cho cac so a,b,c va thoa man \(\frac{ab}{a+b}=\frac{1}{3},\frac{bc}{b+c}=\frac{1}{4},\frac{ca}{c+a}=\frac{1}{5}\)Tinh gia tri bieu thuc P=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Cho cac so a,b,c va thoa man\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=2\)Tinh gia tri bieu thuc \(P=\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\)
Cho a,b,c la cac so duong thoa man a+b+c=9.Tim gia tri nho nhat cua bieu thuc:
\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
Ta có:\(P=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2}\)
\(\Rightarrow P\ge a^2+b^2+c^2+\frac{9}{a^2+b^2+c^2}\)(bđt cauchy-schwarz)
\(P\ge\frac{a^2+b^2+c^2}{81}+\frac{9}{a^2+b^2+c^2}+\frac{80\left(a^2+b^2+c^2\right)}{81}\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\left(a^2+b^2+c^2\right)}{81}\left(AM-GM\right)\)
Sử dụng đánh giá quen thuộc:\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=27\)
\(\Rightarrow P\ge\frac{2}{3}+\frac{80\cdot27}{81}=\frac{82}{3}\)
"="<=>a=b=c=3
Gia su a,b,c la cacso thoa man a+b+c=259 va \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=15\). Khi do gia tri cua bieu thuc \(Q=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)bang
Ta có
\(Q+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{a+c}+\frac{a+c}{a+c}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
\(=259.15\)
\(\Rightarrow Q=259.15-3=3885\)
Cho 2 so thuc a, b thoa man dieu kien ab= 1, a+ b\(\ne\)0. Tinh gia tri bieu thuc :
P= \(\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^3}+\frac{1}{b^3}\right)+\frac{3}{\left(a+b\right)^4}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{6}{\left(a+b\right)^3}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Cho 3 so a,b,c thoa man dieu kien abc=105 va bc+b+1\(\ne\) 0 .
Tinh gia tri cua bieu thuc :
\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)
day du se like , ko neu cau hoi tuong tu
Vì abc=105
=> \(S=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)
\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(b+1+bc\right)}\)
\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}=1\)
Vậy S=1.
Cho cac so a,b,c,d thoa man: \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\)
\(\frac{d}{a+b+c}\). Tinh gia tri bieu thuc:
P=\(\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{b+a}=\frac{d+a}{b+c}\)
Ta có :
\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Leftrightarrow\)\(\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\)\(\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)
Ta thấy các tử bằng nhau suy ra các mẫu bằng nhau
\(\Rightarrow\)\(b+c+d=c+d+a=d+a+b=a+b+c\)
\(\Rightarrow\)\(a=b=c=d\)
\(\Rightarrow\)\(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{b+a}+\frac{d+a}{b+c}=1+1+1+1=4\)
Đề bị nhầm đúng ko bạn ^^
a, cho day ti so bang nhau : \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)
tinh gia tri bieu thuc M: \(\frac{a+b}{c+d}+\frac{b+c}{d+a}+\frac{c+d}{a+b}+\frac{d+a}{b+c}\)
b,cho x= \(1+\frac{1}{2013}+\frac{1}{2013^2}+\frac{1}{2013^3}+....+\frac{1}{2013^{2013}}\)
tinh gia tri bieu thuc: S= (2012x+\(\frac{1}{2013^{2013}}\)) : 2013^2014
cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)
\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)
\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)
\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)
\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cho a,b,c la cac so duong thoa man dieu kien \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Cmr \(\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}\ge\frac{a+b+c}{4}\)