌ X2+2X+2 >=0
NÊN S={-1}
C) X4+7X-8=0
<=> X4-X3+X3-X2+X2-X+8X-8=0
<=> (X-1)(X3+X2+X+8)=0
VÌ X3+X2+X+8>=0
NÊN S={1}
D) 6X4-X3-7X2+X+1=0
<=> 6X4-6X3+5X3-5X2-2X2+2X-X+1=0
<=> (X-1)(6X3+5X2-2X-1)=0
<=> (X-1)(6X3-3X2+8X2-4X+2X-1)=0
<=> (X-1)(2X-1)(3X2_4X+1)=0
<=> (X-1)(2X-1)(3X2-3x-x+1)=0
<=> (X-1)2(2X-1)(3x-1)=0
vậy S={1/3;1/2;1}
= x^2(X-1) - 4(x^2-2x+1)
=x^2(x-1)-4(x-1)^2
=(x-1)(x^2-4x+4)
=(x-1)(x-2)^2
\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
Tìm x
a) ( x - 1 )^3 + 1 + 3x( x - 4 ) = 0
b) x^3 - 6x^2 + 9x = 0
giúp mình với mình cần gấp
mình cảm ơn
b) \(x^3-6x^2+9x=0\)
\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x.\left(x-3\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x=0\)hoặc \(x=3\)
a. ( x - 1 )3 + 1 + 3x ( x - 4 ) = 0
<=> x3 - 3x2 + 3x - 1 + 1 + 3x2 - 12x = 0
<=> x3 - 9x = 0
<=> x ( x2 - 9 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)
b. x3 - 6x2 + 9x = 0
<=> x ( x2 - 6x + 9 ) = 0
<=> x ( x - 3 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Trả lời:
Bài 2:
a, \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-13=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{13}\end{cases}}\)
Vậy ...
b, \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-2000=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2000\end{cases}}\)
Vậy ...
c, \(2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)
Vậy ...
d, \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\-x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
Vậy ...
Trả lời:
Bài 1:
\(C=x-x^2=-\left(x^2-x\right)=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra khi x - 1/2 = 0 <=> x = 1/2
Vậy GTLN của C = 1/4 khi x = 1/2
\(E=4x^2+8x+y^2-4y+32=\left(2x\right)^2+8x+y^2-4y+4+4+24\)
\(=\left[\left(2x\right)^2+8x+4\right]+\left(y^2-4y+4\right)+24=\left(2x+2\right)^2+\left(y-2\right)^2+24\ge24\forall x\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+2=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
Vậy GTNN của E = 24 khi x = - 1; y = 2