=15√20 -3√45+2√5
=15\(\sqrt{4x5}\)-3\(\sqrt{9x5}\)+2√5
=30√5 -9√5+2√5
=23√5
\(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\) =\(\left(150\sqrt{2}-45\sqrt{2}+10\sqrt{2}\right):\sqrt{10}\)
=\(115\sqrt{2}:\sqrt{10}\)
chắc vậy
=15√20 -3√45+2√5
=15\(\sqrt{4x5}\)-3\(\sqrt{9x5}\)+2√5
=30√5 -9√5+2√5
=23√5
\(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\) =\(\left(150\sqrt{2}-45\sqrt{2}+10\sqrt{2}\right):\sqrt{10}\)
=\(115\sqrt{2}:\sqrt{10}\)
chắc vậy
rut gon bieu thuc (5√3+3√5)÷15
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) rut gon bieu thuc gium em a thanks
Rut gon bieu thuc:
(3√2+√10)×√(38-12√5)
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\)\(\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Rut gon bieu thuc
tum x de bieu thuc = \(\frac{-1}{7}\)
rut gon bieu thuc
B=((3/√(1+a))+√(1-a)):((3/√(1+a^2))+1)
Rut Gon bieu thuc
\(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}\)
rut gon bieu thuc (1-2*sin a * cos a)/(sin^2 a - cos^2 a)
Rut gon bieu thuc
A=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\)\(\frac{3}{\sqrt{x}+3}\)