\(\dfrac{x}{15}=\dfrac{3}{5}-\dfrac{-2}{3}\) ok
\(\Rightarrow\dfrac{x}{15}=\dfrac{9}{15}+\dfrac{10}{15}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{19}{15}\)
\(\Rightarrow x=19\)
Những câu hỏi liên quan
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow1^+}\dfrac{x^3+x+1}{x-1}\)
b) \(\lim\limits_{x\rightarrow-1^+}\dfrac{3x+2}{x+1}\)
c) \(\lim\limits_{x\rightarrow2^-}\dfrac{x-15}{x-2}\)
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
Đúng 1
Bình luận (0)
8. dfrac{-5}{9} + dfrac{8}{15} + dfrac{-2}{11} + dfrac{4}{-9} + dfrac{7}{15}9. dfrac{2}{7} + (dfrac{-2}{5} + dfrac{5}{7})10. dfrac{7}{19}. dfrac{8}{11} + dfrac{3}{11}.dfrac{7}{19}+dfrac{-12}{19}11. dfrac{-5}{7}.dfrac{2}{11} + dfrac{-5}{7}.dfrac{9}{11}12. dfrac{-5}{13} + dfrac{5}{7} + dfrac{20}{41} + dfrac{-8}{13} + dfrac{21}{41}Giúp tớ với ạ! Tớ cảm ơn! Các cậu chỉ cần ghi đáp án cuối cùng thôi ạ! Cảm ơn các cậu3
Đọc tiếp
8. \(\dfrac{-5}{9}\) + \(\dfrac{8}{15}\) + \(\dfrac{-2}{11}\) + \(\dfrac{4}{-9}\) + \(\dfrac{7}{15}\)
9. \(\dfrac{2}{7}\) + (\(\dfrac{-2}{5}\) + \(\dfrac{5}{7}\))
10. \(\dfrac{7}{19}\). \(\dfrac{8}{11}\) + \(\dfrac{3}{11}\).\(\dfrac{7}{19}\)+\(\dfrac{-12}{19}\)
11. \(\dfrac{-5}{7}\).\(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\).\(\dfrac{9}{11}\)
12. \(\dfrac{-5}{13}\) + \(\dfrac{5}{7}\) + \(\dfrac{20}{41}\) + \(\dfrac{-8}{13}\) + \(\dfrac{21}{41}\)
Giúp tớ với ạ! Tớ cảm ơn! Các cậu chỉ cần ghi đáp án cuối cùng thôi ạ! Cảm ơn các cậu<3
8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)
9: =2/7-2/5+5/7=1-2/5=3/5
10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)
11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)
Đúng 2
Bình luận (0)
8 = -2/11
9 = 3/5
10 = -5/19
11 = -5/7
11 = 5/13
Đúng 0
Bình luận (0)
\(\dfrac{15}{19}\)x\(\dfrac{38}{5}\)<x<\(\dfrac{67}{15}\)+\(\dfrac{56}{15}\)
C= 2-\(\dfrac{5}{3}\)+\(\dfrac{7}{6}\)-\(\dfrac{9}{10}\)+\(\dfrac{11}{15}\)-\(\dfrac{13}{21}\)+\(\dfrac{15}{28}\)-\(\dfrac{17}{36}\)+\(\dfrac{19}{45}\)
tính C
\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)
\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)
\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)
Đúng 1
Bình luận (0)
Bài 15:a)dfrac{-2}{5}+dfrac{4}{5} . x dfrac{3}{5}b)dfrac{-3}{7} - dfrac{4}{7}:x -2Bài 16a) x - dfrac{10}{3} dfrac{7}{15} . dfrac{3}{5}b) x + dfrac{3}{22} dfrac{27}{121} . dfrac{11}{9}c) dfrac{8}{23} . dfrac{48}{24} - x dfrac{1}{3}d) 1 - x dfrac{49}{65}.dfrac{5}{7}Bài 17: tìm xa) dfrac{62}{7} . x dfrac{29}{9}: dfrac{3}{56}b) dfrac{1}{5} : xdfrac{1}{5}+dfrac{1}{7}bài 18:a)dfrac{2}{5}+dfrac{3}{4}: x dfrac{-1}{2}b)dfrac{5}{7} - dfrac{2}{3} . x dfrac{4}{5}c) dfrac{1}{2}x + dfrac{3}{5}x dfrac{-...
