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Ta có : \(a=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\Leftrightarrow8a=4\sqrt{\sqrt{2}+\frac{1}{8}}-\sqrt{2}\Leftrightarrow8a+\sqrt{2}=4\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(\Leftrightarrow\left(8a+\sqrt{2}\right)^2=16\left(\sqrt{2}+\frac{1}{8}\right)\) \(\Leftrightarrow64a^2+16\sqrt{2}a+2=16\left(\sqrt{2}+\frac{1}{8}\right)\Leftrightarrow64a^2+16\sqrt{2}a+2=16\sqrt{2}+2\)
\(\Leftrightarrow4a^2+\sqrt{2}a=\sqrt{2}\Leftrightarrow4a^2=\sqrt{2}-\sqrt{2}a\)
Đặt \(Y=\sqrt{a^4+a+1}-a^2\) \(\Rightarrow XY=a+1\Leftrightarrow X.\left(-Y\right)=-\left(a+1\right)\) (1)
\(X+\left(-Y\right)=2a^2=\frac{\sqrt{2}-\sqrt{2}a}{2}=\frac{1-a}{\sqrt{2}}\) (2)
Từ (1) và (2) suy ra X và Y là hai nghiệm của phương trình \(t^2+\frac{1-a}{\sqrt{2}}.t-\left(a+1\right)=0\)
Giải phương trình trên được \(t_1=-\sqrt{2}\) ; \(t_2=-\frac{x+1}{\sqrt{2}}\)
Suy ra : \(X=\sqrt{2}\) (vì X > 0)
CM: \(a=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{\sqrt{2}}{8}\Rightarrow a+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)
\(\Leftrightarrow\left(a+\frac{\sqrt{2}}{8}\right)^2=\left(\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\right)^2\)\(\Leftrightarrow a^2+\frac{a\sqrt{2}}{4}+\frac{1}{32}=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\Leftrightarrow a^2+\frac{2\sqrt{a}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)
\(\Leftrightarrow4a^2+\sqrt{2}a-\sqrt{2}=0\)
Theo trên: \(4a^2+\sqrt{2}a-\sqrt{2}=0\Rightarrow a^2=\frac{\sqrt{2}\left(1-a\right)}{4}\Rightarrow a^4=\frac{a^2-2a+1}{8}\)
\(\Rightarrow a^4+a+1=\frac{a^2-2a+1}{8}+a+1=\left(\frac{a+3}{2\sqrt{2}}\right)^2\)
\(B=a^2+\sqrt{a^4+a+1}=a^2+\frac{a+3}{2\sqrt{2}}=\frac{2\sqrt{2}a^2+a+3}{2\sqrt{2}}\)\(=\frac{4a^2+\sqrt{2}a+3\sqrt{2}}{4}=\frac{4\sqrt{2}}{4}=\sqrt{2}\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
ta có :
\(A=\frac{a}{\sqrt{a}-1}-\frac{2a-\sqrt{a}}{a-\sqrt{a}}=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}=\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\sqrt{a}-1\)
mà \(a=3-\sqrt{8}=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\Rightarrow\sqrt{a}=\sqrt{2}-1\Rightarrow A=\sqrt{2}-1-1=\sqrt{2}-2\)
ĐK : a > 0 , a khác 1
\(A=\frac{a}{\sqrt{a}-1}-\frac{\sqrt{a}\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}\)
\(=\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)
Với \(a=3-\sqrt{8}\left(tmđk\right)\)thay vào A ta được :
\(A=\sqrt{3-\sqrt{8}}-1=\sqrt{\left(\sqrt{2}-1\right)^2}-1=\left|\sqrt{2}-1\right|-1=\sqrt{2}-1-1=\sqrt{2}-2\)