Cho mình hỏi tại sao 5\(\sqrt{2}\)= \(\sqrt{50}\)ạ
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KN
21 tháng 7 2019
Có: \(\sqrt{a}\sqrt{b}=\sqrt{âb}\)
\(\Rightarrow\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)
\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)
\(=2\sqrt{6}-1\)
18 tháng 7 2023
\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)
MN
14 tháng 7 2021
Bài 1 :
\(a.\sqrt{x^2-1}\)
\(ĐK:\)
\(x^2-1\ge0\)
\(\Leftrightarrow x^2\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
Bài 2 :
\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)
\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)
\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)
\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)
\(=-27\sqrt{2}-6+4\sqrt{3}\)
2
1
19 tháng 7 2019
\(\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1^2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=|\sqrt{5}-1|\)
\(=\sqrt{5}-1\)
_Vi hạ_
12 tháng 9 2021
=\(\left(3\sqrt{3}-3\sqrt{3}+2\sqrt{6}\right):3\sqrt{3}\)
\(=1-\dfrac{\sqrt{6}}{2}+\dfrac{2\sqrt{2}}{3}\)
=\(\dfrac{6}{6}-\dfrac{3\sqrt{6}}{6}+\dfrac{4\sqrt{2}}{6}\)
=\(\dfrac{6+\sqrt{6}}{6}\)
27 tháng 7 2021
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)
\(\Leftrightarrow A^3=4-3A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)
\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
HP
31 tháng 8 2021
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
DH
1 tháng 11 2017
\(\dfrac{\sqrt{AC^2+\left(v_n.\dfrac{t_1}{2}\right)^2}}{v}=1\)
\(\Rightarrow AC^2+v^2_n.\dfrac{t^2_1}{4}=v^2\)
\(\Rightarrow v_n^2.\dfrac{t^2_1}{4}=v^2-AC^2\)
\(\Rightarrow t_1^2=\dfrac{4\left(v^2-AC^2\right)}{v_n^2}\Rightarrow t_1=\dfrac{2\sqrt{v^2-AC^2}}{v_n}\)
UK
2 tháng 11 2017
Theo tao muốn chuyển đổi vậy thì :
\(\dfrac{\sqrt{AC^2+\left(v_n.\dfrac{t_1}{2}\right)^2}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{AC^2+v_n^2.\dfrac{t_1^2}{4}}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{\dfrac{4AC^2+\left(v_n.t_1\right)^2}{4}}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{4AC^2+\left(v_n.t_1\right)^2}}{2v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow t_1=\dfrac{\sqrt{4AC^2+\left(v_n.t_1\right)^2}}{v}\)
\(\Leftrightarrow t_1v=\sqrt{4AC^2+\left(v_n.t_1\right)^2}\)
\(\Leftrightarrow t_1^2.v^2=4AC^2+v_n^2.t^2_1\)
\(\Leftrightarrow t_1^2\left(v^2-v^2_n\right)=4AC^2\)
\(\Leftrightarrow t_1^2=\dfrac{4AC^2}{v^2-v_n^2}\)
\(\Leftrightarrow t_1=\dfrac{2AC}{\sqrt{v^2-v_n^2}}\)
Are you OK??? :D
ND
Nguyễn Đức Trí
VIP
16 tháng 8 2023
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)
Lời giải:
Bạn xem lại công thức khai phương lớp 9
$5\sqrt{2}=\sqrt{5^2}.\sqrt{2}=\sqrt{25}.\sqrt{2}=\sqrt{25.2}=\sqrt{50}$