Thực hiện phép tính:
\(\left(2\sqrt{2}-\sqrt{5}+3\sqrt{2}\right)\left(\sqrt{18}-\sqrt{20}+2\sqrt{2}\right)\)
\(\left(1+\sqrt{2}-\sqrt{5}\right)\left(1+\sqrt{2}+\sqrt{5}\right)\)
* Thực hiện phép tính.
a.\(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
b.\(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
c.\(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right).\dfrac{1}{2-\sqrt{5}}\)
d.\(\sqrt{\left(2-\sqrt{5}\right)^2-\sqrt{5}}\)
a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)
= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)
= \(-33\sqrt{2}\)
b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)
= \(10-2\sqrt{21}+14\sqrt{21}\)
= \(10+12\sqrt{21}\)
trục căn thức và thực hiện phép tính
1) \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2) \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
(1) thực hiện phép tính:
a) \(\sqrt{5}.\left(\sqrt{20}-3\right)+\sqrt{45}\)
b) \(\sqrt{\left(5-\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
c) \(\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{3-\sqrt{5}}\)
giúp mk vs ạ mai mk học rồi
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
* Thực hiện phép tính:
a. \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
b. \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
c. \(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{2-\sqrt{5}}\)
* Tìm x, biết:
a. \(\sqrt{\left(2x+3\right)^2}=8\)
b. \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
c. \(\sqrt{9x-9}+1=13\)
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
Bài 1: Thực hiện phép tính :
a,\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
b,\(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)
c,\(\left(2\sqrt{18}-5\sqrt{32}+6\sqrt{2}\right):\sqrt{2}\)
a: Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)
\(=0\)
b: Ta có: \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)
\(=5+7-1\)
=11
* Thực hiện phép tính
a. \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(2\left(-5\right)\right)^2}\)
b. \(\dfrac{\sqrt{270}}{\sqrt{30}}-\sqrt{1,8}.\sqrt{20}\)
a, \(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(2\left(-5\right)\right)^2}\)
\(=\left|3-\sqrt{2}\right|+\sqrt{\left(-10\right)^2}\)
\(=3-\sqrt{2}+\left|-10\right|\)
\(=3-\sqrt{2}+10\)
\(=13-\sqrt{2}\)
b, \(\dfrac{\sqrt{270}}{\sqrt{30}}-\sqrt{1,8}.\sqrt{20}\)
\(=\sqrt{9}-\sqrt{1,8.20}\)
\(=3-\sqrt{36}\)
\(=3-6\)
\(=-3\)
Thực hiện phép tính (rút gọn biểu thức)
a)\(\sqrt{\left(3+\sqrt{2}\right)^2}\)-\(\sqrt{\left(3-2\sqrt{2}\right)^2}\)
b) \(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}\)-\(\sqrt{\left(\sqrt{7}+2\sqrt{2}\right)^2}\)
c)\(\sqrt{\left(3+\sqrt{5}\right)^2}\)+\(\sqrt{\left(3-\sqrt{5}\right)^2}\)
d) \(\sqrt{\left(2-\sqrt{3}\right)^2}\)-\(\sqrt{\left(2+\sqrt{3}\right)^2}\)
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
Thực hiện phép tính:
a) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
b) \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
Thực hiện các phép tính (không được ghi mỗi kết quả không, phải giải chi tiết)
A = \(2\sqrt{10}.3\sqrt{8}.2\)
B = \(\sqrt{20}\left(2\sqrt{3}-\sqrt{5}\right)\)
C = \(\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)\)
a: \(=12\sqrt{80}=48\sqrt{5}\)
b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)
c: =20-9=11