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Bài 1: Nhân cả 2 vế cho a+b+c rồi rút gọn được đpcm
Bài 2: Thêm 1 rồi bớt 1 :v (x+y+xy+1-1)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
(x+1)(y+1)=xy+x+y+1 => P=xy+x+y= ( x+1)(y+1)-1
\(\left(x+1\right)=\dfrac{\left(b+c\right)^2-a^2}{2bc}=\dfrac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}\)
\(\left(y+1\right)=\dfrac{a^2-\left(b-c\right)^2+\left(b+c\right)^2-a^2}{\left(b+c+a\right)\left(b+c-a\right)}=\dfrac{4bc}{\left(b+c+a\right)\left(b+c-a\right)}\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)=\dfrac{4bc}{2bc}=2=>xy+x+y=2-1=1\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
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
Xét biểu thức \(x+y+xy+1=\left(x+1\right)\left(y+1\right)\)
Từ giả thiết suy ra \(x+1=\dfrac{\left(b+c\right)^2-a^2}{2bc};y+1=\dfrac{4bc}{\left(b+c\right)^2-a^2}\)
Do đó \(\left(x+1\right)\left(y+1\right)=2\Rightarrow xy+x+y+1=2\Rightarrow xy+x+y=1\)
A = x + y + xy
A = x( y + 1) + y
A = \(\dfrac{b^2+c^2-a^2}{2bc}\left(\dfrac{a^2-b^2+2bc-c^2}{\left(b+c\right)^2-a^2}+1\right)+\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)
A = \(\dfrac{b^2+c^2-a^2}{2bc}.\dfrac{4bc}{\left(b+c\right)^2-a^2}+\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)
A= \(\dfrac{2\left(b^2+c^2-a^2\right)+a^2-b^2+2bc-c^2}{\left(b+c\right)^2-a^2}\)
A = \(\dfrac{b^2+2bc+c^2-a^2}{\left(b+c\right)^2-a^2}=\dfrac{\left(b+c\right)^2-a^2}{\left(b+c\right)^2-a^2}=1\)