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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:...
\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
<=> \(\frac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\frac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\) 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
1. Ta có: \(x^2-2xy-x+y+3=0\)
<=> \(x^2-2xy-2.x.\frac{1}{2}+2.y.\frac{1}{2}+\frac{1}{4}+y^2-y^2-\frac{1}{4}+3=0\)
<=> \(\left(x-y-\frac{1}{2}\right)^2-y^2=-\frac{11}{4}\)
<=> \(\left(x-2y-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)=-\frac{11}{4}\)
<=> \(\left(2x-4y-1\right)\left(2x-1\right)=-11\)
Th1: \(\hept{\begin{cases}2x-4y-1=11\\2x-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-3\end{cases}}\)
Th2: \(\hept{\begin{cases}2x-4y-1=-11\\2x-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
Th3: \(\hept{\begin{cases}2x-4y-1=1\\2x-1=-11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)
Th4: \(\hept{\begin{cases}2x-4y-1=-1\\2x-1=11\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=3\end{cases}}\)
Kết luận:...
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)
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)
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)
\(\sqrt{a^2+3a+5}\ge\frac{5a+13}{6}\Leftrightarrow a^2+3a+5\ge\frac{25a^2+130a+169}{36}\)
\(\Leftrightarrow36a^2+108a+180\ge25a^2+130a+169\Leftrightarrow11a^2-22a+11\ge0\)
\(\Leftrightarrow11\left(a-1\right)^2\ge0\forall a\inℝ\)
Dấu = xảy ra khi a=1
Ta có:
\(\sqrt{a^2+3ab+5b^2}=\sqrt{\left(\frac{25a^2}{36}+\frac{130ab}{36}+\frac{169}{36}\right)+\frac{11}{36}\left(a^2-2ab+b^2\right)}\)
\(=\sqrt{\left(\frac{5a}{6}+\frac{13b}{6}\right)^2+\frac{11}{36}\left(a-b\right)^2}\ge\frac{5a+13b}{6}\)
Tương tự:\(\sqrt{b^2+3bc+5c^2}\ge\frac{5b+13c}{6};\sqrt{c^2+3ca+5a^2}\ge\frac{5c+13a}{6}\)
Khi đó:\(P=\sqrt{a^2+3ab+5b^2}+\sqrt{b^2+3bc+5c^2}+\sqrt{c^2+3ac+5a^2}\)
\(\ge\frac{5a+13b+5b+13c+5c+13a}{6}=\frac{18\left(a+b+c\right)}{6}=3\left(a+b+c\right)=9\)
Dấu "=" xảy ra tại \(a=b=c=1\)
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
\(PT\Leftrightarrow\dfrac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\dfrac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}-3=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\left(18\sqrt{2}-\dfrac{5b\sqrt{2}}{a^2-2b^2}-\dfrac{4b\sqrt{2}}{a^2-2b^2}\right)=0\\ \Leftrightarrow\left(\dfrac{5a}{a^2-2b^2}-\dfrac{4a}{a^2-2b^2}-3\right)+\sqrt{2}\left(18-\dfrac{5b}{a^2-2b^2}-\dfrac{4b}{a^2-2b^2}\right)=0\)
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 \(\left(a;b\right)=\left(3;2\right)\)