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Câu 2:
a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)
b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)
\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)
\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\) \(\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(a=2\left(16-15\right)\)
\(a=2\)
khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)
vậy.....
a) đk: \(\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
Ta có:
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4\sqrt{a}+4a\sqrt{a}-4\sqrt{a}}{a-1}\cdot\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
b) Ta có: \(a=\sqrt{4+\sqrt{15}}\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4^2-\sqrt{15}^2}\)
\(=\sqrt{10}-\sqrt{6}\)
\(\Rightarrow A=\frac{4\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{10}-\sqrt{6}+1\right)}{\sqrt{10}-\sqrt{6}-1}=...\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
a)A=(\(\frac{2}{\sqrt{a}-1}\)+\(\frac{2}{\sqrt{a}+1}\)+4\(\sqrt{a}\)).\(\frac{a-1}{\sqrt{a}}\)=(\(\frac{4\sqrt{a}}{a-1}\)+4\(\sqrt{a}\)).\(\frac{a-1}{\sqrt{a}}\)=\(\frac{4a}{a-1}\)
b)a=(\(\sqrt{\left(4+\sqrt{15}\right).\left(4-\sqrt{15}\right)}\).(\(\sqrt{10}\)-\(\sqrt{6}\))=\(\sqrt{16-15}\).(\(\sqrt{10}\)-\(\sqrt{6}\))=\(\sqrt{10}\)-\(\sqrt{6}\)
Thay vào A rồi tính là xong
a) \(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(=\left[\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right].\left(\frac{a}{\sqrt{a}}-\frac{1}{\sqrt{a}}\right)\)
\(=\left[\frac{a+2\sqrt{a}+1}{a-1}-\frac{a-2\sqrt{a}+1}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right].\frac{a-1}{\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}.a-4\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}}\)
\(=\frac{4\sqrt{a}.a}{a-1}.\frac{a-1}{\sqrt{a}}=4a\)
b) Ta có: \(a=\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}.\left(\sqrt{10}-\sqrt{6}\right)\)
\(=\sqrt{16-15}.\left(\sqrt{10}-\sqrt{6}\right)=\sqrt{10}-\sqrt{6}\)
Thay a vào A ta được: \(A=4.\left(\sqrt{10}-\sqrt{6}\right)=4\sqrt{10}-4\sqrt{6}\)
Bài đầu : \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
b: \(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{a-1}\cdot\dfrac{a+1}{\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}+4a\sqrt{a}-4\sqrt{a}}{a-1}\cdot\dfrac{a+1}{\sqrt{a}}\)
\(=\dfrac{4a\sqrt{a}\left(a+1\right)}{\left(a-1\right)\cdot\sqrt{a}}=\dfrac{4a\left(a+1\right)}{a-1}\)
\(\begin{array}{l} a)A = \left( {\dfrac{{\sqrt a + 1}}{{\sqrt a - 1}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}} + 4\sqrt a } \right).\left( {\sqrt a + \dfrac{1}{{\sqrt a }}} \right)\\ = \left[ {\dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - {{\left( {\sqrt a - 1} \right)}^2}}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\\ = \left[ {\dfrac{{4\sqrt a }}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} + 4\sqrt a } \right].\dfrac{{a + 1}}{{\sqrt a }}\\ = \dfrac{{4\sqrt a + 4\sqrt a \left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}.\dfrac{{a + 1}}{{\sqrt a }}\\ = \dfrac{{4a\sqrt a }}{{a - 1}}.\dfrac{{a + 1}}{{\sqrt a }} = \dfrac{{4a}}{{a - 1}}\left( {a + 1} \right) = \dfrac{{4{a^2} + 4a}}{{a - 1}} \end{array}\)
$b)$Thay $a=\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)$ vào ta được:
$A=\dfrac{4{{\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}^{2}}+4\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}{\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)-1}=12$
$\begin{align}
& a)A=\left( \dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a} \right).\left( \sqrt{a}+\dfrac{1}{\sqrt{a}} \right) \\
& =\left[ \dfrac{{{\left( \sqrt{a}+1 \right)}^{2}}-{{\left( \sqrt{a}-1 \right)}^{2}}}{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}+4\sqrt{a} \right].\dfrac{a+1}{\sqrt{a}} \\
& =\left[ \dfrac{4\sqrt{a}}{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}+4\sqrt{a} \right].\dfrac{a+1}{\sqrt{a}} \\
& =\dfrac{4\sqrt{a}+4\sqrt{a}\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}{\left( \sqrt{a}-1 \right)\left( \sqrt{a}+1 \right)}.\dfrac{a+1}{\sqrt{a}} \\
& =\dfrac{4a\sqrt{a}}{a-1}.\dfrac{a+1}{\sqrt{a}}=\dfrac{4a}{a-1}\left( a+1 \right)=\dfrac{4{{a}^{2}}+4a}{a-1} \\
\end{align}$
$b)$Thay $a=\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)$ vào ta được:
$A=\dfrac{4{{\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}^{2}}+4\left[ \left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right) \right]}{\left( 4+\sqrt{15} \right)\left( \sqrt{10}-\sqrt{6} \right)\left( \sqrt{4-\sqrt{15}} \right)-1}=12$
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
Lời giải:
a) ĐKXĐ: \(a>0; a\neq 1\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+4\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)(\sqrt{a}+1)}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{a+1+2\sqrt{a}-(a+1-2\sqrt{a})+4\sqrt{a}(a-1)}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4\sqrt{a}+4\sqrt{a}(a-1)}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4\sqrt{a}.a}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a(a+1)}{a-1}\)
b)
Ta có:
\(a=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2\)
\(=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(16-15)=2\)
Thay $a=2$ vào biểu thức đã thu gọn:
\(A=24\)
Cái
\(\sqrt{8-2\sqrt{15}}=\sqrt{(\sqrt{3})^2+(\sqrt{5})^2-2\sqrt{3}.\sqrt{5}}=\sqrt{(\sqrt{5}-\sqrt{3})^2}=\sqrt{5}-\sqrt{3}\)
Thường thì những biểu thức căn cồng kềnh bao giờ cũng có hướng khai triển ra chính phương hoặc lập phương nên cứ chịu khó mần là ra thôi, kiểu gì cũng tách ghép được.