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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
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{a}\left(\sqrt{b}+2\right)-3\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}-\frac{\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)-\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+6\sqrt{a}+3\sqrt{ab}+9\sqrt{b}-6\sqrt{a}+a\sqrt{b}+18-3\sqrt{ab}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+9\sqrt{b}+a\sqrt{b}+18}{\left(a-9\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(a+9\right)\left(\sqrt{b}+2\right)}{\left(a-9\right)\left(\sqrt{b}+2\right)}=\frac{a+9}{a-9}\)
a) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\sqrt{\frac{x}{y}}\)
c) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
d) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(\sqrt{1}-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
e) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
a) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
b) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}}{\sqrt{y}}\) (Bạn tự thêm đk)
d) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\) (Bạn tự thêm đk)
\(N=1:\left(\frac{x+2}{\sqrt{x^3}-1}+\frac{\sqrt{x}+1}{x+1+\sqrt{x}}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(N=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}+\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}-\frac{\left(x+1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{x+2+x-1-x-1-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+1+\sqrt{x}\right)}\right)\)
\(N=1:\left(\frac{\sqrt{x}}{\left(x+1+\sqrt{x}\right)}\right)\)
\(N=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
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
Bài 1:
b) Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\frac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}-\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}-\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\frac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Bài 2:
a) Ta có: \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)\)
\(=a-\sqrt{a}-a-\sqrt{a}\)
\(=-2\sqrt{a}\)
b) Ta có: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{ab}-\sqrt{ab}=0\)
d) Ta có: \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)\)
=0
Bài 3:
a) ĐKXĐ: x≥0
Ta có: \(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
\(\Leftrightarrow\frac{\sqrt{27}\cdot\sqrt{x}}{\sqrt{3}}=6\)
\(\Leftrightarrow3\cdot\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{3}=2\)
hay \(x=4\)(thỏa mãn)
Vậy: S={4}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Leftrightarrow x\ge0\)
Ta có: \(\sqrt{x+1}=3-\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(3-\sqrt{x}\right)^2\)
\(\Leftrightarrow x+1=9-6\sqrt{x}+x\)
\(\Leftrightarrow x+1-9+6\sqrt{x}-x=0\)
\(\Leftrightarrow-8+6\sqrt{x}=0\)
\(\Leftrightarrow6\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=\frac{8}{6}=\frac{4}{3}\)
hay \(x=\frac{16}{9}\)(thỏa mãn)
Vậy: \(S=\left\{\frac{16}{9}\right\}\)