ta có : \(\left(\sqrt{7}\right)^2=7\)vì 7 là số nguyên tố nên căn 7 sẽ là số vô tỉ ( mk nghĩ vậy) ^^
ta có : \(\left(\sqrt{7}\right)^2=7\)vì 7 là số nguyên tố nên căn 7 sẽ là số vô tỉ ( mk nghĩ vậy) ^^
Given a triangle ABC. D is a point on AB and E is a point on AC so that DE//BC and \(\frac{BC}{DE}=\sqrt{2}\), F is a point on BC and G is a point on AB so that FG//AC and \(\frac{AC}{FG}=\sqrt{2}\), H is a point on BC and I is a point on AC so that HI//AB and \(\frac{AB}{HI}=\sqrt{2}\). FG meets HI at X, DE meets HI at Y and DE meets FG at Z.
i) Prove that \(DY=ZE\)
ii) Find the exactly value of the ratio \(\frac{YZ}{BC}\)
Circle O has diameters AB and CD perpendicular to each other. AM is any chord intersecting CB at N.I is the center of the circle inscribed in triangle AMB. Prove that M, I, D are collinear points
Let x = y = z, Prove that:
\(\sqrt{\frac{x^2}{y^2+z^2}}+\sqrt{\frac{y^2}{z^2+x^2}}+\sqrt{\frac{z^2}{x^2+y^2}}\ge\frac{3}{2}\)
(By: Wrecking Ball)
Hi vọng sẽ không bị trùng
Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 . Prove that :
\(3\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\ge\left(x+2\right)\left(y+2\right)\left(z+2\right)\)
lamf tis baats ddawngr thwcs :))
Let a,b,c be positive real numbers. Prove that : \(\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\ge\frac{3}{\sqrt[3]{abc}\left(1+\sqrt[3]{abc}\right)}\)
Given an acute triangle ABC, its median AM and its heights BH and CK. The line passes through A and perpendiculars with AM cuts BH and CK at D and E, respectively. Prove that \(\Delta DEM\)is an isosceles triangle.
\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\) bằng
Tính
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{4+\sqrt{7}} -\sqrt{4-\sqrt{7}}\)
c) \(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)