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\(PT\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\dfrac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}-3=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\left(18\sqrt{2}-\dfrac{5b\sqrt{2}}{a^2-2b^2}-\dfrac{4b\sqrt{2}}{a^2-2b^2}\right)=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\sqrt{2}\left(18-\dfrac{5b}{a^2-2b^2}-\dfrac{4b}{a^2-2b^2}\right)=0\)
Vì a,b nguyên mà vế trái có \(\sqrt{2}\) vô tỉ nên 2 biểu thức còn lại phải bằng 0
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}=3\\\dfrac{5b}{a^2-2b^2}+\dfrac{4b}{a^2-2b^2}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{a^2-2b^2}=3\\\dfrac{b}{a^2-2b^2}=2\end{matrix}\right.\left(a,b\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=\dfrac{a}{3}\\b=2\left(a^2-2b^2\right)=2\cdot\dfrac{a}{3}=\dfrac{2}{3}a\end{matrix}\right.\)
\(\Leftrightarrow a^2-\dfrac{8}{9}a^2=\dfrac{a}{3}\Leftrightarrow\dfrac{1}{9}a^2-\dfrac{1}{3}a=0\Leftrightarrow\dfrac{1}{3}a\left(\dfrac{1}{3}a-1\right)=0\\ \Leftrightarrow a=3\left(a\ne0\right)\)
\(\Leftrightarrow b=\dfrac{2}{3}\cdot3=2\left(tm\right)\)
Vậy \(\left(a;b\right)=\left(3;2\right)\)
\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
<=> \(\frac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\frac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\) trục căn thức
<=> \(\frac{5a}{a^2-2b^2}-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4a}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=3\)
Vì a; b nguyên => \(\hept{\begin{cases}\frac{5a}{a^2-2b^2}-\frac{4a}{a^2-2b^2}=3\\-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=0\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{9b}{a^2-2b^2}=18\end{cases}}\)<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{b}{a^2-2b^2}=2\end{cases}}\)
Với b = 0 => loại
Với b khác 0:
=> \(\frac{a}{b}=\frac{3}{2}\Leftrightarrow a=\frac{3}{2}b\)
=> \(\frac{b}{\frac{9}{4}b^2-2b^2}=2\)=> b = 2 => a = 3 thử lại thỏa mãn
Vậy a = 3 và b = 2.
\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18\sqrt{2}\left(a^2-2b^2\right)=3\left(a^2-2b^2\right)\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18a^2\sqrt{2}-36b^2\sqrt{2}=3a^2-6b^2\)
\(\Leftrightarrow\left(18a^2-36b^2-9b\right)\sqrt{2}=3a^2-6b^2-a\)
-Nếu \(18a^2-36b^2-9b\ne0\Rightarrow\sqrt{2}=\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\)
Vì a,b nguyên nên \(\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\inℚ\Rightarrow\sqrt{2}\inℚ\)=> Vô lý vì \(\sqrt{2}\)là số vô tỷ
-Vậy ta có: \(18a^2-36b^2-9b=0\Rightarrow\hept{\begin{cases}18a^2-36b^2-9b=0\\3a^2-6b^2-a=0\end{cases}\Rightarrow\hept{\begin{cases}3a^2-6b^2=\frac{3}{2}b\\3a^2-6b^2=2\end{cases}}\Leftrightarrow a=\frac{3}{2}b}\)
Thay a=\(\frac{3}{2}b\)vào \(3a^2-6b^2-a=0\)
ta có \(3\cdot\frac{9}{4}b^2-6b^2-\frac{3}{2}b=0\Leftrightarrow27b^2-6b=0\Leftrightarrow3b\left(b-2\right)=0\)
Ta có b=0 (loại), b=2 (tm) => a=3
Vậy b=2; a=3
ĐK:\(a\ne0,b\ne0\)
Ta có \(\dfrac{5}{a+b\sqrt{2}}-\dfrac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{\left(a+b\sqrt{2}\right)\left(a-b\sqrt{2}\right)}-\dfrac{4\left(a+b\sqrt{2}\right)}{\left(a+b\sqrt{2}\right)\left(a-b\sqrt{2}\right)}+18\sqrt{2}=3\Leftrightarrow\dfrac{5a-5b\sqrt{2}-4a-4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=3\Leftrightarrow a-9b\sqrt{2}=\left(3-18\sqrt{2}\right)\left(a^2-2b^2\right)\Leftrightarrow a-9b\sqrt{2}=3a^2-6b^2-18a^2\sqrt{2}+36b^2\sqrt{2}\Leftrightarrow a-3a^2+6b^2=9b\sqrt{2}+36b^2\sqrt{2}-18a^2\sqrt{2}\Leftrightarrow a-3a^2+6b^2=9\sqrt{2}\left(b+4b^2-2a^2\right)\)Ta có a,b là số nguyên
