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a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
a)\(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)=1-\sqrt{x^3}\)
b) \(\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)=\sqrt{x^3}+8\)
c)\(\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)=\sqrt{x^3}-\sqrt{y^3}\)
d)\(\left(x+\sqrt{y}\right)\left(x^2+y-x\sqrt{y}\right)=x^3+\sqrt{y^3}\)
C/m tổng quát : \(A=\left(a+1\right)\left(a^2+1\right)\left(a^4+1\right)\left(a^8+1\right)...\left(a^{2^n}+1\right)=\frac{a^{2^{n+1}}-1}{a-1}\)
Có : \(A=\frac{\left(a+1\right)\left(a-1\right)}{a-1}.\frac{\left(a^2+1\right)\left(a^2-1\right)}{a^2-1}.\frac{\left(a^4+1\right)\left(a^4-1\right)}{a^4-1}...\frac{\left(a^{2^n}+1\right)\left(a^{2^n}-1\right)}{a^{2^n}-1}\)
\(=\frac{\left(a^2-1\right)\left(a^4-1\right)\left(a^8-1\right)...\left(a^{2^{n+1}}-1\right)}{\left(a-1\right)\left(a^2-1\right)\left(a^4-1\right)...\left(a^{2^n}-1\right)}=\frac{a^{2^{n+1}}-1}{a-1}\)(đpcm)
Với a = 2 ; n = 11 => \(A=2^{4096}-1\)
A=2^4096-1 nha
HT
k cho mình nha
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