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\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
\(a)\) \(\frac{x-1}{3}=\frac{x+2}{4}\)
\(\Leftrightarrow\)\(\frac{4\left(x-1\right)}{12}=\frac{3\left(x+2\right)}{12}\)
\(\Leftrightarrow\)\(4\left(x-1\right)=3\left(x+2\right)\)
\(\Leftrightarrow\)\(4x-4=3x+6\)
\(\Leftrightarrow\)\(4x-3x=6+4\)
\(\Leftrightarrow\)\(x=10\)
Vậy \(x=10\)
Câu b) thiếu đề nhé bạn
Chúc bạn học tốt ~
Thanks bạn Phùng Minh Quân nha!
Mk chỉnh lại cau b nè:
b, Tìm x € Z để \(B=\frac{2x+9}{x-1}\)có giá trị nguyên.
\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)
\(\Rightarrow3x=\frac{45}{4}\cdot8\)
\(\Rightarrow3x=90\Rightarrow x=30\)
\(c)1+2+3+4+...+x=820\)
Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)
Do đó : \(\frac{(1+x)\cdot x}{2}=820\)
\(\Rightarrow(1+x)\cdot x=820\cdot2\)
\(\Rightarrow(1+x)\cdot x=1640\)
\(\Rightarrow(1+x)\cdot x=40\cdot41\)
Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40
Chúc bạn học tốt :3
\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2
a, (x + 2) + (x + 4) + (x + 6) + ... + (x + 50) = 750
=> x + 2 + x + 4 + x + 6 + ... + x + 50 = 750
=> (x + x + x + ... + x) + (2 + 4 + 6 + ... + 50) = 750
=> 25x + (50 + 2).25 : 2 = 750
=> 25x + 52.25 : 2 = 750
=> 25x + 650 = 750
=> 25x = 100
=> x = 4
a) ( x+x+...+x)+(2+4+6+...+50)= 750
( x*25)+ (50+2)*25:2 = 750
(x*25)+ 650 = 750
x* 25 = 750 - 650 = 100
x = 100 :25 = 4
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x=0+\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}\)
\(\Leftrightarrow x=\frac{6}{11}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\times\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{50}{100}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Leftrightarrow x+1=100\)
\(\Leftrightarrow x=100-1\)
\(\Leftrightarrow x=99\)
a. x + (x - 1) + (x - 2) + ... + (x - 50) = 255
=> x + x - 1 + x - 2 + ... + x - 50 = 255
=> 51.x - (1 + 2 + ... + 50) = 255
=> 51.x - (50 + 1).50:2 = 255
=> 51.x - 1275 = 255
=> 51.x = 255 + 1275
=> 51.x = 1530
=> x = 30
b. 3 + 6 + 9 + ... + x = 630
Số số hạng: (x - 3) : 3 + 1 = x /3 (số hạng)
=> (x + 3).(x/3):2 = 630
=> (x + 3).(x/3) = 1260
=> x.(x +3)/3 = 1260
=> x.(x + 3) = 3780
=> x.(x + 3) = 60.63
=> x.(x + 3) = 60.(60 + 3)
=> x = 60
c. 3.x - x + 1/2.x = 5\(\frac{1}{3}\)
=> x.(3 - 1 + 1/2) = 16/3
=> x.5/2 = 16/3
=> x = 16/3 : 5/2
=> x = 16/3 . 2/5
=> x = 32/15
x+(x-1)+(x-2)+............+(x-50)= 255
=> (x-50)+(x-49)+(x-48)+............+x=255
=> (x+x-50).51:2=255
=> (x+x-50).51=255.2
=> 2x-50=510:51
=> 2x=10+50
=> x=60:2
=> x=30