√(10+2√6-2√10-2√15)
So sánh :
a) A= 10/2^7+ 10/2^6
B= 11/2^7+9/2^6
b) A= -7/ 10^2005+ -15/ 10 ^2006
B= -15/10^2005 + -7/10^2006
Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{5}{2^6}+\frac{10}{2^6}=\frac{15}{2^6}\)
Lại có B = \(\frac{11}{2^7}+\frac{9}{2^6}=\frac{5,5}{2^6}+\frac{9}{2^6}=\frac{14,5}{2^6}\)
Vì \(\frac{15}{2^6}>\frac{14,5}{2^6}\Rightarrow A>B\)
b) Ta có : \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\)
Lại có B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=\frac{-157}{10^{2006}}\)
Vì \(\frac{-85}{10^{2006}}>\frac{-157}{10^{2006}}\Rightarrow A< B\)
nhờ anh chị giải giúp em !
10)\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
11)\(\sqrt{10-2\sqrt{6}+2\sqrt{10}-2\sqrt{15}}\)
√(10-2√6-2√10+2√15)
\(\sqrt{10-2\sqrt{6}-2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2-2\sqrt{2.3}-2\sqrt{2.5}+2\sqrt{3.5}}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(=\sqrt{10-2\cdot\sqrt{2}\cdot\sqrt{3}-2\cdot\sqrt{2}\cdot\sqrt{5}+2\cdot\sqrt{3}\cdot\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
2 nhân căn 15 -2 nhân căn 10 +căn 6 - 3 tất cả trên 2 nhân căn 15- 2 nhân căn 10 +căn 3 + căn 6
Phép tính:
\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{6}=1421411372\)
\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{3}+\sqrt{6}=5602951922\)
P/s: Em ko biết đúng hay sai đâu mới lớp 4 thôi à
\(\sqrt{10+2\sqrt{6+2\sqrt{10+2\sqrt{15}}}}\)
\(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}-\sqrt{6}}\)
\(\dfrac{2\sqrt{15}-2\sqrt{10}-3+\sqrt{6}}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}=-\sqrt{3}-\sqrt{6}+\sqrt{2}+2\)
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`
Rút gọn: 5^2*6^9*10+6^5*2^3*15^3/5^2*6^8*10-2*6^8*10^3
Giúp e vs e đang cần gấp :))
√(10-2√6+2√10-2√15)=?
\(=\sqrt{2+3+5-2\cdot\sqrt{2}\cdot\sqrt{3}+2\sqrt{2}\cdot\sqrt{5}-2\cdot\sqrt{3}\cdot\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}-\sqrt{5}\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{3}\)