Suy ra\(\left\{{}\begin{matrix}a-3a^2+6b^2=0\left(1\right)\\b+4b^2-2a^2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}4a-12a^2+24b^2=0\left(2\right)\\6b+24b^2-12a^2=0\left(3\right)\end{matrix}\right.\)
Trừ (2) cho (3) ta được \(4a-6b=0\Leftrightarrow b=\dfrac{2}{3}a\left(4\right)\)
Thay (4) vào (1) ta có \(a-3a^2+6b^2=0\Leftrightarrow a-3a^2+\dfrac{6.4}{9}a^2=0\Leftrightarrow a-\dfrac{1}{3}a^2=0\Leftrightarrow a^2-3a=0\Leftrightarrow a\left(a-3\right)=0\Leftrightarrow\)\(\left\{{}\begin{matrix}a=0\\a=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=0\left(ktm\right)\\b=1\left(tm\right)\end{matrix}\right.\)
Vậy (a;b)=(3;1)
Tìm các số hữu tỉ a,b thỏa mãn \(\frac{5}{a+b\sqrt{2}}\)- \(\frac{4}{a-b\sqrt{2}}\)+18\(\sqrt{2}\)=3
\(\frac{5\left(a-b\sqrt{2}\right)-4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\)
\(\left(a-9b\sqrt{2}\right)+\left(a^2-2b^2\right)18\sqrt{2}=3\left(a^2-2b\right)\)
\(\sqrt{2}\left[18\left(a^2-2b^2\right)-9b\right]+a=3\left(a^2-2b\right)\)
\(\sqrt{2}\)là số vô tỷ=> \(\hept{\begin{cases}2a^2-4b^2-b=0\\3a^2-6b-a=0\end{cases}\Leftrightarrow}\) (giải hệ này ra a,b)
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
Trục căn thức:
\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
<=> \(\frac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\frac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\)
<=> \(\left(\frac{5a}{a^2-2b^2}-\frac{4a}{a^2-2b^2}-3\right)+\left(18-\frac{5b}{a^2-2b^2}-\frac{4b}{a^2-2b^2}\right)=0\)(1)
Vì a và b là số nguyên nên:
(1) <=> \(\hept{\begin{cases}\frac{5a-4a}{a^2-2b^2}=3\\\frac{5b+4b}{a^2-2b^2}=18\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{b}{a^2-2b^2}=2\end{cases}}\)( a; b khác 0)
<=> \(\hept{\begin{cases}a=\frac{3}{2}b\\\frac{b}{\frac{9}{4}b^2-2b^2}=2\end{cases}}\Leftrightarrow a=3;b=2\)
Vậy:...
Dự đoán xảy ra cực trị khi a = b = c =2. Khi đó P =\(\frac{3\sqrt{2}}{4}\). Ta sẽ chứng minh đó là MAX của P
Ta có: \(\left(\frac{a+b+c}{3}\right)^3-\left(a+b+c\right)\ge abc-\left(a+b+c\right)=2\)
Đặt a + b +c = t>0 suy ra \(\frac{t^3-27t}{27}\ge2\Leftrightarrow t^3-27t\ge54\Leftrightarrow t^3-27t-54\ge0\)
\(\Leftrightarrow\orbr{\begin{cases}t\ge6\\t=-3\left(L\right)\end{cases}}\). Do vậy \(t\ge6\) (em làm tắt xiu nhé,dài quá)
\(P=\Sigma_{cyc}\frac{2}{\sqrt{2}.\sqrt{2\left(a^2+b^2\right)}}\le\sqrt{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
Giờ đi chứng minh \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{3}{4}\)
Em cần suy ra nghĩ tiếp:(
B1:
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18\sqrt{2}\left(a^2-2b^2\right)=3\left(a^2-2b^2\right)\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18a^2\sqrt{2}-36b^2\sqrt{2}=3a^2-6b^2\)
\(\Leftrightarrow18a^2\sqrt{2}-36b^2\sqrt{2}-9b\sqrt{2}=3a^2-6b^2-a\)
\(\Leftrightarrow\left(18a^2-36b^2-9b\right)\sqrt{2}=3a^2-6b^2-a\)
Nếu \(18a^2-36b^2-9b\ne0\Rightarrow\sqrt{2}=\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\)
Vì a,b nguyên nên \(\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\in Q\Rightarrow\sqrt{2}\in Q\)=> Vô lý vì \(\sqrt{2}\)là số vô tỉ.
