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Cho \(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}\)
a, Rút gọn Q
B, Chứng minh Q=\(\frac{b+81}{b-81}\)thì \(\frac{b}{a}\)là một số nguyên chia hết cho 3
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{a}\left(\sqrt{b}+2\right)-3\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}-\frac{\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)-\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+6\sqrt{a}+3\sqrt{ab}+9\sqrt{b}-6\sqrt{a}+a\sqrt{b}+18-3\sqrt{ab}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+9\sqrt{b}+a\sqrt{b}+18}{\left(a-9\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(a+9\right)\left(\sqrt{b}+2\right)}{\left(a-9\right)\left(\sqrt{b}+2\right)}=\frac{a+9}{a-9}\)
\(N=1:\left(\frac{x+2}{\sqrt{x^3}-1}+\frac{\sqrt{x}+1}{x+1+\sqrt{x}}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(N=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}-\frac{\left(x+1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{x+2+x-1-x-1-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}}{\left(x+1+\sqrt{x}\right)}\right)\)
\(N=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
y b
chia 2 ve cho can 2
tc
\(\sqrt{x}+1+\frac{1}{\sqrt{x}}\)
tc \(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)(bdt cosi)
\(\sqrt{x}+1+\frac{1}{\sqrt{x}}\ge3\)
=> dpcm
may mk loi font chu thong cam viet ko co dau
