Cho \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{b+a}\) tinh \(P=2020-\frac{a+b}{c}-\frac{a+c}{b}-\frac{b+c}{c}\)
Những câu hỏi liên quan
Cho a +b +c =2020 và \(\frac{1}{a+b_{ }}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{2019}{2020}\).Tính S =\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
1.Giải phương trình sau: [x-2015] + [2x-2016] x-2017
2. Cho ba số thực a,b,c khác nhau thỏa mãn: a+frac{2020}{b}b+frac{2020}{c}c+frac{2020}{a}. Chứng minh rằng a^2+b^2+c^22020^3
3. Cho a,b,c là số dương thỏa mãn a+b+c9. Chứng minh: frac{1}{a}+frac{1}{b}+frac{1}{c}ge1
4. Chứng minh bất đẳng thức sau vớ a,b,c là các số dương: left(a+b+cright)timesleft(frac{1}{a}+frac{1}{b}+frac{1}{c}right)ge9
5. Cho a 0, b 0, c 0. Chứng minh rằng: frac{bc}{a}+frac{ca}{b}+frac{ab}{c}ge a+b+c
Đọc tiếp
1.Giải phương trình sau: [x-2015] + [2x-2016]= x-2017
2. Cho ba số thực a,b,c khác nhau thỏa mãn: \(a+\frac{2020}{b}=b+\frac{2020}{c}=c+\frac{2020}{a}\). Chứng minh rằng \(a^2+b^2+c^2=2020^3\)
3. Cho a,b,c là số dương thỏa mãn a+b+c=9. Chứng minh: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)
4. Chứng minh bất đẳng thức sau vớ a,b,c là các số dương: \(\left(a+b+c\right)\times\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
5. Cho a >0, b >0, c >0. Chứng minh rằng: \(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)
Cho a,b,c>0. Tìm Max A
\(A=\sqrt[2020]{\frac{a}{a+b}}+\sqrt[2020]{\frac{b}{b+c}}+\sqrt[2020]{\frac{c}{c+a}}\)
Ta chứng minh bổ đề:
Với x,y,z dương thì:
\(8\left(x+y+z\right)\left(xy+yz+zx\right)\le9\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow x\left(y-z\right)^2+y\left(z-x\right)^2+z\left(x-y\right)^2\ge0\)(đúng)
Quay lại bài toán ta có:
\(A^{2020}=\left(\sqrt[2020]{\frac{a}{a+b}}+\sqrt[2020]{\frac{b}{b+c}}+\sqrt[2020]{\frac{c}{c+a}}\right)^{2020}\)
\(=\left(\sqrt[2020]{\frac{a\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}}+\sqrt[2020]{\frac{b\left(b+a\right)}{\left(b+c\right)\left(b+a\right)}}+\sqrt[2020]{\frac{c\left(c+b\right)}{\left(c+a\right)\left(c+b\right)}}\right)^{2020}\)
\(\le\left(1+1+1\right)^{2018}.2.\left(a+b+c\right).\left(\frac{a}{\left(a+b\right)\left(a+c\right)}+\frac{b}{\left(b+c\right)\left(b+a\right)}+\frac{c}{\left(c+a\right)\left(c+b\right)}\right)\)
\(=3^{2018}.\frac{4\left(a+b+c\right)\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\le3^{2018}.\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{3^{2020}}{2}\)
\(\Rightarrow A\le\frac{3}{\sqrt[2020]{2}}\)
Cho a+b+c=0. Tinh Q=\((\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b})(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a})\)
Ta có: a + b + c = 0 => a + b = -c; b + c = -a; a + c = -b
a + b + c = 0 <=> a + b = -c
<=> (a + b)3 = (-c)3
<=> a3 + 3a2b + 3ab2 + b3 = -c3
<=> a3 + b3 + c3 = -3ab(a + b)
<=> a3 + b3 + c3 = 3abc (vì a + b = -c)
Khi đó: Q = \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Q = \(1+\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{c\left(b-c\right)}{a\left(a-b\right)}+1+\frac{b\left(b-c\right)}{a\left(c-a\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}+1\)
Q = \(3+\left(\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}\right)+\left(\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{b\left(b-c\right)}{a\left(c-a\right)}\right)+\left(\frac{c\left(b-c\right)}{a\left(a-b\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}\right)\)
Q = \(3+\frac{ab\left(a-b\right)+ac\left(c-a\right)}{bc\left(b-c\right)}+\frac{ab\left(a-b\right)+bc\left(b-c\right)}{ac\left(c-a\right)}+\frac{bc\left(b-c\right)+ca\left(c-a\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left(ab-b^2+c^2-ac\right)}{bc\left(b-c\right)}+\frac{b\left(a^2-ab+bc-c^2\right)}{ac\left(c-a\right)}+\frac{c\left(b^2-bc+ac-a^2\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]}{bc\left(b-c\right)}+\frac{b\left[b\left(c-a\right)-\left(c-a\right)\left(c+a\right)\right]}{ac\left(c-a\right)}+\frac{c\left[c\left(a-b\right)-\left(a-b\right)\left(a+b\right)\right]}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a-\left(b+c\right)\right]}{bc}+\frac{b\left(b-\left(c+a\right)\right)}{ac}+\frac{c\left[c-\left(a+b\right)\right]}{ab}\)
