\(\dfrac{2}{3}.x-1\dfrac{2}{5}.x=\dfrac{3}{5}\)

\(\dfrac{2}{3}.x-\dfrac{7}{5}.x=\dfrac{3}{5}\)

\(x.\left(\dfrac{2}{3}-\dfrac{7}{5}\right)=\dfrac{3}{5}\)

\(x.\left(\dfrac{10}{15}-\dfrac{21}{15}\right)=\dfrac{3}{5}\)

\(x.\dfrac{-11}{15}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{-11}{15}=\dfrac{3}{5}.\dfrac{-15}{13}\)

\(x=\dfrac{-9}{11}\)

\(\dfrac{6^6+2^7.3^6+2^6.3^7}{4.2^3.3^2.2^2}\)

\(=\dfrac{\left(2^2.3\right)^6+2^7.3^6+2^6.3^7}{2^2.2^3.3^2.2^2}\)\(=\dfrac{2^{12}.3^6+2^7.3^6+2^6.3^7}{2^7.3^2}\)

\(=\dfrac{2^6.3^6.\left(2^6+2+3\right)}{2^7.3^2}\)\(=\dfrac{3^4.69}{2}=\dfrac{5589}{2}\)

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

\(\left(1-x\right)^2=2003.\left(x-1\right)\)

\(\left(1-x\right)^2-2003\left(x-1\right)=0\)

\(\left(1-x\right)^2+2003\left(1-x\right)=0\)

\(\left(1-x\right)\left(1-x+2003\right)=0\)

\(\left(1-x\right)\left(2004-x\right)=0\)

\(TH1:1-x=0\)

\(x=1\)

\(TH2:2004-x=0\)

\(x=2004\)

vậy........

\(A=\left(1-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{2401}-\dfrac{1}{2500}\right)\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{2401}-\dfrac{1}{2500}\)

\(A=1-\dfrac{1}{2500}=\dfrac{2499}{2500}\)

\(15^8.2^4=\left(15^2\right)^4.2^4=225^4.2^4=\left(225.2\right)^4=450^4\\ 27^5:32^3=\left(3^3\right)^5:\left(2^5\right)^3=3^{15}:2^{15}=\left(\dfrac{3}{2}\right)^{15}\)

\(=12,5:18=\dfrac{216}{5}\)

Lời giải:

$(-2,5).\frac{5}{18}=\frac{-5}{2}.\frac{5}{18}=\frac{-25}{36}$

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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