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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
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
từ giả thiết 1 suy ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
lại có 1 + a2 \(\ge\)2a nên \(\frac{1}{1+a^2}\le\frac{1}{2a}\)
do đó \(\frac{3}{2}=\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\le\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
dấu bằng xảy ra khi a = b = c = 1.
vậy S = a + b + c = 3.
\(a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(a^3-abc+b^3-abc+c^3-abc=0\)
\(a^3+b^3+c^3-3abc=0\)
\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca\right)-3ab\left(a+b+c\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca-3ab\right)=0\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-bc-ca-ab\right)=0\)
Mà \(a+b+c\ne0\)
\(\Rightarrow a^2+b^2+c^2-bc-ca-ab=0\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mình làm hơi tắt.
Đến đây bạn tự làm nốt nhé~
\(1.\) Đang duyệt
\(2a.\)
Ta có:
\(P-Q=\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+a^2}-\frac{b^3}{a^2+ab+b^2}-\frac{c^3}{b^2+bc+c^2}-\frac{a^3}{c^2+ac+a^2}\)
\(\Leftrightarrow\) \(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)
\(\Leftrightarrow\) \(P-Q=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}+\frac{\left(b-c\right)\left(b^2+bc+c^2\right)}{b^2+bc+c^2}+\frac{\left(c-a\right)\left(c^2+ac+a^2\right)}{c^2+ac+a^2}\)
\(\Leftrightarrow\) \(P-Q=a-b+b-c+c-a\) (do \(a,b,c\ne0\) )
\(\Leftrightarrow\) \(P-Q=0\)
Vậy, \(P=Q\) \(\left(đpcm\right)\)
\(1.\)
Theo đề bài, ta có:
\(a^3=b^2+b+\frac{1}{3}\) \(\left(1\right)\)
\(b^3=c^3+c^2+\frac{1}{3}\) \(\left(2\right)\)
\(c^3=a^3+a^2+\frac{1}{3}\) \(\left(3\right)\)
Vì \(b^2+b+\frac{1}{3}=\left(b+\frac{1}{2}\right)^2+\frac{1}{12}\ge\frac{1}{12}>0\) nên từ \(\left(1\right)\) \(\Rightarrow\) \(a^3>0\) , tức là \(a>0\)
Tương tự, \(b,c>0\)
Do vai trò hoán vị của các ẩn \(a,b,c\) là như nhau nên có thể giả sử \(a=max\left\{a,b,c\right\}\) hay \(a\ge b\) \(;\) \(a\ge c\)
Do đó,
\(\text{+) }\) Từ \(\left(1\right)\) \(;\) \(\left(3\right)\) , ta có:
\(a^3=b^2+b+\frac{1}{3}\le a^2+a+\frac{1}{3}=c^3\)
Theo đó, \(a^3\le c^3\) hay \(a\le c\)
Mà \(a\ge c\) \(\left(cmt\right)\)
\(\Rightarrow\) \(a=c\) \(\left(\text{*}\right)\)
Lại có:
\(\text{+) }\) Từ \(\left(2\right)\) \(;\) \(\left(3\right)\) , ta có:
\(b^3=c^2+c+\frac{1}{3}=a^2+a+\frac{1}{3}=c^3\) (do \(a=c\) )
nên \(b^3=c^3\) , tức là \(b=c\) \(\left(\text{**}\right)\)
Vậy, từ \(\left(\text{*}\right)\) và \(\left(\text{**}\right)\) , suy ra \(a=b=c\)
Ta có: \(a^2+ab+b^2\)
\(=\left(a+b\right)^2-ab\ge\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3\left(a+b\right)^2}{4}\)
\(\Rightarrow\sqrt{a^2+ab+b^2}\ge\sqrt{\frac{3\left(a+b\right)^2}{4}}=\frac{\sqrt{3}}{2}\left(a+b\right)\)
Tương tự, ta có: \(\sqrt{b^2+bc+c^2}\ge\frac{\sqrt{3}}{2}\left(b+c\right)\)
\(\sqrt{c^2+ca+a^2}\ge\frac{\sqrt{3}}{2}\left(c+a\right)\)
Do đó ta có: \(Q\ge\frac{\sqrt{3}}{2}\left(a+b+b+c+c+a\right)=\sqrt{3}\) ( Do a+b+c=1)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Đặt \(A=\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ca}=\frac{a^2+2ab+b^2}{ab}+\frac{b^2+2bc+c^2}{bc}+\frac{c^2+2ac+c^2}{ca}\)
\(=\frac{a}{b}+2+\frac{b}{a}+\frac{b}{c}+2+\frac{c}{b}+\frac{c}{a}+2+\frac{a}{c}=6+a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{b}+\frac{1}{a}\right)\)
\(\ge6+\frac{4a}{b+c}+\frac{4b}{c+a}+\frac{4c}{a+b}\ge6+2\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+b}\right)+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
\(\ge6+2\cdot\frac{3}{2}+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=9+2\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
Dấu "=" xảy ra <=> a=b=c
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)
\(3,0122015^2=a^2+b^2+c^2+2\left(a^2+b^2+c^2\right)\)
\(3\left(a^2+b^2+c^2\right)=9,073357877\)
\(a^2+b^2+c^2=3,024452626\)