vẫn bình thường bn à 

Oki v hã :">

BÀI 1:

a) 

PT <=>    \(3x-2=7-4\sqrt{3}\)

<=>    \(3x=9-4\sqrt{3}\)

<=>    \(x=3-\frac{4}{\sqrt{3}}\)

b)

pt =>   \(x+1=14-6\sqrt{5}\)

<=>   \(x=13-6\sqrt{5}\)

BÀI 2: 

a)

pt <=>   \(\sqrt{x^2-9}=3\sqrt{x-3}\)

<=>   \(x^2-9=9\left(x-3\right)\)

<=>   \(x^2-9=9x-27\)

<=>   \(x^2-9x+18=0\)

<=>   \(\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

BÀI 2: 

b)

pt <=>   \(\sqrt{x^2-4}=2\sqrt{x+2}\)

<=>   \(x^2-4=4\left(x+2\right)\)

<=>   \(x^2-4=4x+8\)

<=>   \(x^2-4x-12=0\)

<=>   \(\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

BÀI 3:

pt <=>   \(x^2=5\)

<=>   \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)

\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)

\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)

\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)

\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)

Á nhầm nhaaa cái cuối cùng là cộng z2 đó

Ta có :

\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)

tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)

\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)

\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)

Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)

mik rảnh đây bn nói chuyện vs mik đi chán quá

Thám tử lạ mặt lm sao bn biết quản lí đang onl 

Cậu có thể giải hộ mình được không?

\(=\dfrac{x+6\sqrt{x}+5+x-5\sqrt{x}-x+9\sqrt{x}+20}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{x+10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}+5}{\sqrt{x}-5}\)

B2 là đc nhoa =))

Bài 3:

a) \(P=\left(\dfrac{3x-2}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}+2}\left(đk:x\ge0,x\ne4\right)\)

\(=\dfrac{3x-2-\sqrt{x}\left(\sqrt{x}-2\right)-2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{1}\)

\(=\dfrac{3x-2-x+2\sqrt{x}-2x-4\sqrt{x}}{\sqrt{x}-2}=\dfrac{-2\sqrt{x}-2}{\sqrt{x}-2}\)

b) \(P=\dfrac{-2\sqrt{x}-2}{\sqrt{x}-2}=\dfrac{5}{2}\)

\(\Rightarrow-4\sqrt{x}-4=5\sqrt{x}-10\)

\(\Rightarrow9\sqrt{x}=6\Rightarrow\sqrt{x}=\dfrac{6}{9}\Rightarrow x=\dfrac{36}{81}\left(tm\right)\)