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Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
\(\left(a+b+c\right)=\dfrac{1}{2}\Leftrightarrow\left(a+b+c\right)^2=\dfrac{1}{4}\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=\dfrac{1}{4}\)
Ta có: \(ab+bc+ac=\left(a^2+b^2+c^2+2ab+2bc+2ac\right)-\left(a^2+b^2+c^2+ab+bc+ac\right)=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1}{12}\)
\(a^2+b^2+c^2=\dfrac{1}{6}-\left(ab+bc+ac\right)=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)
Suy ra: \(a^2+b^2+c^2=ab+bc+ac\Leftrightarrow a=b=c\)
\(P=\dfrac{3}{2}\)
p/s làm lih tih k chắc đâu:v
Cho mình sửa đề một chút thôi nha mình tin chắc là đề bạn sai rồi
Cho a,b,c thỏa mãn : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tính giá trị biểu thức N = \(\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)
Ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^3=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Ta lại có :
\(N=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
\(\Leftrightarrow N=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=abc\times\dfrac{3}{abc}=3\)
Chúc bạn học tốt =))
bc/a^2 + ac/b^2 + ab/c^2=abc(1/a^3 + 1/b^3 + 1/c^3)
mà 1/a + 1/b + 1/c = 0
=> 1/a + 1/b=-1/c
=> 1/a^3+1/b^3 = (1/a+1/b)^3 - 3.1/a.1/b(1/a+1/b) = -1/c^3 + 3.1/(abc)
=> 1/a^3 + 1/b^3 + 1/c^3=3/(abc)
=> bc/a^2 + ac/b^2 + ab/c^2=3.
Bài 2:
Bài 1:
\(a^2+b^2+c^2=14\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=14\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=14\Rightarrow ab+bc+ac=-7\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=14^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=196-2.49=98\)
a/d vào công thức a^3+b^3+b^3=3abc( khi a+b+c=0)
ta đc 1/a+1/b+1/c=0
=> (1/a)^3+(1/b)^3+(1/c)^3=3. (1/abc)
lại có S=\(\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
=abc (\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\))
=3.\(\dfrac{abc}{abc}\)=1
chúc bạn học tốt ^ ^
Dễ CM : nếu x+y+z=0 thì x^3+y^3+z^3=3xyz
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
\(S=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\\ =abc.\dfrac{1}{abc}=1\)
a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)
b)Ta có: a+b+c=0
\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)
\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)
\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)
\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)
\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)
cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc
a^3+b^3+c^3=3abc
=>a^3+b^3+c^3-3abc=0
=>(a+b)^3-3ab(a+b)+c^3-3abc=0
=>[(a+b)^3+c^3]-3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0
=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0
vì a+b+c=0 nên a^3+b^3+c^3=3abc
thay kết quả vừa chúng minh vào đề bài ta đc
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
chúc bạn học tốt ^ ^
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Ta có: \(P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}\)
=> \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=abc\left(\dfrac{1+1+1}{a^3+b^3+c^3}\right)\)
Mà ta lại có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\dfrac{\Rightarrow1}{a^3}+\dfrac{1}{b^3}=\left(\dfrac{1+1}{a+b}\right)^3-3\dfrac{1}{a}.\dfrac{1}{b}\left(\dfrac{1+1}{a+b}\right)\)
\(\Rightarrow-\dfrac{1^3}{c}+\dfrac{3}{abc}\)
\(\Rightarrow\dfrac{1+1+1}{a^3+b^3+c^3}=\dfrac{3}{abc}\)
\(\dfrac{\Rightarrow bc}{c^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=3\)
Vậy : \(P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}\) = 3