a, A=\(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)b, B= \(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)c, C=\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)d, D= \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)e,E= \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)D,...
Đọc tiếp
a, A=\(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
b, B= \(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
c, C=\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
d, D= \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
e,E= \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
D, D=\(\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)
chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)