Cho 3 so khac nhau va khac 0 thoa man \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\).Khi do gia tri cua \(P=\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\)
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Cho ba so a , b, c thuoc Q khac nhau tung doi mot va khac 0 thoa man \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\). Chung minh \(\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\) khong phu thuoc vao cac so a , b, c
cho 3 so thuc a, b, c khac 0 thoa man: \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
tinh P =\(\dfrac{a^2b+b^2c+c^2a}{a^3+b^3+c^3}\)
Cho a,b,c la cac so nguyen khac 0 thoa man:\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}\dfrac{1}{b^2}\dfrac{1}{c^2}\)
CM a3+b3+c3 chia het cho 3
Chỗ giả thiết vế phải có đúng ko vậy
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cho 3 so a,b,c thoa man dieu kien : \(\left\{{}\begin{matrix}a+b+c=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\)
tinh gia tri cua bieu thuc T=\(a^2+b^2+c^2\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
T = a2 + b2 + c2 = (a + b+ c)2 - 2(ab + bc + ca) = 1 - 0 = 1
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cho tỉ lệ thức \(\dfrac{a}{b}\)chung minh \(\dfrac{a}{a-b}=\dfrac{a}{c-d}\)(giả thiet a khac b ,c khac d va a,b,c khac 0
Thiếu nhé:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)
Ta có điều phải chứng minh
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1, cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\) va a+b+c khac 0 tinh b,c
Theo dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\) (vì \(a+b+c\ne0\))
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c=\pm1\)
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cho 3 so a,b,c khac thuoc Q khac nhau tung doi mot va khac 0 thoa man a/b+c=b/a+c=c/a+b
Chung minh b+c/a+a+c/b+a+b/c khong phu thuoc vao cac gia tri cua a,b,c
1. cho cac so thuc a,,b,c > 0 .Gia tri nho nhat cua bieu thuc T = \(\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)
Áp dụng bđt AM - GM:
\(T=\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}=\left(\dfrac{1}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\right)+\dfrac{8}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{8}{9}.3=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\).
Đẳng thức xảy ra khi a = b = c.
Vậy Min T = \(\dfrac{10}{3}\) khi a = b = c.
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Cho:\(\dfrac{a+b}{a+c}-\dfrac{a-b}{a-c}\)
Với :a khác -c;ac khac
Tinh gia tri :P=\(\dfrac{10b^2+9bc+c^2}{2b^2+bc+2c^2}\)