Thực hiện phép tính (làm chi tiết):
3) \(\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
I
1
O
19 tháng 7 2023
\(\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\\ =\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\\ =\sqrt{10}\cdot\sqrt{3+\sqrt{5}}-\sqrt{2}\cdot\sqrt{3+\sqrt{5}}\\ =\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(5+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\\ =5+\sqrt{5}-\sqrt{5}-1\\ =4\)
25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
27 tháng 7 2021
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
10 tháng 5 2023
a: \(=12\sqrt{80}=48\sqrt{5}\)
b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)
c: =20-9=11
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)
\(=\sqrt{2\cdot\left(4+\sqrt{7}\right)}+\sqrt{2\cdot\left(4-\sqrt{7}\right)}\)
\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)
\(=\sqrt{7}+1+\sqrt{7}-1\)
\(=2\sqrt{7}\)
b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)
\(=-\dfrac{2}{\sqrt{2}}\)
\(=-\sqrt{2}\)
LP
2
15 tháng 7 2019
TL:
\(\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
\(=\frac{8-3\sqrt{7}-8-3\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
\(=\frac{-6\sqrt{7}}{\sqrt{8-3\sqrt{7}}+\sqrt{8+3\sqrt{7}}}\)
15 tháng 7 2019
Cho \(A=\sqrt{8-3\sqrt{7}}-\sqrt{8+3\sqrt{7}}\)
CACH 1 : \(\Rightarrow A\sqrt{2}=\sqrt{16-6\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(\Rightarrow A\sqrt{2}=\sqrt{9-2.3.\sqrt{7}+7}-\sqrt{9+2.3.\sqrt{7}+7}\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(3-\sqrt{7}\right)^2}-\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(\Rightarrow A\sqrt{2}=|3-\sqrt{7}|-|3+\sqrt{7}|\)
\(\Rightarrow A\sqrt{2}=3-\sqrt{7}-3-\sqrt{7}=-2\sqrt{7}=-\sqrt{28}\)
\(\Rightarrow A=-\sqrt{14}\)
CACH 2 : \(A^2=8-3\sqrt{7}+8+3\sqrt{7}-2.\sqrt{8^2-\left(3\sqrt{7}\right)^2}\)
\(\Rightarrow A^2=16-2\sqrt{64-63}=16-2=14\)
\(\Rightarrow A=\sqrt{14}\) hoặc \(A=-\sqrt{14}\)
Mà \(8+3\sqrt{7}>8-3\sqrt{7}\) \(\Rightarrow\sqrt{8+3\sqrt{7}}>\sqrt{8-3\sqrt{7}}\)
Vây A âm \(\Rightarrow A=-\sqrt{14}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
Lời giải:
\(A=\sqrt{8-3\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
$A\sqrt{2}=\sqrt{16-6\sqrt{7}}+\sqrt{8-2\sqrt{7}}$
$=\sqrt{(3-\sqrt{7})^2}+\sqrt{(\sqrt{7}-1)^2}$
$=|3-\sqrt{7}|+|\sqrt{7}-1|$
$=3-\sqrt{7}+\sqrt{7}-1=2$