Rút gọn biểu thức :
\(A=\dfrac{1}{2\sqrt{2}}-\dfrac{2}{\sqrt{2}+1}-\dfrac{7}{2\sqrt{2}+4}\)
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AH
Akai Haruma
Giáo viên
18 tháng 9 2023
Lời giải:
a.
\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)
$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.
$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$
$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.
$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$
$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$
$=1(3+\sqrt{2})=3+\sqrt{2}$
4 tháng 10 2021
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
1) \(A=\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)^2-2\)
\(A=\sqrt{x}\cdot\sqrt{x}+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-x+2\sqrt{x}-1-2\)
\(A=3\sqrt{x}-3\)
Thay \(x=9\) vào A ta có:
\(A=3\cdot\sqrt{9}-3=3\cdot3-3=9-3=6\)
6 tháng 9 2023
a: \(=\dfrac{2\sqrt{2}+3+2\sqrt{2}-3}{8-9}\)
\(=\dfrac{4\sqrt{2}}{-1}=-4\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}-\sqrt{7}\right)+\sqrt{2}\left(2\sqrt{2}+\sqrt{7}\right)}{8-7}\)
\(=4-\sqrt{14}+4+\sqrt{14}=8\)
c: \(=\dfrac{2+\sqrt{5}-2\left(2-\sqrt{5}\right)}{-1}=\dfrac{2+\sqrt{5}-4+2\sqrt{5}}{-1}\)
\(=-3\sqrt{5}+2\)
19 tháng 8 2021
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
19 tháng 8 2021
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
AT
11 tháng 6 2021
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
11 tháng 6 2021
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
17 tháng 1 2022
, \(A=\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
\(=\frac{2\sqrt{7}-10-6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{3}-\frac{5\left(4-\sqrt{7}\right)}{9}\)
\(=\frac{-16+4\sqrt{7}}{4}+\frac{18\sqrt{7}+36-20+5\sqrt{7}}{9}=-4+\sqrt{7}+\frac{23\sqrt{7}+16}{9}\)
b,\(B=\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2\left(\sqrt{6}+2\right)+2\left(\sqrt{6}-2\right)}{2}+\frac{5\sqrt{6}}{6}\)
\(=\frac{12\sqrt{6}+5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
H9
HT.Phong (9A5)
CTVHS
2 tháng 11 2023
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{6}\)
c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)
\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)
\(=\dfrac{3}{4}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)
\(=\sqrt[3]{\dfrac{54}{-2}}\)
\(=\sqrt[3]{-27}\)
\(=\sqrt[3]{\left(-3\right)^3}\)
\(=-3\)
7 tháng 11 2023
a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)
\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)
\(=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}+1-\sqrt{2}+1=2\)
8 tháng 6 2021
Đk:\(a>2\)
\(\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đặt \(b=\sqrt{a-2}\Leftrightarrow a=b^2+2\)
Biểu thức \(\Leftrightarrow\dfrac{b+2}{3}\left(\dfrac{b}{3+b}+\dfrac{b^2+2+7}{11-b^2-2}\right):\left(\dfrac{3b+1}{b^2-3b}-\dfrac{1}{b}\right)\)
\(=\dfrac{b+2}{3}\left[\dfrac{b}{3+b}-\dfrac{b^2+9}{b^2-9}\right]:\left[\dfrac{3b+1}{b\left(b-3\right)}-\dfrac{b-3}{b\left(b-3\right)}\right]\)
\(=\dfrac{b+2}{3}.\dfrac{b\left(b-3\right)-b^2-9}{\left(b-3\right)\left(3+b\right)}:\dfrac{3b+1-\left(b-3\right)}{b\left(b-3\right)}\)
\(=\dfrac{b+2}{3}.\dfrac{-3\left(b+3\right)}{\left(b-3\right)\left(3+b\right)}.\dfrac{b\left(b-3\right)}{2\left(b+2\right)}\)
\(=-\dfrac{b}{2}\)
\(=\dfrac{\sqrt{a-2}}{-2}\)
\(A=\dfrac{1}{2\sqrt{2}}-\dfrac{2}{\sqrt{2}+1}-\dfrac{7}{2\sqrt{2}+4}\)
\(=\dfrac{\sqrt{2}}{4}-2\left(\sqrt{2}-1\right)-\dfrac{7\left(2\sqrt{2}-4\right)}{-8}\)
\(=\dfrac{\sqrt{2}}{4}-2\sqrt{2}+2-\dfrac{7\cdot2\left(\sqrt{2}-2\right)}{-8}\)
\(=-\dfrac{7\sqrt{2}}{4}+2-\dfrac{7\left(\sqrt{2}-2\right)}{-4}\)
\(=-\dfrac{7\sqrt{2}}{4}+2-\dfrac{7\sqrt{2}-14}{-4}\)
\(=-\dfrac{7\sqrt{2}}{4}+2-\dfrac{7\sqrt{2}-14}{4}\)
\(=\dfrac{-7\sqrt{2}+7\sqrt{2}-14}{4}+2\)
\(=-\dfrac{14}{4}+2\)
\(=-\dfrac{3}{2}\)
Ồ! Bn í thật nhanh!