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Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ac=0\)
Đặt \(F=a^2+b^2+c^2\)
Từ \(a+b+c=1\Rightarrow\left(a+b+c\right)^2=1\)
\(\Rightarrow F+2\left(ab+bc+ac\right)=1\)
\(\Rightarrow F+2\cdot0=1\Rightarrow F=1\)
18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)
Ta có ;
\(\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=\frac{\left(a+b\right)\left(a-b\right)+\left(b+c\right)\left(b-c\right)+\left(c+a\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
1)
xét a+b+c = (a+b+c)(\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)) = \(\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=\)
\(\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=Q+a+b+c\)
<=> a+b+c =Q + a+b+c => Q=0
2) = (x+ y)2 + (x+ 1)2 +y(x+ 1) +x + y + 1 =0 <=> (x+ y)(x+ y+ 1) + (x+ 1)(x+ y+ 1) + 1= 0 <=> (x+ y+ 1)(2x+ y+ 1) = -1
=> \(\hept{\begin{cases}x+y+1=1\\2x+y+1=-1\end{cases}}\)hoặc \(\hept{\begin{cases}x+y+1=-1\\2x+y+1=1\end{cases}}\)<=> \(\hept{\begin{cases}x=-2\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=2\\y=-4\end{cases}}\)
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
\(\Rightarrow\sqrt{\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1}=1\)
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Leftrightarrow\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Có :
Q = a.(a/b+c) + b.(b/c+a) + c.(c/a+b)
= a.(a/b+c + 1) + b.(b/c+a + 1) + c.(c/a+b + 1) - (a+b+c)
= a.(a+b+c)/b+c + b.(a+b+c)/c+a + c.(a+b+c)/a+b - (a+b+c)
= (a+b+c).(a/b+c + b/c+a + c/a+b) - (a+b+c)
= (a+b+c)-(a+b+c) = 0
Vậy Q = 0
Tk mk nha
Hạ sách : Nhân hết ra :)))
Ta có :
\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(=x^2+\frac{1}{x^2}+2+y^2+\frac{1}{y^2}+2+x^2y^2+\frac{1}{x^2y^2}+2-\left(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}\right)\left(xy+\frac{1}{xy}\right)\)
\(=x^2+y^2+\frac{1}{x^2y^2}+x^2y^2+\frac{1}{x^2}+\frac{1}{y^2}+6-\left(x^2y^2+1+x^2+\frac{1}{y^2}+y^2+\frac{1}{x^2}+1+\frac{1}{x^2y^2}\right)\)
\(=6-1-1\)
\(=4\)
\(\frac{3}{2}=a\sqrt{1-b^2}+b\sqrt{1-c^2}+c\sqrt{1-a^2}\)
\(\le\frac{a^2+1-b^2}{2}+\frac{b^2+1-c^2}{2}+\frac{c^2+1-a^2}{2}=\frac{3}{2}\)
=> \(\frac{3}{2}\le\frac{3}{2}\)( chỉ xảy ra dấu "=" )
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a^2=1-b^2\\b^2=1-c^2\\c^2=1-a^2\end{cases}}\)=> \(a^2+b^2+c^2=3-\left(a^2+b^2+c^2\right)\)
=> \(B=a^2+b^2+c^2=\frac{3}{2}\)
\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3.\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{3}{abc}\)
\(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
trước hết ta chứng minh: nếu x+y+z=0 thì x3+y3+z3=3xyz
thật vậy, vì x+y+z=0 => z=-(x+y)
=> z3=-[x3+y3+3xy(x+y)]
=> x3+y3+z3=-3xy(x+y)=-3xy(-z)
=> x3+y3+z3=3xyz
áp dụng vào bài đã cho, ta suy ra: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{abc}\)
do đó \(P=\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}=\frac{abc}{c^3}+\frac{bca}{a^3}+\frac{cab}{b^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
vậy P=3
Có : a+b+c=0
<=>a=-(b+c)
<=>a^2=b^2+2bc+c^2
<=>b^2+c^2-a^2=2bc
Tương tự : c^2+a^2-b^2=2ca
a^2+b^2-c^2=2ab
Khi đó : P = 1/2bc + 1/2ca + 1/2ab = a+b+c/2abc = 0
Vậy P = 0
k mk nha