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1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
a: \(\Leftrightarrow\dfrac{x^4+2x^2+1-x^2}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{\left(x^2+1+x\right)\left(x^2+1-x\right)}{x^2}=\dfrac{x^2+x+1}{x}\)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^2}=\dfrac{1}{x}\)
=>x^2=x(x^2-x+1)
=>x(x-x^2+x-1)=0
=>x(-x^2+2x-1)=0
=>x=0(loại) hoặc x=1(nhận)
b: =>3(x+3)^2*(x+2)^2/(x^2-4)^2+68*(x-3)^2*(x-2)^2/(x^2-4)^2-46(x^2-9)(x^2-4)=6(x^2-4)^2
=>3(x^2+5x+6)^2+68(x^2-5x+6)^2-46(x^4-13x^2+36)=6(x^4-8x^2+16)
=>\(x\simeq28,4\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)
\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)
=>\(-9x^4+81x^2-36=0\)
=>9x^4-81x^2+36=0
=>x^4-9x^2+4=0
=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)
=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).....\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2+2^2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right).......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).......\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right)........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
Ta có: \(16a^4+4=16a^4+2.4a^2.2+4-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a+2\right)^2-16a^2\)
\(=\left(4a^2-4a+2\right).\left(4a^2+4a+2\right)\)
\(=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( a \(\in\) N* )
Do đó: \(16a^4+4=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( * )
Thay a lần lượt bằng 1, 2, 3, ..., 2014, ta có:
\(16.1^4+4=\left[\left(2.1-1\right)^2+1\right].\left[\left(2.1+1\right)^2+1\right]=\left(1^2+1\right).\left(3^2+1\right)\)
\(16.2^4+4=\left[\left(2.2-1\right)^2+1\right].\left[\left(2.2+1\right)^2+1\right]=\left(3^2+1\right).\left(5^2+1\right)\)
\(16.3^4+4=\left[\left(2.3-1\right)^2+1\right].\left[\left(2.3+1\right)^2+1\right]=\left(5^2+1\right).\left(7^2+1\right)\)
\(16.4^4+4=\left[\left(2.4-1\right)^2+1\right].\left[\left(2.4+1\right)^2+1\right]=\left(7^2+1\right).\left(9^2+1\right)\)
\(......\)
\(16.2005^4+4=\left[\left(2.2005-1\right)^2+1\right].\left[\left(2.2005+1\right)^2+1\right]=\left(4009^2+1\right).\left(4011^2+1\right)\)
\(16.2006^4+4=\left[\left(2.2006-1\right)^2+1\right].\left[\left(2.2006+1\right)^2+1\right]=\left(4011^2+1\right).\left(4013^2+1\right)\)
Đặt \(T=\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)...\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)...\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{16.\left(1^4+\dfrac{1}{4}\right).16\left(3^4+\dfrac{1}{4}\right)...16\left(2005^4+\dfrac{1}{4}\right)}{16.\left(2^4+\dfrac{1}{4}\right).16\left(4^4+\dfrac{1}{4}\right)...16\left(2006^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow T=\dfrac{\left(16.1^4+4\right).\left(16.3^4+4\right)...\left(16.2005^4+4\right)}{\left(16.2^4+4\right).\left(16.4^4+4\right)...\left(16.2006^4+4\right)}\)
\(\Leftrightarrow T=\dfrac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right)...\left(4009^2+1\right).\left(4011^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right)...\left(4011^2+1\right).\left(4013^2+1\right)}\)
\(\Leftrightarrow T=\dfrac{1^2+1}{4013^2+1}\)
\(\Leftrightarrow T=\dfrac{2}{4013^2+1}\)
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