lớp 9 có học bài này à

ko có radian thì ghi \(2>\frac{\left(p-2\right)q}{p}\) cho rồi 

\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)

\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)

b. ta có \(x=\frac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)

vậy \(P=\frac{4}{\sqrt{4}-1}=4\)

c.\(P=\frac{x}{\sqrt{x}-1}=\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2+2=4\)

vậy \(\sqrt{P}\ge2\)

\(x^4+\sqrt{x^2+2}=2\)

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 ...

ai kb vs mình ko

\(M=\left(x-1\right)4+\left(3-x\right)4+6\left(x^2-4x+3\right)2+2013\)

\(\Leftrightarrow M=\left(x-1+3-x\right)4+12\left(x^2-4x+4\right)-12+2013\)

\(\Leftrightarrow M=12\left(x-2\right)^2+2019\)

Mà \(12\left(x-2\right)^2\ge0\forall x\in R\)

\(\Rightarrow GTNN=2019\)

\(M=\left(x-1\right)^4+\left(3-x\right)^4+6\left(x^2-4x+3\right)^2+2013\)

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:

\(a=0\)

\(\Rightarrow x-2=0\)

\(\Leftrightarrow x=2\)

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