Tìm y:
y X\(\frac{52}{26}=\frac{17}{13}\)
\(4\frac{1}{2}:y=\frac{2}{7}\)
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a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
Ta có:
\(\left(\right. \frac{13 \frac{2}{9} - 15 \frac{2}{3}}{18 \frac{3}{7} - 17 \frac{1}{4}} \cdot \frac{30^{2} - 5^{4}}{25 - 12 \cdot 5^{2}} \left.\right) \cdot x = \frac{\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17}}{4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17}}\)
Bước 1: Đổi hỗn số về phân số
Bước 2: Tính toán từng phần
Ta có:
\(\frac{119}{9} - \frac{47}{3} = \frac{119 - 141}{9} = \frac{- 22}{9}\) \(\frac{129}{7} - \frac{69}{4} = \frac{516 - 483}{28} = \frac{33}{28}\) \(30^{2} - 5^{4} = 900 - 625 = 275\) \(25 - 12 \cdot 25 = 25 - 300 = - 275\)
Khi đó:
\(\left(\right. \frac{- 22}{9} \div \frac{33}{28} \cdot \frac{275}{- 275} \left.\right) = \left(\right. \frac{- 22}{9} \cdot \frac{28}{33} \cdot \left(\right. - 1 \left.\right) \left.\right) = \frac{616}{297}\)
Bước 3: Tính vế phải
Tử số:
\(\frac{2}{11} + \frac{3}{13} + \frac{4}{15} + \frac{5}{17} = \frac{35494}{36465}\)
Mẫu số:
\(4 \frac{1}{11} + \frac{5}{13} + \frac{9}{15} + \frac{13}{17} = \frac{149645}{36465}\)
→ Vế phải:
\(\frac{35494}{36465} \div \frac{149645}{36465} = \frac{35494}{149645}\)
Bước 4: Giải phương trình
\(\frac{616}{297} \cdot x = \frac{35494}{149645} \Rightarrow x = \frac{35494}{149645} \cdot \frac{297}{616} = \frac{813}{7118}\)
Vậy:
\(\boxed{x = \frac{813}{7118}}\)
hỉu không =]]]
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
Lời giải :
Do \(VT\ge0\forall x;y\)nên ta có hệ :
\(\hept{\begin{cases}\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x=0\\1,5-\frac{11}{17}+\frac{23}{13}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{9}\\y=\frac{-377}{782}\end{cases}}\)
Vậy...
\(y.\frac{52}{26}=\frac{17}{13}\)
\(y=\frac{17}{13}:\frac{52}{26}\)
\(y=\frac{17}{26}\)
\(4\frac{1}{2}:y=\frac{2}{7}\)
\(\frac{9}{2}:y=\frac{2}{7}\)
\(y=\frac{9}{2}:\frac{2}{7}\)
\(y=\frac{63}{4}\)
\(y\times\frac{52}{26}=\frac{17}{13}\)
\(y=\frac{17}{13}:\frac{52}{26}\)
\(y=\frac{221}{338}\)
\(4\frac{1}{2}:y=\frac{2}{7}\)
\(y=4\frac{1}{2}:\frac{2}{7}\)
\(y=\frac{63}{4}\)