Cho \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1.\)Tính giá trị biểu thức \(Q=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\).
Cho ba số a,b,c khác 0 thỏa mãn: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\) .
Tính giá trị của biểu thức \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
Cho các số a, b, c thỏa mãn điều kiện\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
Tính giá trị biểu thức:\(P=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
Các cao nhân giúp với!!!!!!!!!! Thanks for all
Ta có:\(a+b+c\ne0\)vì nếu \(a+b+c=0\)thế vào giả thiết ta có:
\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\Leftrightarrow-3=1\)(vô lí)
Khi \(a+b+c\ne0\)ta có:
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).\left(a+b+c\right)=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{a.\left(b+c\right)}{b+c}+\frac{b.\left(c+a\right)}{c+a}+\frac{b^2}{c+a}+\frac{c.\left(a+b\right)}{a+b}+\frac{c^2}{a+b}=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)\(\Rightarrow P=0\)
Học tốt
\(P=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(< =>P=a\left(\frac{a}{b+c}+1-1\right)+b\left(\frac{b}{a+c}+1-1\right)+c\left(\frac{c}{a+b}+1-1\right)\)
\(< =>P=a\left(\frac{a+b+c}{b+c}-1\right)+b
\left(\frac{a+b+c}{a+c}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)\)
\(< =>P=\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{a+c}+\frac{c\left(a+b+c\right)}{a+b}-\left(a+b+c\right)\)
\(< = >P=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
\(< =>P=a+b+c-a-b-c=0\)
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 ba số a,b,c thỏa mãn a+b+c =6 và \(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\).Tính giá trị của biểu thức \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c=a}\)
Ta có :
\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}=\frac{3}{2}\)
\(\Leftrightarrow\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}+1=\frac{3}{2}+3\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)
\(\Leftrightarrow6.\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{9}{2}\)
\(\Rightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{9}{2}:6=\frac{3}{4}\)
Vậy \(P=\frac{3}{4}\)
Cho \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\).Tính giá trị của biểu thức \(A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\) bằng bao nhiêu ?
Ta có : \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
=> \(\frac{a^2+ab+ac}{b+c}+\frac{b^2+ab+bc}{a+c}+\frac{c^2+ac+bc}{a+b}=a+b+c\)
=> \(\frac{a^2}{b+c}+\frac{ab+ac}{b+c}+\frac{b^2}{a+c}+\frac{ab+bc}{a+c}+\frac{c^2}{a+b}+\frac{ac+bc}{a+b}=a+b+c\)
=> \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}+a+b+c-a-b-c=0\)
=> \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)
Cấm ai được chép
Naruto lục đạo Bạn cần xét hai trường hợp :
\(a+b+c=0\)và \(a+b+c\ne0\)nhé!
Cho a,b,c khác 0\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\), Tính giá trị biểu thức A= \(\frac{a^2+bc}{a^2+2bc}+\frac{b^2+ca}{b^2+2ca}+\frac{c^2+ab}{c^2+2ab}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\\ \)
\(\Rightarrow bc=-ab-ac,ca=-ab-bc,ab=-bc-ca\)
\(\Rightarrow\frac{a^2+bc}{a^2+2bc}=\frac{a^2+bc}{a^2+bc+bc}=\frac{a^2+bc}{a^2+bc-ca-ab}=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}\)
Làm tương tự. có: \(\frac{b^2+ca}{b^2+2ca}=\frac{b^2+ca}{b^2+ca-ab-bc}=\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}\)
\(\frac{c^2+ab}{c^2+2ab}=\frac{c^2+ab}{c^2+ab-ca-bc}=\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(\Rightarrow A=\frac{a^2+bc}{\left(a-b\right).\left(a-c\right)}+\frac{b^2+ca}{\left(a-b\right).\left(c-b\right)}+\frac{c^2+ab}{\left(b-c\right).\left(a-c\right)}\)
\(=\frac{\left(a^2+bc\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{\left(b^2+ca\right).\left(a-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}+\frac{\left(c^2+ab\right).\left(a-b\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
Sau đó bạn thực hiện tiếp nhé.
Bài 1: Cho \(a,b,c\ge0:a^2+b^2+c^2=3\). CMR: \(a^4b^4+b^4c^4+c^4a^4\le3\)
Bài 2: Cho \(a,b,c\ge0\). CMR: \(a^2+b^2+c^2+2abc+1\ge2\left(ab+bc+ca\right)\)
Bài 3: Cho \(a,b,c\ge0:a^2+b^2+c^2=a+b+c\). CMR: \(a^2b^2+b^2c^2+c^2a^2\le ab+bc+ca\)
Bài 4: Cho \(a,b,c\ge0\). CMR: \(4\left(a+b+c\right)^3\ge27\left(ab^2+bc^2+ca^2+abc\right)\)
Bài 5: Cho \(a,b,c\ge0:a+b+c=3\).CMR: \(\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}+\frac{1}{2ab^2+1}\ge1\)
Cho a,b,c thỏa
\(\text{}\frac{a}{b-c}+\frac{b}{c-a}_+\frac{c}{a-b}=0\)Tính giá trị biểu thức
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Đặt \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=A\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
<=> \(\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
<=> \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(b-c\right)\left(c-a\right)}+\frac{c}{\left(b-c\right)\left(a-b\right)}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{b}{\left(a-b\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)^2}=0\)
<=> \(A+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(a-b\right)\left(c-a\right)}=0\)
<=> \(A+\frac{\left(a+b\right)\left(a-b\right)+\left(c-a\right)\left(c+a\right)+\left(c+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A+\frac{a^2-b^2+c^2-a^2+b^2-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A=0\)
=> ....
17 Cho a,b khác 0 thỏa mãn a+b=1.CM:
\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)
18 Cho\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
Tính giá trị biểu thức A=\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{z^2}\)
19Cho a,b,c đôi một khác nhau và \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
Tính giá trị P=\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-c\right)^2}\)
18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)
Ta có ;
\(\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=\frac{\left(a+b\right)\left(a-b\right)+\left(b+c\right)\left(b-c\right)+\left(c+a\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
17. Xét vế trái ;
\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{a}{\left(b-1\right)\left(b^2+b+1\right)}+\frac{b}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{a}{-a\left(b^2+b+1\right)}+\frac{b}{-b\left(a^2+a+1\right)}=\frac{-1}{b^2+b+1}+\frac{-1}{a^2+a+1}\)
\(=\frac{-\left(a^2+a+1+b^2+b+1\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}=\frac{-\left[\left(a+b\right)^2-2ab+3\right]}{a^2b^2+ab\left(a+b\right)+a^2+b^2+ab+2}\)\(=\frac{2\left(ab-2\right)}{a^2b^2+\left(a^2+2ab+b^2\right)+2}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)
Cho \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\left(1\right)\)và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
Tính giá trị biểu thức \(A=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=0\)
b, Tính \(\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)