DN
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PN
29 tháng 7 2020
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
6 tháng 7 2017
\(\left(\frac{x+3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\frac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)^2}\)
\(=\frac{-3\cdot\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\)
\(=\frac{-3}{-\left(\sqrt{x}-2\right)}=\frac{3}{\sqrt{x}-2}\)
Chúc bạn học giỏi
Kết bạn với mình nha
BA
24 tháng 11 2019
\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)
\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)
24 tháng 5 2017
\(C=\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{\frac{4-2\sqrt{3}}{2}}.\left[\sqrt{2}.\left(\sqrt{3}+\sqrt{1}\right)\right]\)
\(=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
24 tháng 5 2017
\(D=\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(8+2\sqrt{2}\right).\left(3+\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}.\left(2+3\sqrt{2}\right)}{2}+\frac{\sqrt{2}.\left(1+\sqrt{2}\right)}{1-2}\)
\(=\frac{24+14\sqrt{2}+4}{7}-\frac{2\sqrt{2}+6}{2}-\frac{\sqrt{2}+2}{1}\)
\(=\frac{28+14\sqrt{2}}{7}-\sqrt{2}-3-\sqrt{2}-2\)
\(=4+2\sqrt{2}-2\sqrt{2}-5\)
\(=-1\)
18 tháng 5 2019
\(a)\)\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}-x}{1-\sqrt{x}}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(\sqrt{x}-x\sqrt{x}\right)+\left(1-x\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\left(\sqrt{x}-1\right)\left[\frac{\left(1-x\right)\left(1+\sqrt{x}\right)}{1-\sqrt{x}}\right]\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(P=\frac{\left(x-1\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}=\frac{-\left(1-x\right)\left(1-\sqrt{x}\right)}{1-x}=\sqrt{x}-1\)
\(b)\)\(P=\sqrt{9+4\sqrt{2}}-1=\sqrt{8+4\sqrt{2}+1}-1=\sqrt{\left(2\sqrt{2}+1\right)^2}-1=2\sqrt{2}\)
\(c)\) Ta có : \(\frac{2}{P}=\frac{2}{\sqrt{x}-1}\)
Để P nguyên thì \(\frac{2}{\sqrt{x}-1}\) nguyên hay \(2⋮\left(\sqrt{x}-1\right)\)\(\Rightarrow\)\(\left(\sqrt{x}-1\right)\inƯ\left(2\right)\)
Mà \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)\(\Rightarrow\)\(x\in\left\{\sqrt{2};0;\sqrt{3}\right\}\)
Do x là số chính phương nên \(x=0\)
Vậy để \(\frac{2}{P}\) là số nguyên thì \(x=0\)
NM
Nguyễn Minh Quang
Giáo viên
9 tháng 9 2021
\(\frac{3}{\sqrt{7}-1}+\frac{3}{\sqrt{7}+1}=\frac{3\left[\sqrt{7}+1+\sqrt{7}-1\right]}{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}=\frac{6\sqrt{7}}{6}=\sqrt{7}\)
\(\frac{3}{\sqrt{X}-1}-\frac{2}{\sqrt{X}+1}+\frac{X-7}{X-1}=\frac{3\left(\sqrt{X}+1\right)-2\left(\sqrt{X}-1\right)+X-7}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{X+\sqrt{X}-2}{\left(\sqrt{X}+1\right)\left(\sqrt{X}-1\right)}=\frac{\sqrt{X}+2}{\sqrt{X}+1}\)
VT
9 tháng 9 2021
TÍNH GIÁ TRỊ BIỂU THỨC:
\(\frac{3}{\sqrt{7}-1}\) + \(\frac{3}{\sqrt{7}+1}\)= \(\frac{3\left(\sqrt{7}+1\right)+3\left(\sqrt{7}-1\right)}{\left(\sqrt{7}-1\right)\left(\sqrt{7}+1\right)}\)= \(\frac{3\sqrt{7}+3+3\sqrt{7}-3}{6}\)=\(\frac{6\sqrt{7}}{6}\)=\(\sqrt{7}\)
RÚT GỌN BIỂU THỨC:
\(\frac{3}{\sqrt{X}-1}\)-\(\frac{2}{\sqrt{X}+1}\)+\(\frac{X-7}{X-1}\)
= \(\frac{3\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)-\(\frac{2\left(\sqrt{X}-1\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)+\(\frac{X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{3\sqrt{X}+3-2\sqrt{X}+2+X-7}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{X+\sqrt{X}-2}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\left(\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-1\right)\left(\sqrt{X}+1\right)}\)
= \(\frac{\sqrt{X}-2}{\sqrt{X}-1}\)
CHÚC EM HỌC TỐT!
KT
26 tháng 7 2018
\(A=4\sqrt{32}+2\sqrt{50}-8\sqrt{2}-2\sqrt{98}\)
\(=4\sqrt{16.2}+2\sqrt{25.2}-8\sqrt{2}-2\sqrt{49.2}\)
\(=16\sqrt{2}+10\sqrt{2}-8\sqrt{2}-14\sqrt{2}=4\sqrt{2}\)
\(B=\frac{1}{\sqrt{6}+\sqrt{10}}-\frac{1}{\sqrt{6}-\sqrt{10}}\)
\(=\frac{\sqrt{10}-\sqrt{6}}{\left(\sqrt{6}+\sqrt{10}\right)\left(\sqrt{10}-\sqrt{6}\right)}+\frac{\sqrt{6}+\sqrt{10}}{\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{6}+\sqrt{10}\right)}\)
\(=\frac{\sqrt{10}-\sqrt{6}}{4}+\frac{\sqrt{10}+\sqrt{6}}{4}\)
\(=\frac{2\sqrt{10}}{4}=\frac{\sqrt{10}}{2}=\sqrt{2,5}\)