Rút gọn biểu thức sau:
\(\frac{a^3}{ab-ac}-\frac{b^3}{bc-ba}-\frac{c^3}{ca-cb}\)
(p/s: mình đang cần gấp nha!)
Tìm 3 số a;b;c khác 0 thoả mãn:
\(\frac{ab+ac}{2}\)=\(\frac{bc+ba}{3}\)=\(\frac{ca+cb}{4}\)và a+b+c=69.
Mn giúp mình nhanh với, cần gấp ạ T^T
Cho 3 số a,b,c, đôi một khác nhau và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\).Rút gọn các biểu thức sau
a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
b) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
GIÚP MIK VỚI MIK ĐANG CẦN GẤP!
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=> \(\frac{ab+bc+ac}{abc}=0\)
=> \(ab+bc+ac=0\)
=> \(\hept{\begin{cases}ab=-bc-ac\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
a) \(N=\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ac}+\frac{ab}{c^2+2ab}\)
\(=\frac{bc}{a^2-ab-ac+bc}+\frac{ca}{b^2-ab-bc+ac}+\frac{ab}{c^2-ac-bc+ab}\)
\(=\frac{bc}{a\left(a-b\right)-c\left(a-b\right)}+\frac{ca}{b\left(b-a\right)-c\left(b-a\right)}+\frac{ab}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{bc}{\left(a-b\right)\left(a-c\right)}-\frac{ca}{\left(a-b\right)\left(b-c\right)}+\frac{ab}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{ca^2-c^2a}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{b^2c-bc^2-ca^2+c^2a+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(c^2a-bc^2\right)-\left(ca^2-b^2c\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
b) \(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab-ac+bc}+\frac{b^2}{b^2-ab-bc+ac}+\frac{c^2}{c^2-bc-ac+ab}\)
\(=\frac{a^2}{a\left(a-b\right)-c\left(a-b\right)}+\frac{b^2}{b\left(b-a\right)-c\left(b-a\right)}+\frac{c^2}{c\left(c-b\right)-a\left(c-b\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{b^2a-b^2c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\frac{a^3b-ab^3+b^3c-cb^3+c^3a-ca^3}{a^2b-ab^2+b^2c-cb^2+c^2a-ca^2}\)
. Rút gọn phân thức trên. Mình giải mãi cũng không ra. Mấy bạn giúp mình nha. Cảm ơn mấy bạn nè <3
Nhận xét: \(b^3c-cb^3=0;b^2c-cb^2=0.\).Nên phân thức trở thành:
\(\frac{a^3b-ab^3+c^3a-ca^3}{a^2b-ab^2+c^2a-ca^2}=\frac{a^3\left(b-c\right)-a\left(b^3-c^3\right)}{a^2\left(b-c\right)-a\left(b^2-c^2\right)}\)
\(=\frac{a\left(b-c\right)\left\{a^2-\left(b^2-bc+c^2\right)\right\}}{a\left(b-c\right)\left\{a-\left(b+c\right)\right\}}\)
\(=\frac{a^2-\left(b^2-bc+c^2\right)}{a-\left(b+c\right)}=\frac{a^2-\left(b+c\right)^2+3bc}{a-\left(b+c\right)}\)
\(=a+b+c+\frac{3bc}{a-b-c}\).
Rút gọn biểu thức
\(\frac{\text{a^3+b^3+c^3-3abc}}{\text{a^2+b^2+c^2-ab-bc-ca}}\)
a^3+b^3+c^3-3abc
<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc
<=>[(a+b)^3 +c^3] -3ab.(a+b+c)
<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)
<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...
<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)
thay vào và rút gọn ta được:\(a+b+c\)
\(\frac{abc}{a^3+c^3+b^3}+\frac{2}{3}\ge\frac{\left(ab+bc+ca\right)}{a^2+b^2+c^2}\)
giúp mình nha,mình cần gấp,cảm ơn các bạn.
dùng bất đẳng thức chebyshev được không ạ?
Bạn ơi đề bài có điều kiện a, b, c không vậy. Hay là a, b, c bất kì?
Với a, b, c >0
\(\frac{abc}{a^3+b^3+c^3}+\frac{2}{3}\ge\frac{ab+bc+ac}{a^2+b^2+c^2}\) (1)
<=> \(1-\left(\frac{abc}{a^3+b^3+c^3}+\frac{2}{3}\right)\le1-\frac{ab+bc+ac}{a^2+b^2+c^2}\)
\(\Leftrightarrow\frac{1}{3}-\frac{abc}{a^3+b^3+c^3}\le\frac{a^2+b^2+c^2-ab-ac-bc}{a^2+b^2+c^2}\)
\(\Leftrightarrow\frac{a^3+b^3+c^3-3abc}{3\left(a^3+b^3+c^3\right)}\le\frac{a^2+b^2+c^2-ab-ac-bc}{a^2+b^2+c^2}\)
\(\Leftrightarrow\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{3\left(a^3+b^3+c^3\right)}\le\frac{a^2+b^2+c^2-ab-ac-bc}{a^2+b^2+c^2}\)
\(\Leftrightarrow\left(a^2+b^2+c^2-ab-ac-bc\right)\left(\frac{1}{a^2+b^2+c^2}-\frac{a+b+c}{3\left(a^3+b^3+c^3\right)}\right)\ge0\)(2)
Ta có: \(a^2+b^2+c^2-ab-ac-bc=\frac{1}{2}\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\ge0\)
Với a,b, c>0
(1) <=> \(\frac{1}{a^2+b^2+c^2}\ge\frac{a+b+c}{3\left(a^3+b^3+c^3\right)}\)
\(\Leftrightarrow3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2a^3+2b^3+2c^3-ab^2-ac^2-ba^2-bc^2-ca^2-cb^2\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+a^2\left(a-c\right)+b^2\left(b-a\right)+b^2\left(b-c\right)+c^2\left(c-a\right)+c^2\left(c-b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(b+c\right)\left(b-c\right)^2+\left(a+c\right)\left(a-c\right)^2\ge0\)Luôn đúng với mọi a, b, c dương
Vậy (1) đúng
"=" xảy ra <=> a=b=c
Rút gọn phân thức sau:
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
b) \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
Làm hộ mk 2 câu này nha! Mình cám mơn nhìu ạ!
Cho \(a;b;c\)khác không và\(a+b+c\)khác \(0\)thoả mãn \(a+b+c=\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}\). Tính giá trị của biểu thức \(P=\frac{2}{b+c}+\frac{3}{a+c}+\frac{4}{a+b}\)
a+b+c=\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}\) tim a,b,c khac 0
\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}\)thì \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\)