Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: a2 + b2 = (a + b)2 - 2ab = 62 - 2.4 = 28
a4 + b4 = (a2 + b2)2 - 2a2b2 = 282 - 2.42 = 752
Giả sử \(c=max\left\{a,b,c\right\}\)
BĐT \(\Leftrightarrow a^4+b^4+c^4\ge\frac{a+b+c}{3}\left(a^3+b^3+c^3\right)\)
\(\Leftrightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(VT-VP=\frac{1}{8}\left[\left(b+c-2a\right)^2\left\{3a^2+\left(a+b+c\right)^2\right\}+3\left(5b^2+6bc+5c^2-2ab-2ac\right)\right]\ge0\)
Ta có :
\(a^2+b^2+\left(a+b\right)^2=a^2+b^2+\left(a^2+b^2+2ab\right)\)
\(=2\left(a^2+b^2+ab\right)=2.7=14\)
\(a^4+b^4+\left(a+b\right)^4=a^4+b^4+a^4+C_4^1a^3b+C_4^2a^2b^2+C_4^3ab^3+b^4\)
\(=2a^4+2b^4+4a^3b+6a^2b^2+4ab^3\)
\(=2\left(a^4+b^4+3a^2b^2+2ab^3+2a^3b\right)\)
\(=2\left[\left(a^2\right)^2+\left(b^2\right)^2+\left(ab\right)^2+2a^2b^2+2\left(ab\right)b^2+2\left(ab\right)a^2\right]\)
\(=2.\left(a^2+b^2+ab\right)^2=2.7^2=98\)
\(\Rightarrow M=\frac{a^2+b^2+\left(a+b\right)^2}{a^4+b^4+\left(a+b\right)^4}=\frac{14}{98}=\frac{1}{7}\)
Vậy ...
Chia cả tử và mẫu của các phân số cho a khác 0 ta được:
\(A=\frac{a+b}{a-b}+\frac{a-b}{a+b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}+\frac{\frac{a}{b}-1}{\frac{a}{b}+1}=\frac{\left(\frac{a}{b}+1\right)^2+\left(\frac{a}{b}-1\right)^2}{\left(\frac{a}{b}-1\right)\left(\frac{a}{b}+1\right)}=\frac{2.\left(\frac{a}{b}\right)^2+2}{\left(\frac{a}{b}\right)^2-1}\)
\(\Rightarrow A.\left(\frac{a}{b}\right)^2-A=2.\left(\frac{a}{b}\right)^2+2\Rightarrow A.\left(\frac{a}{b}\right)^2-2.\left(\frac{a}{b}\right)^2=A+2\)
\(\Rightarrow\left(A-2\right).\left(\frac{a}{b}\right)^2=A+2\Rightarrow\left(\frac{a}{b}\right)^2=\frac{A+2}{A-2}\)
ta có: \(B=\frac{\left(\frac{a}{b}\right)^4+1}{\left(\frac{a}{b}\right)^4-1}+\frac{\left(\frac{a}{b}\right)^4-1}{\left(\frac{a}{b}\right)^4+1}\)
\(\Rightarrow B=\frac{\left(\frac{A+2}{A-2}\right)^2+1}{\left(\frac{A+2}{A-2}\right)^2-1}+\frac{\left(\frac{A+2}{A-2}\right)^2-1}{\left(\frac{A+2}{A-2}\right)^2+1}=\frac{\left(A+2\right)^2+\left(A-2\right)^2}{\left(A+2\right)^2-\left(A-2\right)^2}+\frac{\left(A+2\right)^2-\left(A-2\right)^2}{\left(A+2\right)^2+\left(A-2\right)^2}\)
\(\Rightarrow B=\frac{2.A^2+8}{8.A}+\frac{8.A}{2.A^2+8}=\frac{\left(2A^2+8\right)^2+64.A^2}{8.A\left(2A^2+8\right)}=\frac{\left(A^2+4\right)^2+16.A^2}{4.A\left(A^2+4\right)}\)
Chia cả tử và mẫu của các phân số cho a khác 0 ta được:
$A=\frac{a+b}{a-b}+\frac{a-b}{a+b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}+\frac{\frac{a}{b}-1}{\frac{a}{b}+1}=\frac{\left(\frac{a}{b}+1\right)^2+\left(\frac{a}{b}-1\right)^2}{\left(\frac{a}{b}-1\right)\left(\frac{a}{b}+1\right)}=\frac{2.\left(\frac{a}{b}\right)^2+2}{\left(\frac{a}{b}\right)^2-1}$A=a+ba−b +a−ba+b =ab +1ab −1 +ab −1ab +1 =(ab +1)2+(ab −1)2(ab −1)(ab +1) =2.(ab )2+2(ab )2−1
$\Rightarrow A.\left(\frac{a}{b}\right)^2-A=2.\left(\frac{a}{b}\right)^2+2\Rightarrow A.\left(\frac{a}{b}\right)^2-2.\left(\frac{a}{b}\right)^2=A+2$⇒A.(ab )2−A=2.(ab )2+2⇒A.(ab )2−2.(ab )2=A+2
$\Rightarrow\left(A-2\right).\left(\frac{a}{b}\right)^2=A+2\Rightarrow\left(\frac{a}{b}\right)^2=\frac{A+2}{A-2}$⇒(A−2).(ab )2=A+2⇒(ab )2=A+2A−2
ta có: $B=\frac{\left(\frac{a}{b}\right)^4+1}{\left(\frac{a}{b}\right)^4-1}+\frac{\left(\frac{a}{b}\right)^4-1}{\left(\frac{a}{b}\right)^4+1}$B=(ab )4+1(ab )4−1 +(ab )4−1(ab )4+1
$\Rightarrow B=\frac{\left(\frac{A+2}{A-2}\right)^2+1}{\left(\frac{A+2}{A-2}\right)^2-1}+\frac{\left(\frac{A+2}{A-2}\right)^2-1}{\left(\frac{A+2}{A-2}\right)^2+1}=\frac{\left(A+2\right)^2+\left(A-2\right)^2}{\left(A+2\right)^2-\left(A-2\right)^2}+\frac{\left(A+2\right)^2-\left(A-2\right)^2}{\left(A+2\right)^2+\left(A-2\right)^2}$⇒B=(A+2A−2 )2+1(A+2A−2 )2−1 +(A+2A−2 )2−1(A+2A−2 )2+1 =(A+2)2+(A−2)2(A+2)2−(A−2)2 +(A+2)2−(A−2)2(A+2)2+(A−2)2
$\Rightarrow B=\frac{2.A^2+8}{8.A}+\frac{8.A}{2.A^2+8}=\frac{\left(2A^2+8\right)^2+64.A^2}{8.A\left(2A^2+8\right)}=\frac{\left(A^2+4\right)^2+16.A^2}{4.A\left(A^2+4\right)}$⇒B=2.A2+88.A +8.A2.A2+8 =(2A2+8)2+64.A28.A(2A2+8) =(A2+4)2+16.A24.A(A2+4)