1+1+1+1+1+1+1+2+2+2+2+2+2+2+3+3+3+3+3+3+3+3+122+123+124
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(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-5.25)
=(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-125)
=(1/2+2/3+3/4+4/5+...+122/123+123/124).0=0
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\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)
\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)
\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)
<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)
<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)
<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)
<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)
<=>x+125=0
<=>x=-125
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A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
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1 - 1/2 + 1/3 - 1/4 + .. + 1/121 - 1/122 + 1/123
= ( 1 + 1/3 + ... + 1/121 + 1/123 ) - ( 1/2 + 1/4 + .. + 1/122 )
Đến bước này ta sẽ cùng cộng 2 vế với : 1/2 + 1/4 + .. + 1/122
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/121 + 1/122 + 1/123 ) - 2. ( 1/2 + 1/4 + .. + 1/122 )
= ( 1 + 1/2 + 1/3 + ...+ 1/122 +1 /123 ) - ( 1 + 1/2 + ...+ 1/61 )
= 1/62 + 1/63 + ..+1/122 + 1/123
Chúc học giỏi !!
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bằng 292 nhé