16x^9-x^5=0
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\(\left(1-x\right)^2=2003.\left(x-1\right)\)
\(\left(1-x\right)^2-2003\left(x-1\right)=0\)
\(\left(1-x\right)^2+2003\left(1-x\right)=0\)
\(\left(1-x\right)\left(1-x+2003\right)=0\)
\(\left(1-x\right)\left(2004-x\right)=0\)
\(TH1:1-x=0\)
\(x=1\)
\(TH2:2004-x=0\)
\(x=2004\)
vậy........
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\(A=\left(1-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{2401}-\dfrac{1}{2500}\right)\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{2401}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2500}=\dfrac{2499}{2500}\)
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\(15^8.2^4=\left(15^2\right)^4.2^4=225^4.2^4=\left(225.2\right)^4=450^4\\ 27^5:32^3=\left(3^3\right)^5:\left(2^5\right)^3=3^{15}:2^{15}=\left(\dfrac{3}{2}\right)^{15}\)
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Lời giải:
$(-2,5).\frac{5}{18}=\frac{-5}{2}.\frac{5}{18}=\frac{-25}{36}$
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Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}$
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{20-18}{18.19.20}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}$
$=\frac{1}{1.2}-\frac{1}{19.20}=\frac{189}{380}$
$\Rightarrow A=\frac{189}{760}$
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`#3107`
\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{7}=\dfrac{19}{8}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{19}{8}-\dfrac{3}{7}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{109}{56}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\left(\sqrt{\dfrac{109}{56}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=\sqrt{\dfrac{109}{56}}\\2x-\dfrac{1}{2}=-\sqrt{\dfrac{109}{56}}\end{matrix}\right.\)
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16x9 - x5 = 0
<=>x5(16x4-1)=0
<=>x5(16x4-4x2 +4x2-1)=0
<=>x5[4x2(4x2-1) +(4x2-1)]=0
<=>x5(4x2-1)(4x2+1)=0
<=>x= 0 hoặc 4x2-1 = 0 (loại 4x2+1=0 vì nó >0)
<=> x= 0 hoặc x=+- 1/2