cho a, b, c khac 0 va doi mot khac nhau thoa man: \(a^2\left(b+c\right)=b^2\left(a+c\right)=2013\)
tinh gia tri cua bthuc \(H=c^2\left(a+b\right)\)
cho a,b,c khac nhau tung doi mot. Tinh
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{-a^2}{\left(a-b\right)\left(c-a\right)}+\frac{-b^2}{\left(b-c\right)\left(a-b\right)}+\frac{-c^2}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(-a^2\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-b^2\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-a^2b+ca^2-b^2c+ab^2-c^2a+bc^2}{-a^2b-c^2a+ca^2-b^2c+ab^2+bc^2}=1\)
Vậy \(P=1.\)
cho 3 so thuc abc khac 0 va mot doi so khac nhau thoa man
a2 . ( b+ c ) = b2 . ( a + c ) = 2018
Tinh gia tri bieu thuc H = c2 . ( a+b)
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
Cho a,b,c khac 0 thoa man: \(\frac{2a+b+c}{a}\)=\(\frac{2b+c+a}{b}\)=\(\frac{2c+a+b}{c}\)
Tinh gia tri cua bieu thuc: P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
GIUP MINH VOI NHA!
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\frac{2a+b+c}{a}=\frac{2b+c+a}{b}=\frac{2c+a+b}{c}=\frac{2a+b+c+2b+c+a+2c+a+b}{a+b+c}=\frac{4\left(a+b+c\right)}{a+b+c}=4\)
\(\Rightarrow\frac{2a+b+c}{a}=4\Rightarrow2a+b+c=4a\Rightarrow b+c=4a-2a=2a\)
\(\frac{2b+c+a}{b}=4\Rightarrow2b+c+a=4b\Rightarrow c+a=4b-2b=2b\)
\(\frac{2c+a+b}{c}=4\Rightarrow2c+a+b=4c\Rightarrow a+b=4c-2c=2c\)
Suy ra \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)
Vậy P=8
Cho hỏi tớ sai chỗ nào ạ :>?Góp ý giúp nha?
Cho 3 so a,b,c khac 0 va doi mot khac nhau thoa man a^2.(b+c)=b^2.(a+c)=2015 Tinh c^2.(a+b)
Cho a,b,c la 3 so doi mot khac nhau va \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\frac{\Leftrightarrow a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\) Nhân hai vế với \(\frac{1}{b-c}\)
Tương tự ta có:\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng (1),(2),(3) ta được đpcm
Cho a,b,c doi mot khac nhau va\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR: \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
cho a^3+b^3+c^3=3abc va a+b+c khac 0 . tinh gia tri bieu thuc N=\(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
Ta có: a3+b3+c3=3abc <=> a3+b3+c3-3abc=0
<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)
<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)
=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
cho a^3+b^3+c^3=3abc va a+b+c khac 0 . tinh gia tri bieu thuc \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
- Ta có : \(a^3+b^3+c^3=3abc\)
=> \(a^3+b^3+c^3-3abc=0\)
=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
=> \(a^2+b^2+c^2-ab-bc-ac=0\)
=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)
=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)
=> \(a-b=b-c=c-a=0\)
=> \(a=b=c\)
- Thay a = b = c vào biểu thức N ta được :
\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)