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a/ \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
Mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
\(\Leftrightarrow x+101=0\)
\(\Leftrightarrow x=-101\)
Vậy...
b/ Đặt :
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+.........+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+....+\dfrac{10^2-9^2}{9^2.10^2}\)
\(=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+....+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)
\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(=1-\dfrac{1}{10^2}< 1\)
\(\Leftrightarrow A< 1\left(đpcm\right)\)
Vậy...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A=\left|x-10\right|+\left|5-x\right|\)
\(\Leftrightarrow A\ge\left|x-10+5-x\right|\)
\(\Leftrightarrow A\ge5\)
Dấu "=" xảy ra
\(\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\5\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\5\le x\le10\end{matrix}\right.\)
Vậy..
a) S = 1.2 + 2.3 + 3.4 + ... + 99.100
S có thể được viết lại thành:
S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)
= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)
Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:
S = n(n+1)(2n+1)/6
Với n = 99, ta có:
S = 99.100.199/6 = 331650
Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:
S = n(n+1)/2
Với n = 98, ta có:
S = 98.99/2 = 4851
Do đó, S = 331650 - 4851 = 326799
b) B = 4924.12517.28−530.749.45529.162.748
B có thể được viết lại thành:
B = (4924.12517.28) / (530.749.45529.162.748)
B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)
B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529
B = 108 / 45529
c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101
C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101
C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)
C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)
C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)
d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018
D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^
A=\(\dfrac{7^2-1}{7^4}+\dfrac{7^2-1}{7^8}+...+\dfrac{7^2-1}{7^{100}}=\left(7^2-1\right)\left(\dfrac{1}{7^4}+\dfrac{1}{7^8}+...+\dfrac{1}{7^{100}}\right)=48\cdot B\)Dễ dàng tính được B( nhân hết với 7 mũ 4 roi trừ đi, chia ra là xong) ra đpcm.
Lên lớp 11 thì ta có dạng tổng quát luôn này(tức là nếu n quá lớn thì có thể coi là xảy ra dấu bằng) \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^n}-\dfrac{1}{7^{n+2}}< \dfrac{1}{50}\)
Đặt \(A=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\)
Ta có:
\(\dfrac{A}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\)
\(\Rightarrow A+\dfrac{A}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}+\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}+\dfrac{1}{7^{102}}\right)\)
\(\Rightarrow\dfrac{50A}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}\)
\(\Rightarrow A< \dfrac{1}{50}\)
=> ĐPCM.
\(\text{Đặt:}S=\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\Rightarrow49S=1-\frac{1}{7^2}+.....-\frac{1}{7^{98}}\Rightarrow49S+S=50S=\left(1-\frac{1}{7^2}+\frac{1}{7^4}-....-\frac{1}{7^{98}}\right)+\left(\frac{1}{7^2}-\frac{1}{7^4}+....-\frac{1}{7^{100}}\right)=1-\frac{1}{7^{100}}< 1\Rightarrow S< \frac{1}{50}\left(\text{đpcm}\right)\)
Đặt \(S=\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{4n-2}}-\dfrac{1}{7^{4n}}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\)
\(\Rightarrow\dfrac{S}{7^2}=\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\)
\(\Rightarrow S+\dfrac{S}{7^2}=\left(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}\right)+\left(\dfrac{1}{7^4}-\dfrac{1}{7^6}+...+\dfrac{1}{7^{100}}-\dfrac{1}{7^{102}}\right)\)
\(\Leftrightarrow\dfrac{50S}{49}=\dfrac{1}{7^2}-\dfrac{1}{7^{102}}< \dfrac{1}{7^2}=\dfrac{1}{49}< \dfrac{1}{50}\)
\(\Leftrightarrow S< \dfrac{1}{50}\)
Vậy \(\dfrac{1}{7^2}-\dfrac{1}{7^4}+...+\dfrac{1}{7^{98}}-\dfrac{1}{7^{100}}< \dfrac{1}{50}\) (Đpcm)
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
Ta có : B = 3/5 . 5/7 . 7/9 . ... . 97/99
=> B = 3.5.7.9. ... .97/5.7.9. ... .99
=> B = 3/99
Vậy B = 3/99
A = 1 - 2 + 3 - 4 +...+97 - 98 + 99 - 100
A = 1 + ( -2 + 3) +...+ ( -98 + 99 ) - 100
A = 1 + 1 + ... + 1 - 100
A = 50 - 100
A = -50
B = \(\frac{7}{19.29}\) + \(\frac{7}{29.39}\) + \(\frac{7}{39.49}\) + \(\frac{7}{49.59}\) + \(\frac{7}{59.69}\)
B = 7. ( \(\frac{1}{19.29}\) + \(\frac{1}{29.39}\) + \(\frac{1}{39.49}\) + \(\frac{1}{49.59}\) + \(\frac{1}{59.69}\))
B= 7. \(\frac{1}{10}\)( \(\frac{10}{19.29}\)+ \(\frac{10}{29.39}\)+ \(\frac{10}{39.49}\)+\(\frac{10}{49.59}\)+\(\frac{10}{59.69}\))
B = 7 . \(\frac{1}{10}\) ( \(\frac{1}{19}\) - \(\frac{1}{69}\) )
B = 7 . \(\frac{1}{10}\) . \(\frac{50}{1311}\)
B = \(\frac{7}{10}\) . \(\frac{50}{1311}\)
B = \(\frac{35}{1311}\)
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