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\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(6x-42=7y-42\)
\(6x=7y\Leftrightarrow x=\dfrac{7}{6}y\)
\(x=-4:\left(7-6\right).7=-28\)
\(y=-28-4=-24\)
b tương tự
Giải:b)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\) nên \(6\left(x-7\right)=7\left(y-6\right)\)
Do đó \(6x-42=7y-42\) nên \(6x=7y\)
Suy ra \(6x-6y=y\) hay \(6\left(x-y\right)=y\)
Nên 6.(-4) = y
Vậy y = -24, x = \(\dfrac{7.\left(-24\right)}{6}\)= -28
c)
\(\dfrac{x+3}{y+5}=\dfrac{3}{5}\) nên \(5\left(x+3\right)=3\left(y+5\right)\)
Do đó \(5x+15=3y+15\) nên \(5x=3y\)
Suy ra \(5x+5y=3y+5y\)
\(5\left(x+y\right)=8y\)
\(5.16=8y\)
Nên \(y=\dfrac{5.16}{8}=\dfrac{80}{8}=10\)
Vậy y = 10, x = 16 - 10 =6
Đặt A = \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
\(\Rightarrow2A=\)\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{15}=\dfrac{14}{15}\)
\(\Rightarrow A=\dfrac{14}{15}:2=\dfrac{7}{15}\)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
a) Để phân số \(\dfrac{12}{n}\) có giá trị nguyên thì :
\(12⋮n\)
\(\Leftrightarrow n\inƯ\left(12\right)\)
\(\Leftrightarrow n\in\left\{-1;1;-12;12;-2;2;-6;6;-3;3;-4;4\right\}\)
Vậy \(n\in\left\{-1;1;-12;12;-2;2-6;6;-3;3;-4;4\right\}\) là giá trị cần tìm
b) Để phân số \(\dfrac{15}{n-2}\) có giá trị nguyên thì :
\(15⋮n-2\)
\(\Leftrightarrow x-2\inƯ\left(15\right)\)
Tới đây tự lập bảng zồi làm típ!
c) Để phân số \(\dfrac{8}{n+1}\) có giá trị nguyên thì :
\(8⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(8\right)\)
Lập bảng rồi làm nhs!
\(4\dfrac{1}{3}.\dfrac{4}{9}+13\dfrac{2}{3}.\dfrac{4}{9}\)\(=\dfrac{4}{9}\left(4\dfrac{1}{3}+13\dfrac{2}{3}\right)=\dfrac{4}{9}.18=8\)
\(5\dfrac{1}{4}.\dfrac{3}{8}+10\dfrac{3}{4}.\dfrac{3}{8}=\dfrac{3}{8}\left(5\dfrac{1}{4}+10\dfrac{3}{4}\right)=\dfrac{3}{8}.16=6\)
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\left(\dfrac{1}{2}+\dfrac{1}{3}\right):2\)
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\dfrac{5}{6}:2\)
\(\dfrac{1,11+0,19-1,3.2}{0,269+0,094}-x=\dfrac{5}{12}\)
\(\dfrac{1,3-2,6}{0,363}-x=\dfrac{5}{12}\)
\(\dfrac{-1,3}{0,363}-x=\dfrac{5}{12}\)
\(\dfrac{-1300}{363}-x=\dfrac{5}{12}\)
\(x=\dfrac{-1300}{363}-\dfrac{5}{12}\)
\(x=\dfrac{-1935}{484}\)
\(A=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{-5}{7}\)
\(=\dfrac{-5}{7}+\dfrac{-5}{7}+1=\dfrac{-3}{7}\)
\(B=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(=\dfrac{7}{10}.\dfrac{5}{28}.20=\dfrac{5}{2}.\)
Ta có :
A = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)
= \(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}+1\)
= \(\left(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}\right)+1\)
= \(\dfrac{5}{7}.\left(\dfrac{-2}{11}-\dfrac{9}{11}+1\right)+1\)
= \(\dfrac{5}{7}.0+1\)
= 1
B = \(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
= \(\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
= \(\left(\dfrac{7}{10}.20\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\dfrac{5}{28}\)
= 14.1.\(\dfrac{5}{28}\)
= \(\dfrac{5}{2}\)
Vậy A = 1
B = \(\dfrac{5}{2}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)
\(=3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}\right)\)
\(=3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)
\(=1-\dfrac{1}{43}\)
\(=\dfrac{42}{43}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\)
\(=\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{13}{56}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}\cdot\left(-2\right)^2\\ =\dfrac{6}{7}+\dfrac{5}{8}\cdot\dfrac{1}{5}-\dfrac{3}{16}\cdot4\\ =\dfrac{6}{7}+\left(\dfrac{5}{8}\cdot\dfrac{1}{5}\right)-\left(\dfrac{3}{16}\cdot4\right)\\ =\dfrac{6}{7}+\left(\dfrac{1\cdot1}{8\cdot1}\right)-\left(\dfrac{3}{4}\cdot1\right)\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}\\ =\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{42}{56}\\ =\dfrac{48+7-42}{56}\\ =\dfrac{13}{56}\)