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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
\(x>\dfrac{1}{2}\sqrt{1}-\dfrac{\sqrt{2}}{8}>0\)
\(x^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)+\dfrac{1}{32}-\dfrac{\sqrt{2}}{8}\sqrt{\sqrt{2}+\dfrac{1}{8}}\)
\(x^2=\dfrac{1}{16}+\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{8}\left(2x+\dfrac{\sqrt{2}}{4}\right)\)
\(x^2=\dfrac{1}{16}+\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{2}}{4}x-\dfrac{1}{16}=\dfrac{\sqrt{2}}{4}\left(1-x\right)\)
\(\Rightarrow x^4=\dfrac{1}{8}\left(x^2-2x+1\right)\)
\(\Rightarrow x^4+x+1=\dfrac{1}{8}\left(x^2-2x+1\right)+x+1=\dfrac{\left(x+3\right)^2}{8}\)
\(\Rightarrow A=x^2+\sqrt{\dfrac{\left(x+3\right)^2}{8}}=\dfrac{\sqrt{2}}{4}\left(1-x\right)+\dfrac{\sqrt{2}}{4}\left(x+3\right)=\sqrt{2}\)
`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}`
`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}`
\(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\\ \Leftrightarrow a+\dfrac{\sqrt{2}}{8}=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}\\ \Leftrightarrow\left(a+\dfrac{\sqrt{2}}{8}\right)^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)\\ \Leftrightarrow a^2+\dfrac{a\sqrt{2}}{4}+\dfrac{1}{32}=\dfrac{\sqrt{2}}{4}+\dfrac{1}{32}\\ \Leftrightarrow a^2=\dfrac{\sqrt{2}-a\sqrt{2}}{4}=\dfrac{\sqrt{2}\left(1-a\right)}{4}\\ \Leftrightarrow a^4=\dfrac{a^2-2a+1}{8}\\ \Leftrightarrow a^4+a^2+1=\dfrac{a^2-2a+1}{8}+a^2+1=\dfrac{9a^2-2a+9}{8}\)
\(\Leftrightarrow a^2+\sqrt{a^4+a^2+1}=a^2+\dfrac{\sqrt{9a^2-2a+9}}{2\sqrt{2}}=\dfrac{2a^2\sqrt{2}+\sqrt{9a^2-2a+9}}{2\sqrt{2}}\)