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\(\left\{{}\begin{matrix}3a-2b=27\\6a+8b=26\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{67}{9}\\b=-\frac{7}{3}\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\sqrt{\frac{1}{x+4}}=4-\sqrt{\frac{5}{x+5}}\)
\(\Leftrightarrow2-\sqrt{\frac{1}{x+4}}+2-\sqrt{\frac{5}{x+5}}=0\)
\(\Leftrightarrow\frac{4-\frac{1}{x+4}}{2+\sqrt{\frac{1}{x+4}}}+\frac{4-\frac{5}{x+5}}{2+\sqrt{\frac{5}{x+5}}}=0\)
\(\Leftrightarrow\frac{4x+15}{\left(x+4\right)\left(2+\sqrt{\frac{1}{x+4}}\right)}+\frac{4x+15}{\left(x+5\right)\left(2+\sqrt{\frac{5}{x+5}}\right)}=0\)
\(\Leftrightarrow4x+15=0\Rightarrow x=-\frac{15}{4}\)
Thay \(x=\sqrt{2};y=\sqrt{3}\)ta có:
\(\left\{{}\begin{matrix}2\sqrt{2}-a\sqrt{3}=b\\a\sqrt{2}+b\sqrt{3}=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}a\sqrt{3}=2\sqrt{2}-b\\a\sqrt{2}+b\sqrt{3}=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\\\sqrt{2}\cdot\dfrac{2\sqrt{2}-b}{\sqrt{3}}+b\sqrt{3}=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\\4-b\sqrt{2}+3b=\sqrt{3}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}b\left(\sqrt{2}-3\right)=4-\sqrt{3}\\a=\dfrac{2\sqrt{2}-b}{\sqrt{3}}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}b=\dfrac{4-\sqrt{3}}{\sqrt{2}-3}\\a=\dfrac{2\sqrt{2}-\dfrac{4-\sqrt{3}}{\sqrt{2}-3}}{\sqrt{3}}=\dfrac{4-6\sqrt{2}-4+\sqrt{3}}{\sqrt{3}\left(\sqrt{2}-3\right)}=\dfrac{\sqrt{3}-6\sqrt{2}}{\sqrt{3}\left(\sqrt{2}-3\right)}=\dfrac{1-2\sqrt{6}}{\sqrt{2}-3}\end{matrix}\right.\)
Dùng máy tính casio thực hiện quy trình bấm phím liên tục như sau:
X=X+1:B=2B:A=A+B (X=1;C=1;A=3)
Thì ta được ax5+by5=33
Ấn lt bn xẽ biết ax2015+by2015