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a,ĐK : x \(\ne\)3/7
\(\frac{24}{7x-3}=-\frac{4}{25}\Leftrightarrow600=-28x+12\Leftrightarrow-28x=588\Leftrightarrow x=-21\)
b, ĐK : x;y \(\ne\)6
Xét : \(\frac{4}{x-6}=-\frac{12}{18}\Leftrightarrow72=-12x+72\Leftrightarrow x=0\)
Xét : \(\frac{y}{24}=-\frac{12}{18}\Leftrightarrow18y=-288\Leftrightarrow y=-16\)
\(\frac{24}{7.x-3}=-\frac{4}{25}\)
24.25=7.x-3.-4
600=7.x-3.-4
7.x-3.-4=600
7.x-3=600:-4
7.x-3=-150
7.x=-150+3
7.x=-147
x=-147:7
x=-21
vậy x=-21
a, \(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}=(\frac{19}{24}-\frac{7}{24})-\frac{1}{2}-\frac{1}{3}\)
\(=\frac{12}{24}-\frac{1}{2}-\frac{1}{3}\)
\(=\frac{1}{2}-\frac{1}{2}-\frac{1}{3}=0-\frac{1}{3}=-\frac{1}{3}\)
\(B=\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}=(\frac{7}{12}-\frac{5}{12})+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\)
\(=\frac{1}{6}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\)
\(=1+\frac{1}{4}-\frac{3}{7}=\frac{23}{28}\)
b, Thay thế A = \(-\frac{1}{3}\)và B = \(\frac{23}{28}\)ta có :
\(-\frac{1}{3}-x=\frac{23}{28}\)
\(\Rightarrow x=-\frac{1}{3}-\frac{23}{28}=-\frac{28}{84}-\frac{69}{84}=\frac{-28-69}{84}=\frac{-97}{84}\)
dễ mà bạn đây là bài cơ bản lớp 6 dấy
câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)
b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)
c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)
d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)
a) \(x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x.\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow x=-\frac{1}{2}\)
b) \(\frac{7}{9}:\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(\Leftrightarrow2+\frac{3}{4}x=\frac{21}{8}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{5}{8}\)
\(\Leftrightarrow x=\frac{5}{6}\)
c) \(\frac{3}{5}.\left(3x-3,7\right)=-\frac{57}{10}\)
\(\Leftrightarrow3x-3,7=-\frac{19}{2}\)
\(\Leftrightarrow3x=-\frac{29}{5}\)
\(\Leftrightarrow x=-\frac{29}{15}\)
1,\(a,\frac{7}{12}+\frac{13}{32}\)
\(=\frac{56}{96}+\frac{39}{96}\)
\(=\frac{95}{96}\)
\(b,\frac{-18}{24}+\frac{25}{30}\)
\(=\frac{-3}{4}+\frac{5}{6}\)
\(=\frac{-18}{24}+\frac{20}{24}=\frac{2}{24}=\frac{1}{12}\)
2,\(a,\frac{2}{5}+\frac{-3}{7}=\frac{x}{70}\)
\(=>\frac{28}{70}-\frac{30}{70}=\frac{x}{70}\)
\(=>-\frac{2}{70}=\frac{x}{70}\)
\(=>x=-2\)
\(b,\frac{5}{6}+\frac{-19}{30}=\frac{1}{x}\)
\(=>\frac{25}{30}-\frac{19}{30}=\frac{1}{x}\)
\(=>\frac{6}{30}=\frac{1}{x}\)
\(=>\frac{1}{5}=\frac{1}{x}\)
\(=>x=5\)
1.
a) \(\frac{7}{12}+\frac{13}{32}=\frac{56}{96}+\frac{39}{96}=\frac{95}{96}\)
b) \(\frac{-18}{24}+\frac{25}{30}=\frac{-18}{24}+\frac{20}{24}=\frac{2}{24}=\frac{1}{12}\)
a) \(\left(x+\frac{1}{4}\right)^2+\frac{11}{25}=\frac{18}{25}\)
\(\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{7}{25}\)
\(\Rightarrow\) Không có x
Bài1
a) 25/42 - 20/63 =5/18
b) 9/50 - 13/75 - 1/6 = -4/25
c) 2/15 - 2/65 - 4/39 = 0
Bài2
a) x + 7/12 =17/18-1/9 b) 29/30 - (18/23 + x)=7/69
x + 7/12 = 5/6 18/23 + x =29/30 - 7/69
x =5/6 - 7/12 18/23 +x = 199/230
x = 1/4 x = 199/230 - 18/23
x= 19/230
a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
câu a đề có sai ko bn
a, \(\frac{x}{4}=\frac{24}{28}\)
\(\Rightarrow28x=4.24=96\)
\(x=96\div28\)
\(x=\frac{24}{7}\)
b, \(x-\frac{7}{18}=3\frac{5}{12}\)
\(x-\frac{7}{18}=\frac{41}{12}\)
\(x=\frac{41}{12}+\frac{7}{18}\)
\(x=\frac{137}{36}\)