Sửa đề: \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)

\(\Leftrightarrow\left|\sqrt{x-1}+2\right|+\left|\sqrt{x-1}-3\right|=5\)

=>\(\sqrt{x-1}+\left|\sqrt{x-1}-3\right|=3\)

TH1: x>=10

Pt sẽ là \(\sqrt{x-1}+\sqrt{x-1}-3=3\)

=>2 căn x-1=6

=>x-1=9

=>x=10

TH2: 1<=x<10

Pt sẽ là \(\sqrt{x-1}+3-\sqrt{x-1}=3\)

=>3=3(nhận)

Câu 1:

Có: \(8-4\sqrt{3}=8-2\sqrt{12}=6+2-2\sqrt{6.2}=(\sqrt{6}-\sqrt{2})^2\)

\(\Rightarrow \sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}\)

Do đó:

\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}\)

\(=\sqrt{(\sqrt{6})^2-(\sqrt{2})^2}=\sqrt{6-2}=2\)

Câu 2:

\(16-5\sqrt{7}=\frac{32-10\sqrt{7}}{2}=\frac{32-2\sqrt{175}}{2}=\frac{25+7-2\sqrt{25.7}}{2}=\frac{(5-\sqrt{7})^2}{2}\)

\(\Rightarrow \sqrt{16-5\sqrt{7}}=\frac{5-\sqrt{7}}{\sqrt{2}}\)

Do đó:

\(\sqrt{16-5\sqrt{7}}(5\sqrt{2}+\sqrt{14})+\frac{6}{3+\sqrt{10}}=\frac{5-\sqrt{7}}{\sqrt{2}}.\sqrt{2}(5+\sqrt{7})+\frac{6(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})}\)

\(=(5-\sqrt{7})(5+\sqrt{7})+\frac{18-6\sqrt{10}}{3^2-10}=25-7+(-18+6\sqrt{10})\)

\(=6\sqrt{10}\)

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........

e , \(\sqrt{11^2-\left(6\sqrt{2}\right)^2}\)

g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath

\(x=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}=\dfrac{3}{3}=1\)

\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)

\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}\cdot\sqrt{2}\left(3+\sqrt{7}\right)\\ =\sqrt{\dfrac{2\left(3+\sqrt{7}\right)^2}{8+3\sqrt{7}}}=\sqrt{\dfrac{32+12\sqrt{7}}{8+3\sqrt{7}}}\\ =\sqrt{\dfrac{4\left(8+3\sqrt{7}\right)}{8+3\sqrt{7}}}=\sqrt{4}=2\)

\(x=\dfrac{\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}.\left(\sqrt{5}+2\right)=\dfrac{\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}.\left(\sqrt{5}+2\right)=\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{5-4}{3}=\dfrac{1}{3}\) Thay : \(x=\dfrac{1}{3}\) vào A , ta được :

\(A=\left(\dfrac{3}{27}+\dfrac{8}{9}-\dfrac{3}{3}+1\right)^{2012}=1^{2012}=1\)

Vậy ,...

\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)

\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)

\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)

https://hoc24.vn/hoi-dap/question/405366.html

\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)

= 5 - 3

= 2

\(A=\frac{\left(\sqrt{3}+1\right)\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}=\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}=\frac{44\left(\sqrt{3}-1\right)}{22}=2\sqrt{3}-2\)

\(\frac{4\left(11\sqrt{3}-11\right)}{25^2-\left(\sqrt{3}\right)^2}\) Thay Bằng:

\(\frac{4\left(11\sqrt{3}-11\right)}{5^2-\left(\sqrt{3}\right)^2}\)

Cảm ơn bạn! 

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