Bài 6:

a: \(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)

\(\dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35}\)

b: \(\dfrac{-2}{21}=\dfrac{-2\cdot8}{21\cdot8}=\dfrac{-16}{168}\)

\(\dfrac{5}{-24}=\dfrac{-5}{24}=\dfrac{-5\cdot7}{24\cdot7}=\dfrac{-35}{168}\)

c: \(-\dfrac{7}{12}=\dfrac{-7\cdot3}{12\cdot3}=\dfrac{-21}{36}\)

\(\dfrac{11}{-18}=\dfrac{-11}{18}=\dfrac{-11\cdot2}{18\cdot2}=\dfrac{-22}{36}\)

d: \(\dfrac{-15}{45}=\dfrac{-1}{3}=\dfrac{-1\cdot5}{3\cdot5}=\dfrac{-5}{15}\)

\(\dfrac{-12}{60}=\dfrac{-12:12}{60:12}=\dfrac{-1}{5}=\dfrac{-1\cdot3}{5\cdot3}=\dfrac{-3}{15}\)

e: \(\dfrac{3}{20}=\dfrac{3\cdot3}{20\cdot3}=\dfrac{9}{60}\)

\(\dfrac{4}{30}=\dfrac{4\cdot2}{30\cdot2}=\dfrac{8}{60}\)

\(\dfrac{7}{15}=\dfrac{7\cdot4}{15\cdot4}=\dfrac{28}{60}\)

f: \(\dfrac{-5}{16}=\dfrac{-5\cdot3}{16\cdot3}=\dfrac{-15}{48}\)

\(\dfrac{11}{24}=\dfrac{11\cdot2}{24\cdot2}=\dfrac{22}{48}\)

\(\dfrac{-21}{56}=\dfrac{-21:7}{56:7}=\dfrac{-3}{8}=\dfrac{-3\cdot6}{8\cdot6}=\dfrac{-18}{48}\)

Xét tứ giác MEHF có \(\widehat{MEH}+\widehat{MFH}=90^0+90^0=180^0\)

nên MEHF là tứ giác nội tiếp

Số cái bắt tay tất cả là:

60x59:2=30x59=1770(cái)

Ta có: \(A=1000-\left\{\left(-5\right)^3\cdot\left(-2\right)^3-11\cdot\left[7^2-5\cdot2^3+8\left(11^2-121\right)\right]\right\}\)

\(=1000-\left\{\left(-125\right)\cdot\left(-8\right)-11\cdot\left(49-40\right)\right\}\)

\(=1000-1000+11\cdot9\)

=0+99

=99

Ta có: \(A=1000-\left\{\left(-5\right)^3\cdot\left(-2\right)^3-11\cdot\left[7^2-5\cdot2^3+8\left(11^2-121\right)\right]\right\}\)

\(=1000-\left\{\left(-125\right)\cdot\left(-8\right)-11\cdot\left(49-40\right)\right\}\)

\(=1000-1000+11\cdot9\)

=0+99

=99

Sau 10 phút, lượng nước trong bể đã chiếm:

\(\dfrac{10}{40}=\dfrac{1}{4}\left(bể\right)\)

Tổng, hiệu tỉ lệ nghịch với 35;210

=>35(x+y)=210(x-y)

=>x+y=6(x-y)

=>x+y=6x-6y

=>-5x=-7y

=>\(\dfrac{x}{7}=\dfrac{y}{5}=k\)(k>0)

=>x=7k; y=5k

Hiệu và tích lần lượt tỉ lệ nghịch với 210;12

=>\(210\left(x-y\right)=12\cdot x\cdot y\)

=>\(210\left(7k-5k\right)=12\cdot7k\cdot5k\)

=>\(210\cdot2k=420k^2\)

=>\(k^2=k\)

=>\(k^2-k=0\)

=>k(k-1)=0

mà k>0

nên k-1=0

=>k=1

=>\(x=7\cdot1=7;y=5\cdot1=5\)

Bạn gửi lại ảnh đi bạn!

a: Để \(\dfrac{3}{n-3}\) là số nguyên thì \(n-3\inƯ\left(3\right)\)

=>\(n-3\in\left\{1;-1;3;-3\right\}\)

=>\(n\in\left\{4;2;6;0\right\}\)

b: Để \(-\dfrac{3}{n-1}\) là số nguyên thì \(n-1\inƯ\left(-3\right)\)

=>\(n-1\in\left\{1;-1;3;-3\right\}\)

=>\(n\in\left\{2;0;4;-2\right\}\)

c: Để \(\dfrac{4}{3n+1}\) là số nguyên thì \(3n+1\inƯ\left(4\right)\)

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(3n\in\left\{0;-2;1;-3;3;-5\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

mà n nguyên

nên \(n\in\left\{0;1;-1\right\}\)

\(15p=\dfrac{15}{60}\left(giờ\right)=\dfrac{1}{4}\left(giờ\right)\)

\(90phút=\dfrac{90}{60}\left(giờ\right)=\dfrac{3}{2}giờ\)