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B> \(\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\)
\(\Leftrightarrow\left(x+\sqrt{x^2+2013}\right)\left(y+\sqrt{y^2+2013}\right)\)\(\left(x-\sqrt{x^2+2013}\right)=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2013\right)\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow-2013\left(y+\sqrt{y^2+2013}\right)\)\(=2013\left(x-\sqrt{x^2+2013}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+2013}=-x+\sqrt{x^2+2013}\)
Chứng minh tương tự: \(x+\sqrt{x^2+2013}=-y+\sqrt{y^2+2013}\)
cộng vế theo vế ta được: \(x+y=-x-y\)
\(\Leftrightarrow x+y=0\Leftrightarrow x=-y\Leftrightarrow x^{2013}=-y^{2013}\)
\(\Leftrightarrow x^{2013}+y^{2013}=0\)
a,Ta có x =...
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)-\sqrt{3}\left(\sqrt{\sqrt{3+1}-1}\right)}{\left(\sqrt{\sqrt{3}+1}\right)\left(\sqrt{\sqrt{3}-1}\right)}\)
x = \(\frac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}+1-1}\)
x = \(\frac{\sqrt{3}.2}{\sqrt{3}}\)
x = 2
sau đó thay x=2 vào A nhé.
A=2014 !!!
a)= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
= \(-1+\sqrt{100}\)
= -1 +10
=9
b)Ta có\(\left(\sqrt{n+1}-\sqrt{n}\right)\cdot\left(\sqrt{n+1}+\sqrt{n}\right)\)=n+1-n=1 (1)
Lại có:\(\frac{1}{\sqrt{n+1}+1}\cdot\left(\sqrt{n+1}+1\right)=1\)(2)
Từ (1) và (2)=>\(\left(\sqrt{n+1}-1\right)=\frac{1}{\sqrt{n+1}+1}\)
\(\sqrt{\left(2x-1\right)^2=3}\)
<=> |2x - 1| = 3
*) Với x >= 1/2
=> 2x - 1 = 3
<=> 2x = 4
<=> x = 2 (TM)
*) Với x < 1/2
=> -2x + 1 = 3
-2x = 2
x = -1 (TM)
Vậy x = 2 hoặc x = -1
\(\sqrt{\left(2x-1\right)^2}=3=\sqrt{9}\)
\(\Leftrightarrow\left(2x-1\right)^2=9\Leftrightarrow\left(2x-1\right)^2-3^2=0\)
\(\Leftrightarrow\left(2x-1-3\right)\left(2x-1+3\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x+2\right)=0\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
=>\(x-1=0\Leftrightarrow x=1\)
=>\(x+1=0\Leftrightarrow x=-1\)
Vậy S = {-1;1}
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
Xét \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}a>0\)
Ta có: \(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
Vì a>0, D>0 nên \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng ta có: \(D=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)=100-\frac{1}{100}=99,99\)
<=>\(\left(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}\right)^2=\left(\sqrt{2\left(\sqrt{2}-1\right)}\right)^2\)
<=>\(\sqrt{2}+1+\sqrt{2}-1-2\left(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\right)=2\left(\sqrt{2}-1\right)\)
<=>\(2\sqrt{2}-2=2\sqrt{2}-2\left(dpcm\right)\)
¬¬¬¬¬¬hoc tot ¬¬¬¬¬¬¬