Toán 9 Căn bậc hai và căn thức bậc hai

ĐKXĐ:........
d) $\sqrt{x-2\sqrt{x-1}} = 2$
=> $x-2\sqrt{x-1} = 4$
$2\sqrt{x-1} = x-4$
$4(x-1) = x^{2}-8x+16$
=> $x^{2}-12x+20=0$
$(x-2)(x-10)=0$
=>....

Bài 5:

a) [tex]A=3+\sqrt{2x^2-4x+3}=3+\sqrt{2(x-1)^2+1} \geq 3+\sqrt{1}=4[/tex]

Dấu "=" xảy ra khi [tex]x=1[/tex]

b) [tex]B=\sqrt{x^2+y^2-2xy+2x-2y+5}+2y^2-8y+2018[/tex]

[tex]=\sqrt{(x-y)^2+2(x-y)+1+4}+2(y^2-4y+4)+2010[/tex]

[tex]=\sqrt{(x-y+1)^2+4}+2(y-2)^2+2010 \geq \sqrt{4}+2010=2012[/tex]

Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x=y-1 & \\y=2 & \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} x=1 & \\y=2 & \end{matrix}\right.[/tex]

Bài 6:

Thay [TEX]ab+bc+ca=1[/TEX] vào biểu thức:

[tex]A=\sqrt{(a^2+ab+bc+ca)(b^2+ab+bc+ca)(c^2+ab+bc+ca)}[/tex]

[tex]=\sqrt{(a+b)(c+a)(b+c)(a+b)(c+a)(b+c)}[/tex]

[tex]=\sqrt{[(a+b)(b+c)(c+a)]^2}=(a+b)(b+c)(c+a) \in Q[/tex]

Bài 4:
a) [TEX]x^2-11=(x-\sqrt{11})(x+\sqrt{11})[/TEX]
b)[TEX]x-5=[/TEX][TEX](\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})[/TEX]
c)[TEX]3+4x=3-4(-x)=[/TEX][TEX](\sqrt{3}-2\sqrt{-x})(\sqrt{3}+2\sqrt{-x})[/TEX]
d)[TEX]5-7x^2=(\sqrt{5}-\sqrt{7}x)(\sqrt{5}+\sqrt{7}x)[/TEX]
e)[TEX]x^2-2\sqrt{2}x+2=(x-\sqrt{2})^2[/TEX]