Cho: \(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\). Tính: \(X=a^2+\sqrt{a^4+a^2+1}\)
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
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 9 2021
Tham khảo:
Tính giá trị biểu thức A = \(x^2+\sqrt{x^{^4}+x+1}\) với x =\(\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{\sqrt{2}}{... - Hoc24
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 8 2021
\(x>\dfrac{1}{2}\sqrt{1}-\dfrac{\sqrt{2}}{8}>0\)
\(x^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)+\dfrac{1}{32}-\dfrac{\sqrt{2}}{8}\sqrt{\sqrt{2}+\dfrac{1}{8}}\)
\(x^2=\dfrac{1}{16}+\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{8}\left(2x+\dfrac{\sqrt{2}}{4}\right)\)
\(x^2=\dfrac{1}{16}+\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}x-\dfrac{1}{16}=\dfrac{\sqrt{2}}{4}\left(1-x\right)\)
\(\Rightarrow x^4=\dfrac{1}{8}\left(x^2-2x+1\right)\)
\(\Rightarrow x^4+x+1=\dfrac{1}{8}\left(x^2-2x+1\right)+x+1=\dfrac{\left(x+3\right)^2}{8}\)
\(\Rightarrow A=x^2+\sqrt{\dfrac{\left(x+3\right)^2}{8}}=\dfrac{\sqrt{2}}{4}\left(1-x\right)+\dfrac{\sqrt{2}}{4}\left(x+3\right)=\sqrt{2}\)
5 tháng 9 2023
a: \(=\dfrac{2\left(\sqrt{2}+1\right)}{2-1}-\sqrt{\dfrac{3}{4}:\dfrac{3}{2}}+2\sqrt{2}\)
\(=2\sqrt{2}+2+2\sqrt{2}-\sqrt{\dfrac{1}{2}}\)
\(=4\sqrt{2}+2-\dfrac{\sqrt{2}}{2}=\dfrac{7}{2}\sqrt{2}+2\)
b: \(B=\left|3x\right|+x+\sqrt{x}\)
\(=3x+x+\sqrt{x}=4x+\sqrt{x}\)
17 tháng 5 2021
`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
17 tháng 5 2021
`a)A=\sqrt{4+2sqrt3}`
`=\sqrt{3+2sqrt3+1}`
`=sqrt{(sqrt3+1)^2}`
`=sqrt3+1`
`B)1/(2-sqrt3)+1/(2+sqrt3)`
`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`
`=2+sqrt3+2-sqrt3`
`=4`
`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`
`đk:x>=3`
`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`
`<=>2sqrt{x-3}=8`
`<=>sqrt{x-3}=4`
`<=>x-3=16`
`<=>x=19`
Vậy `S={19}`
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}}{3}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2-\sqrt{5}-2\)
\(=-\sqrt{5}\)
\(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\\ \Leftrightarrow a+\dfrac{\sqrt{2}}{8}=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}\\ \Leftrightarrow\left(a+\dfrac{\sqrt{2}}{8}\right)^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)\\ \Leftrightarrow a^2+\dfrac{a\sqrt{2}}{4}+\dfrac{1}{32}=\dfrac{\sqrt{2}}{4}+\dfrac{1}{32}\\ \Leftrightarrow a^2=\dfrac{\sqrt{2}-a\sqrt{2}}{4}=\dfrac{\sqrt{2}\left(1-a\right)}{4}\\ \Leftrightarrow a^4=\dfrac{a^2-2a+1}{8}\\ \Leftrightarrow a^4+a^2+1=\dfrac{a^2-2a+1}{8}+a^2+1=\dfrac{9a^2-2a+9}{8}\)
\(\Leftrightarrow a^2+\sqrt{a^4+a^2+1}=a^2+\dfrac{\sqrt{9a^2-2a+9}}{2\sqrt{2}}=\dfrac{2a^2\sqrt{2}+\sqrt{9a^2-2a+9}}{2\sqrt{2}}\)