Tam giac ABC co chu vi bang 1, ba canh a,b,c thoa man dang thuc
\(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{3}{2}\)
Chung minh tam giac ABC deu
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( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
\(-2x\left(-3x+2\right)-\left(x+2\right)^2\)
\(=6x^2-4x-\left(x^2+4x+4\right)\)
\(=6x^2-4x-x^2-4x-4\)
\(=5x^2-8x-4\)
( 2x + 1 )3 - ( 3x + 2 )2 = ( 2x - 5 )( 4x2 + 10x + 25 ) + 6x( 2x + 1 ) - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - ( 9x2 + 12x + 4 ) = 8x3 - 125 + 12x2 + 6x - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - 9x2 - 12x - 4 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 - 8x3 - 3x2 - 6x + 125 = 0
⇔ -12x + 122 = 0
⇔ -12x = -122
⇔ x = 61/6
( 3x - 2 )( 9x2 + 6x + 4 ) - ( 2x - 5 )( 2x + 5 ) = ( 3x - 1 )3 - ( 2x + 3 )2 + 9x( 3x - 1 )
⇔ 27x3 - 8 - ( 4x2 - 25 ) = 27x3 - 27x2 + 9x - 1 - ( 4x2 + 12x + 9 ) + 27x2 - 9x
⇔ 27x3 - 8 - 4x2 + 25 = 27x3 - 1 - 4x2 - 12x - 9
⇔ 27x3 - 4x2 + 17 - 27x3 + 4x2 + 12x + 10 = 0
⇔ 12x + 27 = 0
⇔ 12x = -27
⇔ x = -27/12 = -9/4
\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-3=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
vậy......
Ta có: \(9x^2+y^2+2z^2-18x+4z-6y+20=0\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Vậy \(\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\)
Tam giác ABC có ba cạnh a,b,c và có chu vi bằng 1
=> \(a+b+c=1\)
=> \(\hept{\begin{cases}b+c=1-a\\a+c=1-b\\a+b=1-c\end{cases}}\)
Do đó ta viết lại đề bài thành \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)
Ta sẽ chứng minh \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)
Thật vậy, ta có :
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(a+c\right)\right]\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(\ge\frac{1}{2}\cdot3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\cdot\frac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}}-3\)( bất đẳng thức Cauchy )
\(=\frac{1}{2}\cdot9-3=\frac{3}{2}\)
Đẳng thức xảy ra <=> a = b = c
=> Tam giác ABC đều ( đpcm )
Đặt \(\hept{\begin{cases}b+c=x\\a+c=y\\a+b=z\end{cases}}\)Với (x,y,z>0) và \(a=\frac{y+z-x}{2};b=\frac{x+z-y}{2};c=\frac{x+y-z}{2}\)
Ta có \(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{x}{y}\right)+\frac{1}{2}\left(\frac{z}{x}+\frac{x}{z}\right)+\frac{1}{2}\left(\frac{z}{y}+\frac{y}{z}\right)-\frac{3}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
Dấu ''='' xảy ra khi và chỉ khi \(x=y=z\)
Với x = y = z thì \(a=b=c\)
=> \(\Delta ABC\) đều