Rút gọn biểu thức: \(P=\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}+\frac{\left(\sqrt{5}-1\right).\sqrt[3]{2+\sqrt{5}}}{\sqrt{28}-10\sqrt{3}+\sqrt{3}}\)
Giúp mk nha!
Rút gọn biểu thức
A= \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
rút gọn biểu thức
a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
c) \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
d) \(\frac{3}{3+2\sqrt{3}}+\frac{3}{3-2\sqrt{3}}\)
e) \(\sqrt{20}-15\sqrt{\frac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}\)
a) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)
\(=3.2+4=6+4=10\)
b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)
c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)
d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)
e) Tự làm.
Cái câu c đánh nhầm:
\(=5^2-2.5.\sqrt{3}+3=\left(5-\sqrt{3}\right)^2\) nha!
a) Rút gọn biểu thức:\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}\)
b) Tìm giá trị nhỏ nhất của biểu thức B=\(x^2-x\sqrt{3}+1\)
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
mấy bạn 2k2 giúp mình với mk cần gấp, thanks nhiều
Rút gọn biểu thức
a) A=\(\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}-1}-\frac{10+2\sqrt{5}}{\sqrt{5+}1}+\sqrt{5}-1\)
b) B=\(\sqrt{\left(1-\sqrt{2014}\right)^2}.\sqrt{2015+2\sqrt{2014}}\)
c) C=\(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}.\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
Rút gọn các biểu thức sau :
a,\(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
b \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right).\sqrt{5}\)
c,\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
d \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
e \(\sqrt{2+\sqrt{3}}\times\sqrt{2-\sqrt{3}}\)
g \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
h \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\times\sqrt{6}\)
i \(\left(1+\sqrt{2}+\sqrt{3}\right)\times\left(1+\sqrt{2}-\sqrt{3}\right)\)
k \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
l \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
m \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
tu lam di cau nao kho thi hoi hoi vay ko ai tra loi cho dau
cau e)
\(A=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)(suy ra A>=0)
\(A^2=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)
\(A^2=1\)
A=1
(bai toan co nhieu cach)
cau m)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\)
\(=1\)
cau G)
\(=\frac{5\sqrt{7}}{\sqrt{35}}-\frac{7\sqrt{5}}{\sqrt{35}}+\frac{2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{5}{\sqrt{5}}-\frac{7}{\sqrt{7}}+2\sqrt{2}\)
\(=\sqrt{5}-\sqrt{7}+2\sqrt{2}\)
* rút gọn biểu thức:
a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)
c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)
= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)
= \(\sqrt{15}.0\)
= 0
b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)
= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)
= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)
= \(\sqrt{5}.\sqrt{5}\)
= 5
c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
= \(\sqrt{5}+\sqrt{3}\)
d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)
= \(\left(2+\sqrt{5}-2+\sqrt{5}\right)\left(2+\sqrt{5}+2-\sqrt{5}\right)\)
= \(2\sqrt{5}.4\)
= \(8\sqrt{5}\)
e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)
= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)
= \(\sqrt{3}.\frac{20}{3}\)
= \(\frac{20\sqrt{3}}{3}\)
a, 320+160−2115
b, (15−3+15+3).5
c, (53+35):15
d, (2+5)2−(2+5)2
e, 1348+375−27−10113
Rút gọn biểu thức
1)\(\frac{15}{3\sqrt{20}}\)
2) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{2}-\sqrt{5}}\)
3) \(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}\)
4) \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
5) \(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right)\sqrt{5}\)
6)\(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
7) \(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\right):2\sqrt{5}\)
8)\(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
9) \(2\sqrt{3}\left(2\sqrt{6}-\sqrt{3}+1\right)\)
10) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
11) \(\sqrt{\sqrt{10}+1}.\sqrt{\sqrt{10}-1}\)
12) \(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
13) \(\sqrt{\frac{3}{4}}+\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{12}}\)
14) \(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}\right)\sqrt{6}\)
15 ) \(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
16) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
17) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
Rút gọn
G = \(\frac{3-2\sqrt{3}}{\sqrt{3}}+\frac{6}{3+\sqrt{3}}\)
H= \(\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}\)
i = \(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
K = \(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
N= \(\left(1-\frac{\sqrt{3}-1}{2}\right):\left(\frac{\sqrt{3}-1}{2}+2\right)\)