Đọc tiếp
Bài 15:
a)\(\dfrac{-2}{5}\)+\(\dfrac{4}{5}\) . x =\(\dfrac{3}{5}\)
b)\(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\):x = -2
Bài 16
a) x - \(\dfrac{10}{3}\) = \(\dfrac{7}{15}\) . \(\dfrac{3}{5}\)
b) x + \(\dfrac{3}{22}\)= \(\dfrac{27}{121}\) . \(\dfrac{11}{9}\)
c) \(\dfrac{8}{23}\) . \(\dfrac{48}{24}\) - x = \(\dfrac{1}{3}\)
d) 1 - x = \(\dfrac{49}{65}\).\(\dfrac{5}{7}\)
Bài 17: tìm x
a) \(\dfrac{62}{7}\) . x = \(\dfrac{29}{9}\): \(\dfrac{3}{56}\)
b) \(\dfrac{1}{5}\) : x=\(\dfrac{1}{5}\)+\(\dfrac{1}{7}\)
bài 18:
a)\(\dfrac{2}{5}\)+\(\dfrac{3}{4}\): x =\(\dfrac{-1}{2}\)
b)\(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\)x + \(\dfrac{3}{5}\)x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\).x-x = \(\dfrac{-9}{14}\)
bài 19: tính
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+ \(\dfrac{1}{2018.2019}\)
bài 20:tìm x
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2008}{2009}\)
bài 21: tìm x
\(\dfrac{x+1}{99}\)+\(\dfrac{x+2}{98}\)\(\dfrac{x+3}{97}\)\(\dfrac{x+4}{96}\)=-4
bài 22 : so sánh các phân số sau:
a) \(\dfrac{-1}{5}\)+\(\dfrac{4}{-5}\)và 1
b) \(\dfrac{3}{5}\) và \(\dfrac{2}{3}\)+\(\dfrac{-1}{5}\)
c)\(\dfrac{3}{2}\)+\(\dfrac{-4}{3}\) và \(\dfrac{1}{10}\)+\(\dfrac{-4}{5}\)
d) \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{1}{6}\) và 2
\(A.\dfrac{-15}{28}x\dfrac{7}{25}\\ B.\dfrac{-5}{14}x\dfrac{7}{-3}\\ C.\dfrac{-1}{5}-\dfrac{7}{15}x\dfrac{9}{35}\\ D.\dfrac{-3}{4}-(\dfrac{-1}{2})^2\\ E.\dfrac{-4}{5}-\dfrac{-4}{5}x\dfrac{15}{16}\\F.(\dfrac{3}{4}+\dfrac{-7}{2})x(\dfrac{2}{11}+\dfrac{12}{22})\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Đúng 1
Bình luận (0)
\(\dfrac{-3}{4}x\dfrac{4}{7}+\dfrac{-3}{5}x\dfrac{3}{7}-\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}x\dfrac{5}{6}-\dfrac{7}{12}\right)x\left(\dfrac{-12}{7}\right)\)
\(1,4x\dfrac{15}{19}x\left(\dfrac{4}{5}x\dfrac{-2}{3}\right)+\dfrac{7}{15}\)
Tính nha
a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)
\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)
b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)
\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)
c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)
\(=\dfrac{-56}{95}+\dfrac{7}{15}\)
\(=\dfrac{-7}{57}\)
Đúng 0
Bình luận (0)
20 tháng 9 2023 lúc 17:35
\(a,\dfrac{15-x}{14}=\dfrac{5}{7}\)
\(b,\dfrac{x}{9}=\dfrac{17}{3}\)
\(c,15-\dfrac{x}{2}=\dfrac{3}{7}\)
Tìm x
\(\dfrac{15-x}{14}=\dfrac{5}{7}\\ \left(15-x\right)\times7=5\times14\\ \left(15-x\right)\times7=70\\ 15-x=70:7\\ 15-x=10\\ x=15-10\\ x=5\)
