\(\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}\)
Thực hiện phép tính:
1) \(A=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
2) \(B=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
3, \(C=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ac}{\left(b-a\right)\left(b-c\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
4) \(D=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
chứng minh
\(2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3=\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)
VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)
\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)
\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)
\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)
\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)
\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)
\(VT=\dfrac{2a^3-a^2b-a^2c-ab^2-ac^2+2b^3-b^2c-bc^2+2c^3}{(a+b)(b+c)(c+a)} \)
\(\\=\dfrac{a^3+a^2b-2a^2b-2ab^2+ab^2+b^3+b^3+b^2c-2b^2c-2bc^2+bc^2+c^3+c^3+c^2a-2c^a+2ca^2-ca^2+a^3}{(a+b)(b+c)(c+a)}\)
\(\\=\dfrac{(a-b)^2(a+b)+(b-c)^2(b+c)+(c-a)^2(c+a)}{(a+b)(b+c)(c+a)}\)
\(\\\Rightarrow VT=\dfrac{(a-b)^2}{(c+a)(b+c)}+\dfrac{(b-c)^2}{(c+a)(a+b)}+\dfrac{(c-a)^2}{(a+b)(b+c)}=VP\)từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Cho 3 số a,b,c đôi 1 khác nhau. CMR:
\(\dfrac{b-c}{\left(a-b\right).\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right).\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right).\left(c-b\right)}=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}\)
`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`
`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`
`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`
`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`
`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`
Biến đổi tương đương thôi em, dễ mà =)
\(1,Cho.a,b,c\ge1.CMR:\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\ge\left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\)
2, Cho a,b,c>0.CMR:
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ac+b^2}+\dfrac{c+a}{ab+c^2}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
1.
BĐT cần chứng minh tương đương:
\(\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\)
Ta có:
\(\left(ab-1\right)^2=a^2b^2-2ab+1=a^2b^2-a^2-b^2+1+a^2+b^2-2ab\)
\(=\left(a^2-1\right)\left(b^2-1\right)+\left(a-b\right)^2\ge\left(a^2-1\right)\left(b^2-1\right)\)
Tương tự: \(\left(bc-1\right)^2\ge\left(b^2-1\right)\left(c^2-1\right)\)
\(\left(ca-1\right)^2\ge\left(c^2-1\right)\left(a^2-1\right)\)
Do \(a;b;c\ge1\) nên 2 vế của các BĐT trên đều không âm, nhân vế với vế:
\(\left[\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\right]^2\ge\left[\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\right]^2\)
\(\Rightarrow\left(ab-1\right)\left(bc-1\right)\left(ca-1\right)\ge\left(a^2-1\right)\left(b^2-1\right)\left(c^2-1\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Câu 2 em kiểm tra lại đề có chính xác chưa
2.
Câu 2 đề thế này cũng làm được nhưng khá xấu, mình nghĩ là không thể chứng minh bằng Cauchy-Schwaz được, phải chứng minh bằng SOS
Không mất tính tổng quát, giả sử \(c=max\left\{a;b;c\right\}\)
\(\Rightarrow\left(c-a\right)\left(c-b\right)\ge0\) (1)
BĐT cần chứng minh tương đương:
\(\dfrac{1}{a}-\dfrac{a+b}{bc+a^2}+\dfrac{1}{b}-\dfrac{b+c}{ac+b^2}+\dfrac{1}{c}-\dfrac{c+a}{ab+c^2}\ge0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)+a\left(c-b\right)}{a^3+abc}+\dfrac{c\left(a-b\right)}{b^3+abc}+\dfrac{a\left(b-c\right)}{c^3+abc}\ge0\)
\(\Leftrightarrow c\left(b-a\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{b^3+abc}\right)+a\left(c-b\right)\left(\dfrac{1}{a^3+abc}-\dfrac{1}{c^3+abc}\right)\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)\left(b^3-a^3\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c^3-a^3\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)
\(\Leftrightarrow\dfrac{c\left(b-a\right)^2\left(a^2+ab+b^2\right)}{\left(a^3+abc\right)\left(b^3+abc\right)}+\dfrac{a\left(c-b\right)\left(c-a\right)\left(a^2+ac+c^2\right)}{\left(a^3+abc\right)\left(c^3+abc\right)}\ge0\)
Đúng theo (1)
Dấu "=" xảy ra khi \(a=b=c\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Tính:
\(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
Lời giải:
\(-A=\frac{a^2}{(a-b)(c-a)}+\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(c-a)(b-c)}\)
\(=\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}=\frac{a^2b+b^2c+c^2a-(ab^2+bc^2+ca^2)}{-[(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)]}=-1\)
$\Rightarrow A=1$
Cho \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
Chứng minh rằng: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)
Từ \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
\(=>\dfrac{a}{b-c}+1+\dfrac{b}{c-a}+1+\dfrac{c}{a-b}+1=3\)
\(=>\dfrac{a}{b-c}-\dfrac{b}{a-c}-\dfrac{c}{b-a}=0\)
\(=>\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
Nhân cả 2 vế với \(\dfrac{1}{b-c}\) ta được
\(\dfrac{a}{\left(b-c\right)^2}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\)
Tương tự ta có:
\(\dfrac{b}{\left(c-a\right)^2}=\dfrac{c^2-bc+bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right)\)
\(\dfrac{c}{\left(a-b\right)^2}=\dfrac{a^2-ca+cb-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng theo vế (1);(2);(3) ta có ĐPCM
CHÚC BẠN HỌC TỐT.........
Giá trị của biểu thức: A=\(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
Ta có: \(A=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\dfrac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a^2b-a^2c-ab^2+b^2c+ac^2-bc^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab+c^2\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab+c^2-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{c^2+ab-c}{\left(a-c\right)\left(b-c\right)}\)