\(\sqrt{3-\sqrt{5}}\div\sqrt{2}\)

">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 8 2023

ko bt

2 tháng 8 2023

\(\sqrt[]{3-\sqrt[]{5}}:\sqrt[]{2}\)

\(=\sqrt[]{3-\sqrt[]{5}}.\dfrac{1}{\sqrt[]{2}}\)

\(=\dfrac{\sqrt[]{3-\sqrt[]{5}}}{\sqrt[]{2}}\)

\(=\sqrt[]{\dfrac{3-\sqrt[]{5}}{2}}\)

18 tháng 10 2019

Căn bậc haiCăn bậc hai

ta có

26 tháng 9 2016

1/ Điều kiện xác định \(x\ge0\)

\(\frac{\sqrt{x}-1}{2}-\frac{\sqrt{x}+2}{3}=\sqrt{x}-1\)

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

\(\Leftrightarrow-\frac{5}{6}\sqrt{x}=\frac{1}{6}\Leftrightarrow\sqrt{x}=-\frac{1}{5}\) (vô lí)

Vậy pt vô nghiệm

2/ \(x-\left(\sqrt{x}-4\right)\left(\sqrt{x}-5\right)=-38\)

\(\Leftrightarrow x-\left(x-9\sqrt{x}+20\right)+38=0\)

\(\Leftrightarrow9\sqrt{x}=-18\Leftrightarrow\sqrt{x}=-2\) (vô lí)

Vậy pt vô nghiệm.

26 tháng 9 2016

1)\(\frac{\sqrt{x}-1}{2}-\frac{\sqrt{x}+2}{3}=\sqrt{x}-1\)

Đặt \(a=\sqrt{x}-1\) ta  đc:

\(\frac{a}{2}-\frac{a+3}{3}=a\)\(\Leftrightarrow\frac{a-6}{6}=a\)

\(\Leftrightarrow a-6=6a\)\(\Leftrightarrow a=-\frac{6}{5}\)

\(\Leftrightarrow\sqrt{x}-1=-\frac{6}{5}\)

\(\Leftrightarrow\sqrt{x}=-\frac{1}{5}\)

=>vô nghiệm (vì \(\sqrt{x}\ge0>-\frac{1}{5}\))

 

\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)

\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)

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

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

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

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

\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)

\(=10\)

\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)

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

\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)

\(=\sqrt{3}-1\)

\(a=\dfrac{4\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}}{2}\)

\(=2\sqrt{\sqrt{5}-\sqrt{5}+1}=2\)

\(P=\left(2^5-7\cdot2^2-3\right)^{81}+19=1+19=20\)

15 tháng 10 2016

1/\(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)

\(\Rightarrow2A=\left(\sqrt{24-x^2}+\sqrt{8-x^2}\right)\left(\sqrt{24-x^2}-\sqrt{8-x^2}\right)\)

\(\Leftrightarrow2A=16\Rightarrow A=8\)

2/ ĐKXĐ : \(x\ge5\)

\(\sqrt{x-2}+\sqrt{x-5}=\sqrt{x+3}\)

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

\(\Leftrightarrow2x+2\sqrt{x-2}.\sqrt{x-5}-7=x+3\)

\(\Rightarrow2\sqrt{x-2}.\sqrt{x-5}=10-x\)

\(\Leftrightarrow4\left(x-2\right)\left(x-5\right)=x^2-20x+100\)

\(\Leftrightarrow3x^2-8x-60=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)

\(x\ge5\) nên x = 6 thỏa mãn đề bài.

27 tháng 7 2016

Có: \(\frac{P}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\right)\)

\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{6+2\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{6-2\sqrt{5}}}\)

\(=\frac{3+\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\frac{3-\sqrt{5}}{\sqrt{20}+\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\frac{3+\sqrt{5}}{2\sqrt{5}+\sqrt{5}+1}-\frac{3-\sqrt{5}}{2\sqrt{5}+\sqrt{5}-1}\)

\(=\frac{3+\sqrt{5}}{3\sqrt{5}+1}-\frac{3-\sqrt{5}}{3\sqrt{5}-1}\)

\(=\frac{\left(3+\sqrt{5}\right)\left(3\sqrt{5}-1\right)-\left(3-\sqrt{5}\right)\left(3\sqrt{5}+1\right)}{\left(3\sqrt{5}+1\right)\left(3\sqrt{5}-1\right)}\)

\(=\frac{9\sqrt{5}-3+15-\sqrt{5}-9\sqrt{5}-3+15+\sqrt{5}}{9\cdot5-1}\)

\(=\frac{24}{44}=\frac{6}{11}\)

=>P=\(\frac{6}{11}\cdot\sqrt{2}=\frac{6\sqrt{2}}{11}\)

Chính xác 100% mink thử bằng máy tính r

mink làm hơi tắt phần nào k hiểu hói mink nhé

\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)

\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)

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

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

\(=9-2=7\)

\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)

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

\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)

\(=\sqrt{3-\sqrt{3}}\)

\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)

\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)

\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)

\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)

\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)

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

\(=2.6-18\sqrt{2}+16\sqrt{3}\)

\(=12-18\sqrt{2}+16\sqrt{3}\)

18 tháng 8 2019

Trả lời

M=\(\frac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

M.\(\frac{1}{\sqrt{2}}\)\(=\frac{2+\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{2-\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)

M.\(\frac{1}{\sqrt{2}}\)=\(\frac{2+\sqrt{5}}{2+\sqrt{5}+1}+\frac{2-\sqrt{5}}{2-\sqrt{5}-1}\)

M.\(\frac{1}{\sqrt{2}}\)=\(\frac{2+\sqrt{5}}{3+\sqrt{5}}+\frac{2-\sqrt{5}}{1-\sqrt{5}}\)

Phân số bạn làm tiếp nha !

Bài làm nguồn:CHTT , hihi. đg ném đá nha, có ý tốt thoi !