kết pủa phép tính \(\sqrt{\left(\sqrt{7}+3\right)^2}-\sqrt{\left(\sqrt{7}-3\right)^2}\)
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AH
Akai Haruma
Giáo viên
26 tháng 8 2023
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.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
25 tháng 9 2021
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
25 tháng 9 2021
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
28 tháng 6 2021
`A=sqrt{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`
`=5-sqrt3+2-sqrt3`
`=7-2sqrt3`
`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`
`=3-sqrt2-(sqrt2-1)`
`=4-2sqrt2`
`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`
`=3+sqrt7-(sqrt7-2)`
`=5`
`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`
`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`
`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`
`=sqrt3-1+2+sqrt3=1+2sqrt3`
28 tháng 6 2021
\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)
\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)
\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}=\left|\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)
\(=\sqrt{3}-1+2+\sqrt{3}=1+2\sqrt{3}\)
H
24 tháng 12 2023
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
24 tháng 12 2023
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
19 tháng 6 2017
2\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(14-\sqrt{84}+7-\sqrt{84}\)
= 21
26 tháng 9 2021
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: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
26 tháng 9 2021
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
3 tháng 8 2016
a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)
b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)
3 tháng 8 2016
a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
\(=\sqrt{7}+3-3+\sqrt{7}=2\sqrt{7}\)