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Cho a,b,c thỏa mãn a+b+c = 0 và ab+bc+ca =0
Tính giá trị của biểu thức A=(a-1)^2+b^2+c(c+1)
a, Xét : 196 = 14^2 = (a^2+b^2+c^2) = a^4+b^4+c^4+2.(a^2b^2+b^2c^2+c^2a^2)
<=> a^4+b^4+c^4 = 196 - 2.(a^2b^2+b^2c^2+c^2a^2)
Xét : 0 = (a+b+c)^2 = a^2+b^2+c^2+2.(ab+bc+ca)
Mà a^2+b^2+c^2 = 14
<=> 2.(ab+bc+ca) = -14
<=> ab+bc+ca = -7
<=> a^2b^2+b^2c^2+c^2a^2+2abc.(a+b+c) = 49
Lại có : a+b+c = 0
<=> a^2b^2+b^2c^2+c^2a^2 = 49
<=> A = a^4+b^4+c^4 = 196 - 2.49 = 98
Tk mk nha
b) \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\)\(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow\)\(x^2=y^2=z^2=0\)
\(\Leftrightarrow\)\(x=y=z=0\)
Vậy \(D=0\)
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+a+b+c=2+2018\)
\(\Leftrightarrow\frac{a+ab+bc}{b+c}+\frac{b+bc+ab}{c+a}+\frac{c+ac+bc}{a+b}=2020\)
\(\Leftrightarrow a\left(\frac{1+b+c}{b+c}\right)+b\left(\frac{1+a+c}{a+c}\right)+c\left(\frac{1+a+b}{a+b}\right)=2020\left(1\right)\)
Vì \(a+b+c=2018\Rightarrow\hept{\begin{cases}a+b=2018-c\\b+c=2018-a\\c+a=2018-b\end{cases}\left(2\right)}\)
Thay (2) vào (1) ta được:
\(a\left(\frac{2019-a}{b+c}\right)+b\left(\frac{2019-b}{a+c}\right)+c\left(\frac{2019-c}{a+b}\right)=2020\)
\(\Leftrightarrow\frac{2019a-a^2}{b+c}+\frac{2019b-b^2}{a+c}+\frac{2019c-c^2}{a+b}=2020\)
\(\Leftrightarrow\frac{2019a}{b+c}-\frac{a^2}{b+c}+\frac{2019b}{a+c}-\frac{b^2}{a+c}+\frac{2019c}{a+b}-\frac{c^2}{a+b}=2020\)
\(\Leftrightarrow2019\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)
\(\Leftrightarrow4038-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)( vì \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\))
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=2018\)
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1=2019\)
Ta co: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]\\ \)
{Có thể c/m bằng cách ghép--> không thuộc 7 HDT , tuy nhiên cũng nên nhớ }
\(B=\dfrac{\left(a+b+c\right)\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]}{\left[\left(a^2+b^2+c^2\right)-ab-ac-bc\right]}=\left(a+b+c\right)=2016\)
Từ: 1/a + 1/b + 1/c = 0
=> 1/a^3 + 1/b^3 + 1/c^3 = 3. 1/abc
Ta có:
ab/c^2 + bc/a^2 + ac/b^2
= abc/c^3 + abc/a^3 + abc/b^3
= abc. (1/a^3 + 1/b^3 + 1/c^3)
= abc. 3. 1/abc = 3
Cho mình sửa đề một chút nha 
\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)(*)
Theo bài ra , ta có :
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^2=c^2\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\) (1)
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow b+c=-a\)
\(\Leftrightarrow\left(b+c\right)^2=a^2\)
\(\Leftrightarrow b^2+2bc+c^2=a^2\)
\(\Leftrightarrow b^2+c^2-a^2=-2bc\) (2)
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow a+c=-b\)
\(\Leftrightarrow\left(a+c\right)^2=b^2\)
\(\Leftrightarrow a^2+2ac+c^2=b^2\)
\(\Leftrightarrow a^2+c^2-b^2=-2ac\) (3)
Thay (1) , (2) , (3) vào (*) ta được
\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
\(=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}+-\dfrac{1}{2}+-\dfrac{1}{2}=-\dfrac{3}{2}\)
Vậy \(A=-\dfrac{3}{2}\)
Chúc bạn học tốt =))
\(2T=\frac{a^2-2ac+c^2+c^2-2bc+b^2+a^2-2ab+b^2}{\left(a-c\right)\left(a+c\right)-2b\left(a-c\right)}\)
\(2T=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-c\right)\left(a-b+c-b\right)}\)
Theo đề bài ta có:\(\hept{\begin{cases}a-b=4\\b-c=2\end{cases}\Rightarrow}a-c=6\)
\(\Rightarrow2T=\frac{4^2+2^2+6^2}{6\cdot\left(4-2\right)}=\frac{14}{3}\)