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\(-\frac{9}{11}\cdot\frac{3}{8}-\frac{9}{11}\cdot\frac{5}{8}+\frac{17}{11}=-\frac{9}{11}\left(\frac{3}{8}+\frac{5}{8}\right)+\frac{17}{11}=-\frac{9}{11}\cdot1+\frac{17}{11}=1\)
\(\frac{2}{1.3}+....+\frac{2}{53.55}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{53}-\frac{1}{55}=1-\frac{1}{55}=\frac{54}{55}\)
\(x+5-\frac{1}{2}=3\frac{1}{2}\)
\(x+5=3.5+0.5=4\)
\(x=4-5=-1\)
\(3^{x+1}=27=3^3\)
\(x+1=3\)
vậy x=2
a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)
\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)
\(\Leftrightarrow-4\left(x-6\right)=3x\)
\(\Leftrightarrow-4x+24=3x\)
\(\Leftrightarrow24=3x+4x\)
\(\Leftrightarrow7x=24\)
\(\Leftrightarrow x=\frac{24}{7}\)
b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{15}{8}\)
\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)
\(\Rightarrow3x=\frac{45}{4}\cdot8\)
\(\Rightarrow3x=90\Rightarrow x=30\)
\(c)1+2+3+4+...+x=820\)
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
\(a)\) \(\frac{x-1}{3}=\frac{x+2}{4}\)
\(\Leftrightarrow\)\(\frac{4\left(x-1\right)}{12}=\frac{3\left(x+2\right)}{12}\)
\(\Leftrightarrow\)\(4\left(x-1\right)=3\left(x+2\right)\)
\(\Leftrightarrow\)\(4x-4=3x+6\)
\(\Leftrightarrow\)\(4x-3x=6+4\)
\(\Leftrightarrow\)\(x=10\)
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) x + \(\frac{1}{6}\)=\(-\frac{3}{8}\)
x=\(\frac{-3}{8}-\frac{1}{6}\)
x=\(\frac{-13}{24}\)
b)\(\frac{1}{2}\)x + \(\frac{1}{8}\)x = \(\frac{3}{4}\)
\(\left(\frac{1}{2}+\frac{1}{8}\right)\)x = \(\frac{3}{4}\)
\(\frac{5}{8}\)x = \(\frac{3}{4}\)
x = \(\frac{3}{4}:\frac{5}{8}\)
x =\(\frac{6}{5}\)
c) 2 - l3/4-xl=7/12
l3/4-xl=17/12
nếu 3/4 - x > 0
thì 3/4 - x = 17/12
x= -2/3
nếu 3/4 - x <0
thì 3/4 - x = -17/12
x = 13/6
d) \(\left(2x-4.5\right):\frac{3}{4}-\frac{1}{3}=1\)
\(\left(2x-4.5\right):\frac{3}{4}=\frac{4}{3}\)
\(2x-4.5=1\)
2x = 5.5
x = 2.75
a, x + 1/6 = -3/8
x = -3/8 - 1/6
x = -13/24
b, 1/2x + 1/8x = 3/4
x * (1/2+1/8) = 3/4
x * 5/8 = 3/4
x = 3/4 : 5/8
x = 6/5
c, 2 - |3/4-x| = 7/12
|3/4-x| = 2 - 7/12
|3/4-x| = 17/12
=> 3/4 - x = 17/12 hoặc -17/12
# 3/4 - x = 17/12
=> x = -2/3
# 3/4 - x = -17/12
=> x = 13/6
=> x = { -2/3 ; 13/6}
d, (2x - 4,5) : 3/4 -1/3 = 1
(2x - 4,5) : 3/4 = 1 + 1/3
(2x - 4,5 ) : 3/4 = 4/3
2x - 4,5 = 4/3 * 3/4
2x - 4,5 = 1
2x = 1 + 4,5
2x = 5,5
x = 5,5 : 2
x = 2,75
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
a. 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