1 , cho B=\(\dfrac{1}{b^2+c^2-a^2}\)+\(\dfrac{1}{c^2+a^2-b^2}\)+\(\dfrac{1}{a^2+b^2-c^2}\)
rút gọn b biết a+b+c=0
2, tìm GTLN của M = \(\dfrac{2x+1}{x^2+2}\)
3, cho a,b,c là 3 cạnh của tam giác
c/m A=\(\dfrac{a}{a+c-a}\)+\(\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)>= 3
4, tìm nghiệm nguyên dương phương trình
yx2 +yx+y=2
5,
a, cho 3x+y=1. tìm GTLN của A = 3x2+y2
b, cho các số dương a, b,c có tích bằng 1, C/m (a+1)*(b+1)*(c+1)>=8
5)
a)
Có 3x+y = 1
\(\Rightarrow x+x+x+y=1\)
Áp dụng bất đẳng thức bunhiacopxki ta có :
\(\left(x^2+x^2+x^2+y^2\right)\left(1^2+1^2+1^2+1^2\right)\ge\left(x+x+x+y\right)^2\)
\(\Rightarrow3x^2+y^{2^{ }}.4\ge\left(3x+y\right)^2\)
\(\Rightarrow3x^2+y^2\ge\dfrac{1}{4}\)
b)
Áp dụng bất đẳng thức AM - GM ta có :
\(\left[{}\begin{matrix}a^2+1^2\ge2a\\b^2+1^2\ge2b\\c^2+1^2\ge2c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(a+1\right)^2\ge4a^{ }\\\left(b+1\right)^2\ge4b^{ }\\\left(c+1\right)^2\ge4c^{ }\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge4a^{ }.4b.4c^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64a^{ }bc^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64abc\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64\)
\(\Rightarrow\left(a+1\right)^{ }\left(b+1\right)^{ }\left(c+1\right)^{ }\ge8\) \(\left(đpcm\right)\)
3)
Sửa đề \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
Đặt b + c - a = x , a+c-b = y , a+b-c= z
\(\Rightarrow\left[{}\begin{matrix}2a=y+z\\2b=x+z\\2c=x+y\end{matrix}\right.\)
Có :
\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
\(\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Rightarrow\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\)
Áp dụng bất đẳng thức \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\forall a,b>0\)
\(\Rightarrow\) \(\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)
\(\Rightarrow2\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\right)\ge6\)
\(\Rightarrow\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\) \(\left(đpcm\right)\)
1)
Có a+ b+ c = 0
\(\Rightarrow\left[{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(a+b\right)^2=c^2\\\left(a+c\right)^2=b^2\\\left(b+c\right)^2=a^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a^2+2ab+b^2=c^2\\a^2+2ac+c^2=b^2\\b^2+2bc+c^2=a^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a^2+b^2-c^2=-2ab\\a^2+c^2-b^2=-2ac\\b^2+c^2-a^2=-2bc\end{matrix}\right.\)
Có \(B=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ac}{a^2+c^2-b^2}\)
\(\Rightarrow B=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ac}{-2ac}\)
\(\Rightarrow B=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)
\(\Rightarrow B=-\dfrac{3}{2}\)
