Cho \(ab+bc+ca=2020\)
CMR \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}=0\)
K
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2 tháng 5 2020
Ta có: \(2020+c^2=ab+bc+ca+c^2=\left(b+c\right)\left(c+a\right)\)
Tương tự => \(2020+a^2=\left(a+b\right)\left(c+a\right)\)
và \(2020+b^2=\left(a+b\right)\left(b+c\right)\)
=> PT = \(\frac{a-b}{\left(b+c\right)\left(c+a\right)}+\frac{b-c}{\left(a+b\right)\left(c+a\right)}+\frac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\frac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) = \(\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) = 0
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 8 2020
\(\left(a+b+c\right)^2=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{a^{2020}+1}{a^{2020}+a^{2020}+a^{2020}+3}=\frac{a^{2020}+1}{3\left(a^{2020}+1\right)}=\frac{1}{3}\)
23 tháng 8 2020
Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\) (áp dụng Bất Đẳng Thức Cosi)
\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)
\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)
Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)
Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)
Dấu "=" xảy ra khi a=b=c
PA
2 tháng 8 2017
\(M=\sqrt{\frac{\left(a^2+2020\right)\left(b^2+2020\right)}{c^2+2020}}\)
\(=\sqrt{\frac{\left(a^2+ab+bc+ac\right)\left(b^2+ab+bc+ac\right)}{c^2+ab+bc+ac}}\)
\(=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)}{\left(c+a\right)\left(c+b\right)}}\)
\(=a+b\) là 1 số hữu tỉ
=> M là 1 số hữu tỉ (đpcm)
LT
Cho a,b,c thõa mãn : a^2 + b^2 +c^2 - ab -bc- ca = 0. Tính: P = (a-b)^2020 + (b-c)^2021 + (c-a)^2022
1
TG
21 tháng 4 2021
\(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Nên PT (1) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)
=> a = b = c
\(P=\left(a-b\right)^{2020}+\left(b-c\right)^{2021}+\left(c-a\right)^{2022}\)
\(=\left(a-a\right)^{2020}+\left(b-b\right)^{2021}+\left(c-c\right)^{2022}\)
= 0
AH
Akai Haruma
Giáo viên
20 tháng 10 2020
Thông thường sẽ tính ra giá trị $T$ cụ thể nhưng bài này thì với $a,b,c$ khác nhau thì giá trị $T$ cũng khác nhau.
Bạn xem lại đề xem có gõ nhầm chỗ nào không?
TG
0
Vì \(ab+bc+ca=2020\)
\(\Rightarrow a^2+2020=a^2+ab+bc+ca\)
\(=\left(a^2+ab\right)+\left(bc+ca\right)=a\left(a+b\right)+c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+c\right)\)
Tương tự ta có: \(b^2+2020=\left(b+a\right)\left(b+c\right)\)
\(c^2+2020=\left(c+b\right)\left(c+a\right)\)
\(\Rightarrow\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)
\(=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(b+a\right)\left(b+c\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(=\frac{\left(a^2-bc\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(b^2-ca\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(c^2-ab\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a^2-bc\right)\left(b+c\right)+\left(b^2-ca\right)\left(c+a\right)+\left(c^2-ab\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a^2b+a^2c-b^2c-bc^2\right)+\left(b^2c+b^2a-c^2a-ca^2\right)+\left(c^2a+c^2b-a^2b-ab^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{a^2b+a^2c-b^2c-bc^2+b^2c+b^2a-c^2a-ca^2+c^2a+c^2b-a^2b-ab^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)( đpcm )
Ta có
\(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)
\(b^2+ab+bc+ac=\left(a+b\right)\left(b+c\right)\)
\(c^2+ab+bc+ac=\left(a+c\right)\left(b+c\right)\)
Thay ab + bc + ac = 2020 vào biểu thức \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)có
\(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)
\(=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(b+c\right)\left(a+c\right)}\)
\(=\frac{\left(a^2-bc\right)\left(b+c\right)+\left(b^2-ca\right)\left(a+c\right)+\left(c^2-ab\right)\left(b+a\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\frac{a^2b+a^2c-b^2c-bc^2+ab^2+b^2c-a^2c-ac^2+ac^2-a^2b+bc^2-ab^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=0\)