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Nếu đề là \(A=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{3}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}\) thì bài mk lm đây
Ta có: \(A=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{3}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{3}\right)}=\frac{3}{11}\)
\(\Rightarrow A=\frac{\frac{3}{4}-\frac{6}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{11}{4}-\frac{22}{10}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow A=\frac{3.\left(\frac{1}{4}-\frac{2}{10}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{4}-\frac{2}{10}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow A=\frac{3}{11}\)
!
\(a,\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}{11\left[\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right]}=\frac{3}{11}\)
Câu b tương tự
#)Giải :
b)\(\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}\right)+\frac{5}{9}:\left(\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{1}{3}+\frac{5}{9}:\frac{1}{4}+\frac{5}{9}:\frac{1}{9}+\frac{5}{9}:\frac{2}{3}\)
\(=\frac{5}{9}:\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{9}+\frac{2}{3}\right)\)
\(=\frac{5}{9}:\frac{49}{36}\)
\(=\frac{20}{49}\)
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}\frac{2}{35}+\frac{2}{63}\)
\(\Rightarrow A=\frac{0,55+\frac{11}{7}+\frac{11}{13}}{0,15+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{42}{63}+\frac{14}{105}+\frac{6}{105}+\frac{2}{63}\)
\(\Rightarrow A=\frac{\frac{11}{20}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{20}+\frac{3}{7}+\frac{11}{13}}-x-\frac{1}{9}=\frac{44}{63}+\frac{20}{105}\)
\(\Rightarrow A=\frac{\frac{1001}{1820}+\frac{2860}{1820}+\frac{1540}{1820}}{\frac{273}{1820}+\frac{780}{1820}+\frac{1540}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{4}{21}\)
\(\Rightarrow A=\frac{\frac{5401}{1820}}{\frac{2593}{1820}}-x-\frac{1}{9}=\frac{44}{63}+\frac{12}{63}\)
\(\Rightarrow A=\frac{5401}{1820}:\frac{2593}{1820}-x-\frac{1}{9}=\frac{56}{63}\)
\(\Rightarrow A=\frac{5401}{1820}.\frac{1820}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow A=\frac{5401}{2593}-x-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=\frac{8}{9}+\frac{1}{9}\)
\(\Rightarrow\frac{5401}{2593}-x=1\)
\(\Rightarrow x=\frac{5401}{2593}-1\)
\(\Rightarrow x=\frac{2808}{2593}\)
Vậy \(x=\frac{2808}{2593}\)
Chúc bn học tốt
0,75-0,6+\(\frac{3}{7}+\frac{3}{\frac{13}{2,75}}\) -2,2+\(\frac{11}{7}\) +\(\frac{11}{13}\)
=\(\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{11}{16}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}\)
= \(\left(\frac{11}{16}+\frac{3}{4}\right)+\left[\left(\frac{-3}{5}\right)-\frac{11}{5}\right]+\left(\frac{3}{7}+\frac{11}{7}\right)+\frac{11}{3}\)
= \(\frac{23}{16}-\frac{14}{5}+2+\frac{11}{3}\)
=\(\left(\frac{23}{16}+2\right)+\left[\left(\frac{-14}{5}\right)+\frac{11}{3}\right]\)
=\(\frac{55}{16}+\frac{13}{15}\)
\(\approx\) 4,3
chúc bạn học good (like nink nha)
Đặt tử = A.Ta có:
\(A=0,75+\frac{3}{7}+\frac{3}{11}=\frac{3}{4}+\frac{3}{7}+\frac{3}{11}\)
\(=3\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{11}\right)\)
Đặt mẫu = B.Ta có:
\(B=2\frac{1}{4}+\frac{9}{11}+\frac{9}{7}=\frac{9}{4}+\frac{9}{11}+\frac{9}{7}\)
\(=9\left(\frac{1}{4}+\frac{1}{11}+\frac{1}{7}\right)\)
Thay A và B vào ta đc: \(\frac{A}{B}=\frac{3\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{11}\right)}{9\left(\frac{1}{4}+\frac{1}{11}+\frac{1}{7}\right)}=\frac{3}{9}=\frac{1}{3}\)
Đề có vấn đề