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a) \(\sqrt{\left(4-\sqrt{15}\right)^{2^{ }}}+\sqrt{15}\)
=\(\left|4-\sqrt{15}\right|+\sqrt{15}\)
= \(4-\sqrt{15}+\sqrt{15}\) ( vì 4 =\(\sqrt{16}\) mà \(\sqrt{16}>\sqrt{15}\) )
=4
b)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
=\(\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\) ( vì \(1< \sqrt{3}< 2\))
= \(2-\sqrt{3}-1+\sqrt{3}\)
=1
a) Ta có: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}\)
=4
b) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
a) \(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(A=\sqrt{\left(2+\sqrt{3}\right)\left(\sqrt{2+\sqrt{3}}+2\right)\left(-\sqrt{2+\sqrt{3}}+2\right)}\)
\(A=\sqrt{1}\)
\(A=1\)
b)\(B=\left(\frac{\sqrt{x}}{\sqrt{xy}-y}-\frac{\sqrt{y}}{\sqrt{xy}-x}\right).\left(x\sqrt{y}-y\sqrt{x}\right)\)
\(B=\frac{\sqrt{xy}}{\sqrt{xy}-y}x\sqrt{y}+\frac{\sqrt{x}}{\sqrt{xy}-y}y\sqrt{x}+\left(-\frac{\sqrt{y}}{\sqrt{xy}-x}\right)^2x\sqrt{y}+y\sqrt{x}\)
\(B=x\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{y}+y\frac{\sqrt{x}}{\sqrt{xy}-y}\sqrt{x}+x\frac{\sqrt{x}}{\sqrt{xy}-x}\sqrt{y}-y\sqrt{x}\frac{\sqrt{y}}{\sqrt{xy}-y}\)
\(B=\frac{-x^{\frac{5}{2}}\sqrt{y}+\sqrt{x}.y^{\frac{5}{2}}}{\left(\sqrt{xy}-y\right)\left(\sqrt{xy}-x\right)}\)
\(B=\frac{\left(\sqrt{x}.y^{\frac{5}{2}}-x^{\frac{5}{2}}\sqrt{y}\right)\left(y+\sqrt{xy}\right)\left(x+\sqrt{xy}\right)}{\left(-y^2+xy\right)\left(-x^2+xy\right)}\)
c) \(C=\sqrt{\left(3-\sqrt{5}\right)^2+\sqrt{6}-2\sqrt{5}}\)
\(C=14-6\sqrt{5}+\sqrt{6}-2\sqrt{5}\)
\(C=14-8\sqrt{5}+\sqrt{6}\)
\(C=\sqrt{14-8\sqrt{5}+\sqrt{6}}\)
a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)