tính :\(\sqrt{7+4\sqrt{5}}+\sqrt{7-4\sqrt{5}}\)
giúp em vs ạ
1) thực hiện phép tính
d)\(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
e) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ạ mk cần gấp
\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+.........+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}\sqrt{36}}\)
giúp mik vs ạ mình cảm ơn ạ!!
đoạn cuối thiếu dấu"+"
\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)
\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)
\(A=\sqrt{36}-\sqrt{4}=6-2=4\)
\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)
\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)
\(=6-2=4\)
\(x-4\sqrt{x}+6=\sqrt{x-3}=\sqrt{5-x}\)
\(\sqrt{x+7}+\sqrt{x+2}=5\)
giúp em với ạ
1) rút gọn
A= \(3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
B= \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
C= \(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
Giúp mk vs ạ mk cần gấp
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)
\(=\dfrac{3}{2}\)
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+3+\sqrt{3}=5\)
Rút gọn biểu thức
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
giúp em với ạ, em cảm ơn!
a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)
b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)
a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}+1\)
=3
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
giúp vs ạ !!!
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+2+2-\sqrt{3}\)
\(=4\)
Còn cách nữa là bình phương
Đag làm thì ấn nhầm trả lời .V
Cách bình phương đây
\(B=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(\Rightarrow B^2=7+4\sqrt{3}+2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+7-4\sqrt{3}\)
\(=14+2\sqrt{49-48}\)
\(=14+2\)
\(=16\)
\(\Rightarrow B=\sqrt{16}=4\)
\(B=\sqrt{\left(2\right)^2+4\sqrt{3}+3}+\sqrt{\left(2\right)^2-4\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(2+\sqrt{3}+2-\sqrt{3}=4\)
1) thực hiện phép tính:
a) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ah mk đang cần gấp lắm
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
\(\sqrt{x-2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}=5}\)
Giúp em câu này với ạ!
ĐKXĐ: \(x\ge2\)
\(\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=5\)
- TH1:
\(2\le x\le6\)
\(\Rightarrow2-\sqrt{x-2}+3-\sqrt{x-2}=5\)
\(\Leftrightarrow\sqrt{x-2}=0\Rightarrow x=2\)
TH2: \(6\le x\le11\)
\(\Rightarrow\sqrt{x-2}-2+3-\sqrt{x-2}=5\Leftrightarrow1=5\) (vô nghiệm)
TH3: \(x>11\)
\(\Rightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=5\)
\(\Leftrightarrow2\sqrt{x-2}=10\Rightarrow\sqrt{x-2}=5\)
\(\Rightarrow x=27\)
Vậy \(\left[{}\begin{matrix}x=2\\x=27\end{matrix}\right.\)
(1) rút gọn biểu thức:
a) A= \(3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
b) B= \(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
c) C= \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
d) D= \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
giúp mk vs ạ mai mk hc rồi
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
Thực hiện phép tính sau:
\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
Giúp em với mọi người oi! em đang rất cần