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Bài 1 rút gọn bc tự làm :
\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)
\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)
\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)
\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)
\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)
Bài 2 )
a ) \(x+\dfrac{1}{x}=3\)
\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)
b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) + x2
= (x2-ax-bx+ab) + (x2-bx-cx+bc) + (x2-cx-ax+ac) + x2
= 4x2 - 2ax - 2bx + ab + bc + ac
Thay a+b+c = 2x, ta được:
M = 4x2 - 2x(a+b+c) + ab + bc + ac
M = 4x2 - 2x.2x + ab + bc + ac
M = ab + bc + ac
Vậy => đcpcm
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
Bài 2 :
Ta có : \(4p(p-a)\)\(=2\left(a+b+c\right)\left(\dfrac{a+b+c}{2}-a\right)\)
=\(2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)
\(=2bc+b^2+c^2-a^2\left(dpcm\right)\)
Vậy :
Bai 2:
Ta có:
\(VP=4p\left(p-a\right)=2p.2p-2a.2p\) (1)
Thay \(a+b+c=2p\) vào (1) ta có:
\(\left(a+b+c\right)^2-2a.\left(a+b+c\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-2a^2-2ab-2ac\)
\(=-a^2+b^2+c^2+2bc=VT\)
Vậy \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
Chúc bạn học tốt!!!