Q = \(3+\frac{a\left(a+a\right)}{bc}+\frac{b\left(b+b\right)}{ac}+\frac{c\left(c+c\right)}{ab}\)
Q = \(3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ab}\)
Q = \(3+\frac{2a^3+2b^3+2c^3}{abc}\)
Q = \(3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)
Q = \(3+\frac{2.3abc}{abc}=3+6=9\)
Bài làm:
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(\Rightarrow M=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)
Đặt \(N=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).N=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-b-c\\b=-c-a\\c=-a-b\end{cases}}\)
Thay vào ta được:
\(N=\frac{c\left(b-c\right)\left(c-a\right)-\left(b+c\right)\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(b-c-a+b\right)+b\left(a-b\right)\left(b-c-c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b-c-a\right)+b\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b+b\right)+b\left(a-b\right)\left(-c-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(c-a\right)-3bc\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(b+c-2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
Mà \(Q=M.N=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}.\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=9\)
Vậy Q = 9
cho a,b,c là các số thực dương thỏa mãn a+b+c=2020
tìm min của Q=\(\frac{a}{b+2020-a}+\frac{b}{c+2020-b}+\frac{c}{a+2020-c}\)
\(Q=\frac{a}{b+2020-a}+\frac{b}{c+2020-b}+\frac{c}{a+2020-c}\)
\(Q=\frac{a}{b+a+b+c-a}+\frac{b}{c+a+b+c-b}+\frac{c}{a+a+b+c-c}\)
\(Q=\frac{a}{2b+c}+\frac{b}{2c+a}+\frac{c}{2a+b}\)
Áp dụng BĐT Cauchy-Schwarz:
\(Q=\frac{a^2}{a\cdot\left(2b+c\right)}+\frac{b^2}{b\cdot\left(2c+a\right)}+\frac{c^2}{c\cdot\left(2a+b\right)}\ge\frac{\left(a+b+c\right)^2}{3\cdot\left(ab+bc+ca\right)}\ge\frac{3\cdot\left(ab+bc+ca\right)}{3\cdot\left(ab+bc+ca\right)}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{2020}{3}\)
2020a hay là 2020-a vậy???
Nếu đề như vậy thì thay 2020 vô các mẫu đc
\(\frac{a}{2b+a}=\frac{a^2}{2ab+a^2}\)
Tương tự sau đó cosi swat là ra nha
Cho \(\frac{a}{b}=\frac{c}{d}\)CMR
1) \(\frac{a^{2020}-b^{2020}}{a^{2020}+b^{2020}}=\frac{^{c^{2020}-d^{2020}}}{c^{2020}+d^{2020}}\)
Ko khó đâu bn ơi
Đặt a/b=c/d=k
=> a=bk và c=dk
Xong thay vào (a^2020-b^2020)/(a^2020+b^2020)=(b^2020.k^2020-b^2020)/(b^2020.k^2020+b^2020)
= (k^2020-1)/(k^2020+1)
Tiếp tục thay vào (c^2020-d^2020)/(c^2020+d^2020)=(d^2020.k^2020-d^2020)/(d^2020.k^2020+d^2020)
= (k^2020-1)/(k^2020+1)
=> đpcm.
26 tháng 12 2020 lúc 21:00
Cho ba số dương a, b, c thỏa mãn điều kiện \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2020\)
Hãy tính giá trị biểu thức \(P=\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right).\frac{1}{a+b+c}\)
Xét \(A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)
\(=a.\frac{a}{b+c}+b.\frac{b}{c+a}+c.\frac{c}{a+b}\)
\(=a.\left(\frac{a}{b+c}+1-1\right)+b.\left(\frac{b}{c+a}+1-1\right)+c.\left(\frac{c}{a+b}+1-1\right)\)
\(=a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{c+a}-b+c.\frac{a+b+c}{a+b}-c\)
\(=\left(a+b+c\right).\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
\(=\left(a+b+c\right).2020-\left(a+b+c\right)\)
\(\Rightarrow P=\frac{A}{a+b+c}=\frac{\left(a+b+c\right).2019}{a+b+c}=2019\)
Vậy...
Cho a,b,c thỏa mãn $\frac{a}{2018}$ =$\frac{b}{2019}$ =$\frac{c}{2020}$
CMR:(a-c)^3=8 $(a-b)^{2}$ (b-c)
Cho a,b,c>0 thỏa mãn ab+bc+ca=2020
Cmr:\(\frac{a-b}{2020+c^2}+\frac{b-c}{2020+a^2}+\frac{c-a}{2020+b^2}\)
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
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