1. Rút gọn \(A=\frac{\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}}{\sqrt{\left(\sqrt{5}+1\right)\cdot\sqrt{6-2\sqrt{5}}}}\)
2.Tính a) \(B=\left(\sqrt[3]{2}+1\right)^3\cdot\left(\sqrt[3]{2}-1\right)^3\)
b)Tìm C=\(a^3b-ab^3\) với \(a=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\); \(b=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
3. Giải \(\left|x^2-x+1\right|-\left|x-2\right|=6\)
Bài 1:
Xét tử số:
\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)
\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)
Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)
\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)
Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$
Bài 2:
a)
$B=(\sqrt[3]{2}+1)^3(\sqrt[3]{2}-1)^3$
$=[(\sqrt[3]{2}+1)(\sqrt[3]{2}-1)]^3$
$=(\sqrt[3]{4}-1)^3$
$=3-3\sqrt[3]{16}+3\sqrt[3]{4}$
b)
Với $a,b$ đã cho ta đặt $\sqrt[3]{2}=x$. Khi đó:
\(a=\frac{6}{2x-2+\frac{2}{x}}=\frac{3x}{x^2-x+1}=\frac{3x(x+1)}{x^3+1}=\frac{3x(x+1)}{2+1}=x(x+1)\)
\(b=\frac{2}{2x+2+\frac{2}{x}}=\frac{x}{x^2+x+1}=\frac{x(x-1)}{x^3-1}=\frac{x(x-1)}{2-1}=x(x-1)\)
Khi đó:
$C=a^3b-ab^3=ab(a^2-b^2)=ab(a-b)(a+b)$
$=x^2(x^2-1)(2x)(2x^2)=4x^5(x^2-1)=8\sqrt[3]{4}(\sqrt[3]{4}-1)$
Bài 3:
Ta biết rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ với mọi $x\in\mathbb{R}$
Do đó:
$|x^2-x+1|-|x-2|=6$
$\Leftrightarrow x^2-x+1-|x-2|=6(*)$
Nếu $x\geq 2$ thì $(*)\Leftrightarrow x^2-x+1-(x-2)=6$
$\Leftrightarrow x^2-2x-3=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x=3$ (do $x\geq 2$)
Nếu $x< 2$ thì $(*)\Leftrightarrow x^2-x+1-(2-x)=6$
$\Leftrightarrow x^2-7=0$
$\Rightarrow x=-\sqrt{7}$ (do $x< 2$)
Vậy........
Đặt \(\sqrt[3]{2}+1=a; \sqrt[3]{2}-1=b\)
Khi đó: \(a+b=2\sqrt[3]{2}; ab=\sqrt[3]{4}-1\)
\((\sqrt[3]{2}+1)^3+(\sqrt[3]{2}-1)^3=a^3+b^3=(a+b)^3-3ab(a+b)\)
\(=(2\sqrt[3]{2})^3-3(\sqrt[3]{4}-1).2\sqrt[3]{2}=16-(12-6\sqrt[3]{2})=4+6\sqrt[3]{2}\)
