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\(A=\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}\)
\(A=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}\)
\(A=\frac{1\cdot5\cdot6}{1\cdot3\cdot5}\)
\(A=2\)
\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}=2\)
3A = 32 + 33 + 34 + ... +32007
=> 3A - A = 2A = 32007 - 31 = 3( 32006-1)
=>A = \(\frac{3\left(3^{2006}-1\right)}{2}\)
Ta có : 2A + 3 = 32007 + 3 - 3
= 32007 = 3x
=> x= 2007
b)
A = \(\frac{1.5.6+2^3.1.5.6+4^3.1.5.6+9^3.1.5.6}{1.3.5+2^3.1.3.5+4^3.1.3.5+9^3.1.3.5}\)= \(\frac{1.5.6\left(1+2^3+4^3+9^3\right)}{1.3.5\left(1+2^3+4^3+9^3\right)}\)=2
<math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>C</mi> <mo>=</mo> <mstyle displaystyle="true" scriptlevel="0"> <mfrac> <mrow class="MJX-TeXAtom-ORD"> <mn>2.</mn> <mo stretchy="false">(</mo> <mn>1.3.5</mn> <mo>+</mo> <mn>2.6.10</mn> <mo>+</mo> <mn>4.12.20</mn> <mo>+</mo> <mn>9.27.45</mn> <mo stretchy="false">)</mo> </mrow> <mrow class="MJX-TeXAtom-ORD"> <mn>1.3.5</mn> <mo>+</mo> <mn>2.6.10</mn> <mo>+</mo> <mn>4.12.20</mn> <mo>+</mo> <mn>9.27.45</mn> </mrow> </mfrac> </mstyle> <mo>=</mo> <mn>2</mn> </math>
C = 1.5.6 + 2.10.12 + 4.20.24 + 9.45.54 1.3.5 + 2.6.10 + 4.12.20 + 9.27.45 C=1.5.6+2.10.12+4.20.24+9.45.541.3.5+2.6.10+4.12.20+9.27.45 C = 2. ( 1.3.5 + 2.6.10 + 4.12.20 + 9.27.45 ) 1.3.5 + 2.6.10 + 4.12.20 + 9.27.45 = 2
a) 3A = 3^2 + 3^3 +...... + 3^2007
2A = (3^2 - 3^2) + (3^3-3^3) +.... + 3^2007 - 3
A = (3^2007 - 3)/2
2A + 3 = (3^2007 - 3)/2 x 2 + 3 = 3^2007
=> x = 2007
Cấu b dễ hơn nhưng hơi dài bạn chỉ lấy thừa số chung là OK
câu 3 :S=(1.5.6+2.10.12+4.20.24+9.45.54):(1.3.5+2.6.10+9.27.45)
=(1.5.6+1.5.6.8+1.5.6.64+1.5.6.729):(1.3.5+1.3.5.8+1.3.5.64+1.3.5.729)
=[1.5.6.(1+8+64+729)]:[1.3.5.(1+8+64+729)]
=(1.5.6):(1.3.5)=2