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\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
bạn viết rõ đề ra nhé
b, \(\left|4x-8\right|=1-x\)ĐK : \(x\le1\)
TH1 : \(4x-8=1-x\Leftrightarrow5x=9\Leftrightarrow x=\dfrac{9}{5}\)( ktm )
TH2 : \(4x-8=x-1\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\)( ktm )
b) Ta có: \(\left|4x-8\right|=1-x\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-8=1-x\left(x\ge2\right)\\4x-8=x-1\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+x=1+8\\4x-x=-1+8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=9\\3x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(loại\right)\\x=\dfrac{7}{3}\left(loại\right)\end{matrix}\right.\)
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
a) x3 -2x2 +5x-4
=x3-x2-x2+x+4x-4
=x2(x-1)-x(x-1)+4(x-1)
=(x2-x+4)(x-1)
b) x3-x2+x+3
=x3+x2-2x2-2x+3x+3
=x2(x+1) -2x(x+1)+3(x+1)
=(x2-2x+3)(x+1)
c) 6x3+x2+x+1
=6x3+ 3x2-2x2-x+2x+1
=6x2(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))
=(6x2-2x+2) (x+\(\frac{1}{2}\))
=2( 3x2-x+1) (x+\(\frac{1}{2}\))
d) 4x3 + 6x2+4x+1
= 4x3+2x2+4x2+2x+2x+1
= 4x2(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))
= 2(2x2 +2x+1)( x+\(\frac{1}{2}\))
e) x6 -9x3+8
\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy dấu \("="\) không xảy ra
Ta có:
\(M=3x\left(x-5y\right)+\left(y-5x\right)\left(-3y\right)-3\left(x^2-y^2\right)-1\)
\(M=3x^2-15xy-3y^2+15xy-3x^2+3y^2\)
\(M=0\left(đpcm\right)\)
\(A=\dfrac{5x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}=\dfrac{1}{x^2}-\dfrac{1}{x}+5=\left(\dfrac{1}{x^2}-\dfrac{1}{x}+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
\(A_{min}=\dfrac{19}{4}\) khi \(\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\)
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