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Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
\(a,\frac{2}{3}\cdot x-\frac{4}{7}=\frac{1}{8}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{1}{8}+\frac{4}{7}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{7}{56}+\frac{32}{56}\)
\(\Leftrightarrow\frac{2}{3}\cdot x=\frac{39}{56}\)
\(\Leftrightarrow x=\frac{39}{56}:\frac{2}{3}=\frac{39}{56}\cdot\frac{3}{2}=\frac{39\cdot3}{56\cdot2}=\frac{117}{112}\)
\(b,\frac{2}{7}-\frac{8}{9}\cdot x=\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{2}{7}-\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{6}{21}-\frac{14}{21}\)
\(\Leftrightarrow\frac{8}{9}\cdot x=\frac{-8}{21}\)
\(\Leftrightarrow x=\frac{-8}{21}:\frac{8}{9}=\frac{-8}{21}\cdot\frac{9}{8}=\frac{-8\cdot9}{21\cdot8}=\frac{-1\cdot3}{7\cdot1}=\frac{-3}{7}\)
Làm nốt hai bài cuối đi nhé
Study well >_<
Mk k chép lại đề bài nha
a)\(\frac{2}{3}.x=\frac{1}{8}+\frac{4}{7}\)
\(\frac{2}{3}.x=\frac{7}{56}+\frac{32}{56}\)
\(\frac{2}{3}.x=\frac{39}{56}\)
\(x=\frac{39}{56}:\frac{2}{3}\)
\(x=\frac{39}{56}.\frac{3}{2}\)
\(x=\frac{117}{112}\)
Mk sợ sai lém!!!
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
a, \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}=6\frac{5}{7}-2\frac{5}{7}-\frac{7}{4}=4-\frac{7}{4}=\frac{16}{4}-\frac{7}{4}=\frac{9}{4}\)
b, (-4/5+4//3)+(-5/4+14/5)-7/3 = -4/5+4/3+-5/4+14/5-7/3=(-4/5+14/5)+(4/3-7/3)+-5/4=2+(-1)+-5/4=1+-5/4=-1/4
Phần b dấu ''/'' là dấu phần nha , mk viết ra phân số thì nó dài quá nên mk viết vậy , t.i.c.k mk hennnn ^^ thanks @@
Nếu thích kb qua face vs mk nick : Akiko Ayano nhé . Kb và ib nch vs nhau nè
\(A,1,1+\frac{2}{3}+0,75+\frac{5}{8}\)
\(=\frac{11}{10}+\frac{2}{3}+\frac{3}{4}+\frac{5}{8}\)
\(=\frac{11}{10}+\frac{2}{3}+\frac{11}{8}\)
\(=\frac{53}{30}+\frac{11}{8}\)
\(=\frac{377}{120}\)
\(b,\frac{5}{11}+\frac{6}{11}:\frac{13}{22}+\frac{7}{3}-\frac{1}{2}\)
\(=\frac{5}{11}+\frac{6}{11}.\frac{22}{13}+\frac{7}{3}-\frac{1}{2}\)
\(=\frac{5}{11}+\frac{12}{13}+\frac{7}{3}-\frac{1}{2}\)
\(=\frac{197}{143}+\frac{7}{3}-\frac{1}{2}\)
\(=\frac{1592}{429}-\frac{1}{2}\)
\(=\frac{2755}{858}\)
Lê Minh Phương tham khảo bài mình nhé
\(a,\frac{9}{-7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-9}{7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-18}{14}< x>\frac{49}{14}\)
\(\Leftrightarrow-18< x>49\)
\(\Leftrightarrow x\in\left\{-17;-16;-15;...;50\right\}\)
Còn bài kia tương tự
\(a,\frac{9}{-7}< x< \frac{7}{2}\)
\(\Rightarrow\frac{9.2}{-7.2}< x< \frac{7.7}{2.7}\)
\(\Rightarrow\frac{-18}{14}< x< \frac{49}{14}\)
\(\text{vì}x\in Z\Rightarrow x=-\frac{14}{14};\frac{0}{14};\frac{14}{14};\frac{28}{14};\frac{42}{14}\)
\(\text{hay }x=\left\{-1;0;1;2;3\right\}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
\(\frac{3}{5}-\frac{1}{9}-\frac{1}{2}-\frac{1}{127}+\frac{7}{18}+\frac{4}{35}+\frac{2}{7}\)=\((\frac{1}{9}-\frac{1}{2}+\frac{7}{18})+(\frac{3}{5}+\frac{2}{7}+\frac{4}{35})+\frac{1}{127}\)=\(0+1+\frac{1}{127}\)=\(1+\frac{1}{127}=\frac{128}{127}\)