\(\vec{978\sqrt[863]{35\sqrt{5}2}}\)= ?
\(y=\frac{1}{x^2+\sqrt{x}}\vec{\frac{^{ }\frac{\sqrt[]{}^2^{ }^{ }\frac{\frac{\vec{\vec{\vec{\vec{\vec{^2^2\sqrt[]{}\frac{\frac{\left(\sqrt[]{}\right)}{ }}{ }}}}}}}{ }}{ }}{ }}{ }}\)
viết lại hẳn hoi đi chứ ,cứ viết kiểu z có ma trả lời
cẩu thả
\(y=\frac{1}{x^2+\sqrt{x}}\int^{ }_{ }^{ }^2_{ }_{ }\vec{\vec{\vec{^2\frac{\frac{\frac{\sqrt[]{}\sqrt[]{}\sqrt[]{}\sqrt[]{}\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{ }{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}}}\)
Cho \(\vec{a} =(-1;3) ,\vec{b} = (2;1). \) Gọi 𝜑 là góc giữa hai vectơ \( \vec {a} và \vec {b}\). Khi đó, cos 𝜑 bằng:
A. \(\dfrac{1}{\sqrt{5}}\)
B. \(-\dfrac{1}{\sqrt{5}}\)
C. \(\dfrac{1}{5\sqrt{2}}\)
D. \(-\dfrac{1}{\sqrt{2}}\)
Nói thiệt là ko biết cái này làm sao lun QwQ
\(cos\varphi=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}=\frac{-1.2+3.1}{\sqrt{\left(-1\right)^2+3^2}.\sqrt{2^2+1^2}}=\frac{1}{5\sqrt{2}}\)
so sánh các số sau : \(a=\dfrac{35}{49};b=\sqrt{\dfrac{5^2}{7^2}};c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}};d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
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: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)
Rút gọn: \(\frac{\sqrt{12-2\sqrt{35}}+\sqrt{8-2\sqrt{15}}+\sqrt{5-2\sqrt{6}}}{\sqrt{12+2\sqrt{35}}-\sqrt{8+2\sqrt{15}}+\sqrt{5-2\sqrt{6}}}\)
1)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
2)\(\sqrt{35+12\sqrt{6}}-\sqrt{35-12\sqrt{6}}\)
3)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
\(y=\frac{1}{x^2+\sqrt{x}}\Delta\theta\lambda\vec{\cos^2\int^{ }_{ }\frac{\sqrt[]{}\sqrt{ }}{ }}\)
đơn giản lắm câu trả lời là tui ko biết nữa....
cos là kiến thức của lớp 9 mà...
So sánh các số sau:
a = \(\frac{35}{49}\)b = \(\sqrt{\frac{5^2}{7^2}}\)c = \(\frac{\sqrt{5^2+\sqrt{35^2}}}{\sqrt{7^2}+\sqrt{49^2}}\)d = \(\frac{\sqrt{5^2-\sqrt{35^2}}}{\sqrt{7^2}-\sqrt{49^2}}\)
So sánh các số sau:
a = 3549 b = √5272 c = √52+√352√72+√492 d = √52−√352√72−√492
=> A < B
bai nay minh chua hoc den nen khong the giai