Vậy ta có: \(18a^2-36b^2-9b=0\Rightarrow\hept{\begin{cases}18a^2-36b^2-9b=0\\3a^2-6b^2-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3a^2-6b^2=\frac{3}{2}b\\3a^2-6b^2=a\end{cases}\Leftrightarrow a=\frac{3}{2}b}\)
Thay \(a=\frac{3}{2}b\)vào \(3a^2-6b^2-a=0\)ta có:
\(3.\frac{9}{4}b^2-6b^2-\frac{3}{2}b=0\Leftrightarrow27b^2-24b^2-6b=0\Leftrightarrow3b\left(b-2\right)=0\)
Ta có: b=0(loại) ; b=2(thoả mãn) . Vậy a=3. KL:...
B2: \(GT\Rightarrow\left[\left(a+b\right)^2-2\left(ab+1\right)\right]\left(a+b\right)^2+\left(1+ab\right)^2=0\)
\(\Leftrightarrow\left(a+b\right)^4-2\left(a+b\right)^2\left(1+ab\right)+\left(1+ab\right)^2=0\)
\(\Leftrightarrow\left[\left(a+b\right)^2-\left(1+ab\right)\right]^2=0\Rightarrow\left(a+b\right)^2-\left(1+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)^2=1+ab\Leftrightarrow\left|a+b\right|=\sqrt{1+ab}\in Q\)( vì a,b thuộc Q)
KL:....
ĐK:\(a\ne0,b\ne0\)
Ta có \(\dfrac{5}{a+b\sqrt{2}}-\dfrac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{\left(a+b\sqrt{2}\right)\left(a-b\sqrt{2}\right)}-\dfrac{4\left(a+b\sqrt{2}\right)}{\left(a+b\sqrt{2}\right)\left(a-b\sqrt{2}\right)}+18\sqrt{2}=3\Leftrightarrow\dfrac{5a-5b\sqrt{2}-4a-4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=3\Leftrightarrow a-9b\sqrt{2}=\left(3-18\sqrt{2}\right)\left(a^2-2b^2\right)\Leftrightarrow a-9b\sqrt{2}=3a^2-6b^2-18a^2\sqrt{2}+36b^2\sqrt{2}\Leftrightarrow a-3a^2+6b^2=9b\sqrt{2}+36b^2\sqrt{2}-18a^2\sqrt{2}\Leftrightarrow a-3a^2+6b^2=9\sqrt{2}\left(b+4b^2-2a^2\right)\)
Ta có a,b là số nguyên
Suy ra \(\left\{{}\begin{matrix}a-3a^2+6b^2=0\left(1\right)\\b+4b^2-2a^2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}4a-12a^2+24b^2=0\left(2\right)\\6b+24b^2-12a^2=0\left(3\right)\end{matrix}\right.\)
Trừ (2) cho (3) ta được \(4a-6b=0\Leftrightarrow b=\dfrac{2}{3}a\left(4\right)\)
Thay (4) vào (1) ta có \(a-3a^2+6b^2=0\Leftrightarrow a-3a^2+\dfrac{6.4}{9}a^2=0\Leftrightarrow a-\dfrac{1}{3}a^2=0\Leftrightarrow a^2-3a=0\Leftrightarrow a\left(a-3\right)=0\Leftrightarrow\)\(\left\{{}\begin{matrix}a=0\\a=3\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}b=0\left(ktm\right)\\b=2\left(tm\right)\end{matrix}\right.\)
Vậy (a;b)=(3;2)