rat xl nhung ta k go~co' dau duok T.Tcau 1:
ta chung minh bang phan chung:
Gia su $\sqrt6$ la so huu ti. Khi do :
$\sqrt6=\frac{a}{b} ( a;b\in N^* ) a,b$ nguyen to cung nhau
$\Leftrightarrow \frac{a^2}{b^2}=6\Leftrightarrow a^2=6b^2 (*)$ vay $a $ chia het cho $6$
Dat $a=6k,k\in Z$
khi do $(*)\Leftrightarrow 36k^2=6b^2\Leftrightarrow b^2=6k^2\Rightarrow b$ chia het cho 6
Vay $a,b$ cung chia het cho $6$ nen khong nguyen to cung nhau ( dieu nay` la trai voi gthiet)
Vay $\sqrt6$ la so vo ti
Cau 2:
DK: $x\geq 1$
khi do pt $\Leftrightarrow (\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-2}})^2=4$
$x-\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1})}=2$
$\Leftrightarrow x-2=\sqrt{(x-2)^2}$
TH1:$ \left\{ \begin{array}{l} x-2\geq 0\\ x-2=x-2 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\geq 2\\ x\in R\end{array} \right.\Leftrightarrow x\geq 2$
th2:$\left\{ \begin{array}{l} x<2\\ x-2=2-x \end{array} \right.\Leftrightarrow $vo nghiem
Vay tap ngiem cua pt da cho la : $S=(2;+\infty )$
Cau 3: $\Leftrightarrow $ $ \left [ \begin{matrix} \left\{ \begin{array}{l} x-2<0\\ (x-1)(4-x)\geq 0 \end{array} \right. (I)\\ \left\{ \begin{array}{l} x-2\geq 0\\ (x-1)(4-x)>(x-2)^2 \end{array} \right. (II)\end{matrix}
\right. $
$(i)\Leftrightarrow \left\{ \begin{array}{l} x<2\\ 1\leq x\leq 4\end{array} \right.\Leftrightarrow 1\leq x<2$
Tap nghiem cua $(I)$ la $S_1=[1;2)$
$(II)\Leftrightarrow \left\{ \begin{array}{l} x\geq 2\\ -x^2+5x-4>x^2-4x+4 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\geq 2\\ 2x^2-9x+8<0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\geq 2\\ \frac{9-\sqrt{17}}{4} <x <\frac{9+\sqrt{17}}{4} \end{array} \right.$
$\Leftrightarrow x\in [2;\frac{9+\sqrt{17}}{4} )$
Tap nghiem cua $(II)$ la $S_2=[2;\frac{9+\sqrt{17}}{4} )$
vay tap nghiem cua bpt da cho : $S=S_1\cup S_2=[1;\frac{9+\sqrt{17}}{4} )$