cho \(\Delta ABC\)có 3 góc nhọn, 3 đường cao AD, BE, CF cắt nhau tại H.
a) Cm BCEF nt và AEHF nt
b) Cm EH.ED=EA.EC
c) Cm H là tâm của đường tròn nt \(\Delta DEF\)
d) AD = 5cm, BD = 3cm, CD = 4cm
\(S_{\Delta BHC}=?cm^2\)
K
Khách
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NB
2
19 tháng 3 2021
\(x^4+\sqrt{x^2+2}=2\)
Đặt t = x2
pt <=> \(t^2+\sqrt{t+2}=2\)
<=> \(\sqrt{t+2}=2-t^2\)( 0 ≤ t ≤ √2 )
Bình phương hai vế
<=> t + 2 = t4 - 4t2 + 4
<=> t4 - 4t2 - t + 2 = 0
<=> t4 - 2t3 + 2t3 - 4t2 - t + 2 = 0
<=> t3( t - 2 ) + 2t2( t - 2 ) - ( t - 2 ) = 0
<=> ( t - 2 )( t3 + 2t2 - 1 ) = 0
<=> ( t - 2 )( t3 + t2 + t2 - 1 ) = 0
<=> ( t - 2 )[ t2( t + 1 ) + ( t - 1 )( t + 1 ) ] = 0
<=> ( t - 2 )( t + 1 )( t2 + t - 1 ) = 0
<=> t - 2 = 0 hoặc t + 1 = 0 hoặc t2 + t - 1 = 0
<=> t = \(\frac{-1+\sqrt{5}}{2}\)( đã loại các nghiệm ktm )
=> \(x^2=\frac{-1+\sqrt{5}}{2}\Leftrightarrow x=\pm\sqrt{\frac{-1+\sqrt{5}}{2}}\)
Vậy ...
W
4
19 tháng 3 2021
\(M=\left(x-1\right)4+\left(3-x\right)4+6\left(x^2-4x+3\right)2+2013\)
\(\Leftrightarrow M=\left(x-1+3-x\right)4+12\left(x^2-4x+4\right)-12+2013\)
\(\Leftrightarrow M=12\left(x-2\right)^2+2019\)
Mà \(12\left(x-2\right)^2\ge0\forall x\in R\)
\(\Rightarrow GTNN=2019\)
19 tháng 3 2021
\(M=\left(x-1\right)^4+\left(3-x\right)^4+6\left(x^2-4x+3\right)^2+2013\)
Đặt \(x-2=a\)
\(\Rightarrow M=\left(a+1\right)^2+\left(a-1\right)^2+6\left(a^2-1\right)^2+2013\)
\(=8a^4+2021\ge2021\)
Dấu = xảy ra khi:
\(a=0\)
\(\Rightarrow x-2=0\)
\(\Leftrightarrow x=2\)
KN
19 tháng 3 2021
Áp dụng BĐT Cô-si: \(\sqrt{\frac{x}{y+z+2x}.\frac{1}{4}}\le\frac{\frac{x}{y+z+2x}+\frac{1}{4}}{2}\le\frac{\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\frac{1}{4}}{2}\)\(\Rightarrow\sqrt{\frac{x}{y+z+2x}}\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\frac{1}{4}\)
Tương tự: \(\sqrt{\frac{y}{z+x+2y}}\le\frac{1}{4}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{4}\); \(\sqrt{\frac{z}{x+y+2z}}\le\frac{1}{4}\left(\frac{z}{y+z}+\frac{z}{z+x}\right)+\frac{1}{4}\)
Cộng theo vế, ta được: \(VT\le\frac{1}{4}.3+\frac{3}{4}=\frac{3}{2}\)
Đẳng thức xảy ra khi x = y = z
19 tháng 3 2021
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
19 tháng 3 2021
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )