1-1/2+1/3-1/4+...+1/121-1/122+1/123=1/62+1/63+...+1/122+1/123
1-1/2+1/3-1/4+...+1/121-1/122+1/123=1/62+1/63+...+1/122+1/123
1-1/2+1/3-1/4+...+1/121-1/122+1/123=1/62+1/63+...+1/122+1/123
Chứng minh rằng:\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{121}-\dfrac{1}{122}+\dfrac{1}{123}=\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{122}+\dfrac{1}{123}\)
Chứng minh rằng
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{121}-\frac{1}{122}+\frac{1}{123}=\frac{1}{62}+\frac{1}{63}+...+\frac{1}{122}\)-\(\frac{1}{123}\)
Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)
\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)
\(\dfrac{x+1}{124}\)+\(\dfrac{x+2}{123}\)=\(\dfrac{x+3}{122}\)+\(\dfrac{x+4}{121}\)
\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)
\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)
\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)
<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)
<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)
<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)
<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)
<=>x+125=0
<=>x=-125
(1+2+3+4+5+..+122+123)
Số số hạng : (123 - 1) : 1 + 1 = 123
Tổng : (123 + 1) x 123 : 2 = 7626
Dãy số có số số hạng là:
(123-1):1+1=123
Tổng dãy số là;
(123+1)x123:2=7626
Đ/S: 7626
Giá trị của X trong phép tính : 486 : x = 4 (dư 2) là bao nhiêu? (1 Point) x = 121 x = 121 (dư 2) x = 122 x = 123
Giá trị của X trong phép tính 486 : X = 4 ( dư 2 ) là : X = 121
tính nhanh:
(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).(125-5.25)
=(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
SO SÁNH : A = 3^123 +1 / 3^125 + 1 và B = 3^122/ 3^124 + 1
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