(5\(\sqrt{7}\)+ 7\(\sqrt{5}\)) / \(\sqrt{35}\)
xin mn giup em vs ak
Những câu hỏi liên quan
\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+.........+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}\sqrt{36}}\)
giúp mik vs ạ mình cảm ơn ạ!!
đoạn cuối thiếu dấu"+"
\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)
\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)
\(A=\sqrt{36}-\sqrt{4}=6-2=4\)
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\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)
\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)
\(=6-2=4\)
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giúp em với ạ
\(\sqrt{5
+2\sqrt{ }6}\)
\(\sqrt{12+2\sqrt{ }35}-\sqrt{12-2\sqrt{ }35}\)
\(\sqrt{16+6\sqrt{ }7}\)
\(\sqrt{31-12\sqrt{ }3}\)
\(\sqrt{27+10\sqrt{ }2}\)
\(\sqrt{14+6\sqrt{ }5}\)
a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)
b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)
c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)
d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)
e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)
f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
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Mn giúp mình vs.. mình đang cần gấp ạ, cảm ơn mn nhiều lắm.. hjx
sqrt{46-6sqrt{ }5}-sqrt{29-12sqrt{ }5}
sqrt{7-3sqrt{ }5}
left(3-sqrt{ }2right)sqrt{11+6sqrt{ }2}
sqrt{4+sqrt{ }15}
left(sqrt{ }5+sqrt{ }7right)sqrt{12-2sqrt{ }35}
sqrt{17-12sqrt[]{}2}
sqrt{14-6sqrt{ }5}
sqrt{27-12sqrt{ }5}
Đọc tiếp
\(\)Mn giúp mình vs.. mình đang cần gấp ạ, cảm ơn mn nhiều lắm.. hjx
\(\sqrt{46-6\sqrt{ }5}-\sqrt{29-12\sqrt{ }5}\)
\(\sqrt{7-3\sqrt{ }5}\)
\(\left(3-\sqrt{ }2\right)\sqrt{11+6\sqrt{ }2}\)
\(\sqrt{4+\sqrt{ }15}\)
\(\left(\sqrt{ }5+\sqrt{ }7\right)\sqrt{12-2\sqrt{ }35}\)
\(\sqrt{17-12\sqrt[]{}2}\)
\(\sqrt{14-6\sqrt{ }5}\)
\(\sqrt{27-12\sqrt{ }5}\)
\(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\\ =\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}+1}-\sqrt{\left(2\sqrt{5}\right)^2-2.3.2\sqrt{5}+3^2}\\ =3\sqrt{5}-1-2\sqrt{5}+3=\sqrt{5}+2\)
Mấy câu sau tương tự.
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so sánh các số sau : \(a=\dfrac{35}{49};b=\sqrt{\dfrac{5^2}{7^2}};c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}};d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
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\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
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Giải giúp mik mấy bài pt vô tỉ vs ak!
1)\(x=\sqrt{5-x}\sqrt{6-x}+\sqrt{6-x}\sqrt{7-x}+\sqrt{7-x}\sqrt{5-x}\)
2)\(2x-1=\sqrt{2-x}\sqrt{10-4x}+\sqrt{5-2x}\sqrt{6-2x}+2\sqrt{3-x}\sqrt{2-x}\)
\(\left(5\sqrt{7}+7\sqrt{5}\right):\sqrt{35}\)
`(5sqrt7+7sqrt5):sqrt{35}`
`=(sqrt{5}.sqrt{7}.sqrt{5}+sqrt{5}.sqrt{7}.sqrt{7}):sqrt{35}`
`=sqrt{5}.sqrt{7}(sqrt5+sqrt7):sqrt{35}`
`=sqrt{35}(sqrt5+sqrt7):sqrt{35}`
`=sqrt5+sqrt7`
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Ta có : \(\left(\sqrt{5}\sqrt{5}\sqrt{7}+\sqrt{7}\sqrt{7}\sqrt{5}\right):\sqrt{35}\)
\(=\left(\sqrt{5}\sqrt{35}+\sqrt{7}\sqrt{35}\right):\sqrt{35}\)
\(=\sqrt{5}+\sqrt{7}\)
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\(y=\frac{\sqrt{2}}{\sqrt{5}+\sqrt{7}}\cdot\sqrt{\frac{7\sqrt{5}+5\sqrt{7}}{7\sqrt{5}-5\sqrt{7}}}\) ai giải giùm câu này vs
m.n giúp mk vs ak
mk xin cảm ơn ak
a.10x-9y =1
15x +21y=36
b.\(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
c.2\(\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)
d.(\(\sqrt{5}+\sqrt{2}\))(\(3\sqrt{2}-1\))
\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