__
\(\dfrac{x}{9}=\dfrac{17}{3}\\ x\times3=17\times9\\ x\times3=153\\ x=153:3\\ x=51\)
__
\(15-\dfrac{x}{2}=\dfrac{3}{7}\\ \dfrac{x}{2}=15-\dfrac{3}{7}\\ \dfrac{x}{2}=\dfrac{102}{7}\\ x\times7=102\times2\\ x\times7=204\\ x=204:7\\ x=\dfrac{204}{7}\)
Đúng 1
Bình luận (0)
Bài 1:
a)
ta có: dfrac{50}{100}dfrac{1}{2};dfrac{-dfrac{4}{13}}{-dfrac{8}{13}}dfrac{1}{2};dfrac{dfrac{2}{15}}{dfrac{4}{15}}dfrac{1}{2};dfrac{-dfrac{2}{17}}{-dfrac{4}{17}}dfrac{1}{2}
dfrac{50}{100}dfrac{dfrac{4}{13}}{dfrac{8}{13}}dfrac{dfrac{2}{15}}{dfrac{4}{15}}dfrac{dfrac{2}{17}}{dfrac{4}{17}}dfrac{50-dfrac{4}{13}+dfrac{2}{15}-dfrac{2}{17}}{100-dfrac{8}{13}+dfrac{4}{15}-dfrac{4}{17}}dfrac{1}{2}
vậy Adfrac{1}{2}
b)
Bdfrac{1}{19}+dfrac{9}{19.29}+dfrac{9}{29.39}+...+dfrac{9}{1999.2009}
Bdfr...
Đọc tiếp
Bài 1:
a)
ta có: \(\dfrac{50}{100}=\dfrac{1}{2};\dfrac{-\dfrac{4}{13}}{-\dfrac{8}{13}}=\dfrac{1}{2};\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{1}{2};\dfrac{-\dfrac{2}{17}}{-\dfrac{4}{17}}=\dfrac{1}{2}\)
\(\dfrac{50}{100}=\dfrac{\dfrac{4}{13}}{\dfrac{8}{13}}=\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{\dfrac{2}{17}}{\dfrac{4}{17}}=\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}=\dfrac{1}{2}\)
vậy \(A=\dfrac{1}{2}\)
b)
\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\\ B=\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{2}{29}-\dfrac{2}{29}+\dfrac{3}{39}-...-\dfrac{199}{1999}+\dfrac{200}{2009}\\ B=\dfrac{200}{2009}\)
Bài 2:
\(\dfrac{a}{b}=\dfrac{b}{3c}=\dfrac{c}{9a}=\dfrac{b+c}{3c+9a}\)
suy ra: \(b=\dfrac{3c\left(b+c\right)}{3c+9a}=\dfrac{3cb+3c^2}{3c+9a}=\dfrac{bc+c^2}{c+3a}\)
\(c=\dfrac{9a\left(b+c\right)}{3c+9a}=\dfrac{9ab+9ac}{3c+9a}=\dfrac{3ab+3ac}{c+3a}\)
giả sử b=c là đúng thì :\(\dfrac{bc+c^2}{c+3a}=\dfrac{3ab+3ac}{c+3a}\)
hay \(bc+c^2=3ab+3ac\\ \Leftrightarrow c^2+bc-3ab-3ac=0\)
\(\Leftrightarrow\left(b+c\right)\left(c-3a\right)=0\Rightarrow c-3a=0\Rightarrow c=3a\)
b) \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}+\dfrac{2}{2014.2016}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{2015}{4032}< 1\)
mà \(1< \dfrac{4}{3}\) nên \(\dfrac{2015}{4032}< \dfrac{4}{3}\)
hay \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}< \dfrac{4}{3}\)
bài 3:
a)\(\left(x-y\right)\left(x+y\right)=x^2-y^2-xy+xy=x^2-y^2\) (đpcm)
b) áp dụng BĐT tam giác, ta có:
\(a+b>c\Rightarrow a+b-c>0\\ b+c>a\Rightarrow b+c-a< 0\\ a+c>b\Rightarrow a-b+c>0\)
suy ra: \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< 0\: \: \: \: \: \: \)
đồng thời \(abc>0\) với mọi a, b, c dương.
nên \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< abc\)
ko tìm dc dấu bằng xảy ra.
