Cho a, b, c \(\ne\)0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\). Tính : \(E=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-a^2c^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}.\)
\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Cho a,b,c khác 0 và thỏa mãn: \(\frac{2ab+1}{2b}=\frac{2bc+1}{c}=\frac{ac+1}{a}\). CMR: a=2b=c hoặc \(4a^2b^2c^2=1\)
Chứng minh rằng nếu a , b , c > 0 thỏa mãn abc = ab + bc + ca thì \(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}<\frac{3}{16}\left(\le\frac{3}{32}\right)\)
Cho a,b,c>0 thỏa mãn \(a+b+c\le3\)
Chứng minh \(\frac{1}{\left(2a+b\right)\left(2c+b\right)}+\frac{1}{\left(2b+c\right)\left(2a+c\right)}+\frac{1}{\left(2c+a\right)\left(2b+a\right)}\ge\frac{3}{\left(a+b+c\right)^2}\)
moi nguoi oi hom truoc minh hoc tap hop cac so TN do thi co cua minh day nhu sau
vd: A={xeN/3<x<9}
thi minh liet ke ra la A=4,5,6,7,8 nhung sua bai lai ko dung
co sua nhu vay A=3,4,5,6,7,8
ko biet hay sai mong ae giup minh
Áp dụng BĐT Cô-si \(ab\le\frac{\left(a+b\right)}{4}^2\)
=> \(\left(2a+b\right)\left(2c+b\right)\le\frac{4\left(a+b+c\right)^2}{4}=\left(a+b+c\right)^2\)
=> \(\frac{1}{\left(2a+b\right)\left(2c+b\right)}\ge\frac{1}{\left(a+b+c\right)^2}\)
Mấy cái kia làm tương tự cậu nhé
Dấu "=" xảy ra khi và chỉ khi a=b=c=1
Chứng minh rằng nếu a, b, c là số dương thỏa mãn a+c=2b thì ta luôc có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)
Cho các số thực dương $a,b,c$ thỏa mãn $a+b+c=1$. Chứng minh rằng $\frac{a}{2a+b^{2}}+\frac{b}{2b+c^{2}}+\frac{c}{2c+a^{2}}\leq \frac{1}{7}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )$
Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)
\(\ge4ab+2ac+a^2\)
\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)
\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)
\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)
CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )
Bài 1.Cho \(x+y+z=0\)
Tính \(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Bài 2. Cho \(a+b+c=1;a^2+b^2+c^2=1;\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
CMR: \(xy+yz+zx=0\)
Bài 3. Cho \(3x-y=2z\)
\(2x+y=7z\)
Tính \(S=\frac{x^2-2xy}{x^2+y^2}\)với \(x,y\ne0\)
Bài 4. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(E=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 5. Cho \(abc\ne0\)thỏa mãn: \(2ab+6bc+2ac=0\)
Tính \(A=\frac{\left(a+2b\right)\left(2b+3c\right)\left(3c+a\right)}{6abc}\)
Bài 6. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(Y=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-c^2a^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}\)
Bài 7. Cho \(\hept{\begin{cases}10a^2-3b^2+5ab=0\\9a^2-b^2\ne0\end{cases}}\)
Tính \(B=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
2. a + b + c = 1 \(\Rightarrow\)( a + b + c )2 = 1 \(\Rightarrow\)a2 + b2 + c2 + 2 ( ab + bc + ac ) = 1 \(\Rightarrow\)ab + bc + ac = 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)
\(\Rightarrow\)x = a ( x + y + z ) ; y = b ( x + y + z ) ; z = c ( x + y + z )
Ta có : xy + yz + xz = ab ( x + y + z )2 + bc ( x + y + z )2 + ac ( x + y + z )2 = ( x + y + z )2 ( ab + bc + ac ) = 0
3. sửa đề : 3x - y = 3z
Ta có : \(\hept{\begin{cases}3x-y=3z\\2x+y=7z\end{cases}\Rightarrow\hept{\begin{cases}\left(3x-y\right)+\left(2x+y\right)=3z+7z\\2x+y=7z\end{cases}\Rightarrow}\hept{\begin{cases}5x=10z\\y=7z-2x\end{cases}\Rightarrow}\hept{\begin{cases}x=2z\\y=3z\end{cases}}}\)
\(\Rightarrow\)\(S=\frac{x^2-2xy}{x^2+y^2}=\frac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8z^2}{13z^2}=\frac{-8}{13}\)
Chứng minh rằng nếu a,b,c là các số dương thỏa mãn a+c=2b thì ta luôn có:
\(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)
Chứng minh rằng với mọi a,b,c dương thỏa mãn \(a+b+c=3\) thì:
\(\frac{a^2b}{2a+b}+\frac{b^2c}{2b+c}+\frac{c^2a}{2c+a}\le\frac{3}{2}\)
có lẽ là AM-GM ngược dấu, bn thử đi nhé giờ mk bận rồi
chẳng biết có sai đề không nữa không dự được dấu bằng xảy ra
