Rút gọn các biểu thức :
a) Rút gọn biểu thức : \(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}\)
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LM
2
B
19 tháng 7 2019
\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{3^2-\sqrt{2}^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\sqrt{3}^2}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)
\(=3-\sqrt{2}-1-\sqrt{3}\)
\(=2-\sqrt{2}-\sqrt{3}\)
NN
18 tháng 10 2020
\(A=\frac{7}{3+\sqrt{2}}+\frac{2}{1-\sqrt{3}}=\frac{7\left(3-\sqrt{2}\right)}{3^2-\left(\sqrt{2}\right)^2}+\frac{2\left(1+\sqrt{3}\right)}{1^2-\left(\sqrt{3}\right)^2}\)
\(=\frac{7\left(3-\sqrt{2}\right)}{9-2}+\frac{2\left(1+\sqrt{3}\right)}{1-3}=\frac{7\left(3-\sqrt{2}\right)}{7}+\frac{2\left(1+\sqrt{3}\right)}{-2}\)
\(=\left(3-\sqrt{2}\right)-\left(1+\sqrt{3}\right)=3-\sqrt{2}-1-\sqrt{3}=2-\sqrt{2}-\sqrt{3}\)
MU
3
5 tháng 7 2019
\(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}\)
\(=\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{3+\sqrt{7}-3+\sqrt{7}}{9-7}\)
\(=\frac{2\sqrt{7}}{2}\)
\(=\sqrt{7}\)
phần B là gì cơ?
28 tháng 7 2017
\(P=\frac{x+2}{\sqrt{x}^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
2,
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}+\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{7\sqrt{7}}{7}\)
\(A=\frac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}+\frac{\left(\sqrt{2}+1\right)}{2-1}-\sqrt{7}\)
\(A=\sqrt{7}-\sqrt{2}+\sqrt{2}+1-\sqrt{7}=1\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
10 tháng 11 2019
\(\frac{\sqrt{3}+\sqrt{7}}{\sqrt{3}-\sqrt{7}}+\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)+\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)^2+\left(\sqrt{3}-\sqrt{7}\right)^2}{3-7}\)
\(=\frac{3+2\sqrt{3}.\sqrt{7}+7+3-2\sqrt{3}.\sqrt{7}+7}{-4}\)
\(=\frac{3+7+3+7}{-4}\)
\(=\frac{20}{-4}=-5\)
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!
HT
3 tháng 9 2020
Ta đặt: \(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\right)^2\)
<=> \(A^2=\sqrt{7}-\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\sqrt{7}+\sqrt{3}\)
<=> \(A^2=2\sqrt{7}-2\sqrt{7-3}\)
<=> \(A^2=2\sqrt{7}-2\sqrt{4}=2\left(\sqrt{7}-2\right)\)
=> \(A=\sqrt{2\left(\sqrt{7}-2\right)}\)
Thay vào ta được:
\(\frac{\sqrt{2\left(\sqrt{7}-2\right)}}{\sqrt{\sqrt{7}-2}}=\sqrt{2}\)
\(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{3-\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{3+\sqrt{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)
a, \(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt[]{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)