\(3\dfrac{1}{14}-x=\dfrac{1}{11}\\ x=3\dfrac{1}{4}-\dfrac{1}{11}\\ x=\dfrac{3\times4+1}{4}-\dfrac{1}{11}\\ x=\dfrac{13}{4}-\dfrac{1}{11}\\ x=\dfrac{143}{44}-\dfrac{4}{44}\\ x=\dfrac{139}{44}\)

a) \(\dfrac{-14}{12}+0,65-\left(\dfrac{-7}{42}-0,35\right)\\ =\dfrac{-7}{6}+0,65+\dfrac{7}{42}+0,35\\ =\left(-\dfrac{7}{6}+\dfrac{7}{42}\right)+\left(0,65+0,35\right)\\ =\left(-\dfrac{7}{6}+\dfrac{1}{6}\right)+1\\ =\dfrac{-6}{6}+1=-1+1=0\)

b) \(\left(\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}\right)-\left(-\dfrac{3}{7}+1-\dfrac{13}{8}\right)\\ =\dfrac{7}{8}-\dfrac{5}{2}+\dfrac{4}{7}+\dfrac{3}{7}-1+\dfrac{13}{8}\\ =\left(\dfrac{7}{8}+\dfrac{13}{8}-\dfrac{5}{2}\right)+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-1\\ =\left(\dfrac{20}{8}-\dfrac{20}{8}\right)+\dfrac{7}{7}-1\\ =0+1-1=0\)

c) \(\dfrac{1}{2}-\dfrac{43}{101}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\\ =\dfrac{1}{2}-\dfrac{43}{101}-\dfrac{1}{3}-\dfrac{1}{6}\\ =\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)-\dfrac{43}{101}\\ =\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)-\dfrac{43}{101}\\ =0-\dfrac{43}{101}=-\dfrac{43}{101}\)

\(\dfrac{1}{a^{150}}< \left(\dfrac{1}{5}\right)^{225}\\ \Rightarrow\left(\dfrac{1}{a}\right)^{150}< \left(\dfrac{1}{5}\right)^{225}\\ \Rightarrow\left[\left(\dfrac{1}{a}\right)^2\right]^{75}< \left[\left(\dfrac{1}{5}\right)^3\right]^{75}\\ \Rightarrow\left(\dfrac{1}{a}\right)^2< \left(\dfrac{1}{5}\right)^3\\ \Rightarrow\dfrac{1}{a^2}< \dfrac{1}{5^3}\\ \Rightarrow a^2>5^3\\ \Rightarrow a^2>125\)

Mà: a là số nguyên dương nhỏ nhất

Vậy: a = 12

g) \(\left(-\dfrac{1}{5}\right)^2+\dfrac{1}{5}-2\times\left(-\dfrac{1}{2}\right)^3-\dfrac{1}{2}\\ =\dfrac{1}{25}+\dfrac{5}{25}-2\times\left(-\dfrac{1}{8}\right)-\dfrac{1}{2}\\ =\dfrac{6}{25}+\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{100}\)

h) \(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)^2:2020^0=\left(\dfrac{36}{12}-\dfrac{3}{12}+\dfrac{8}{12}\right)^2:1\\ =\left(\dfrac{41}{12}\right)^2=\dfrac{1681}{144}\)

i) \(2^2.9:\left(3\dfrac{4}{5}+0,2\right)=4.9:\left(3+\dfrac{8}{10}+\dfrac{2}{10}\right)\\ =36:\left(3+\dfrac{10}{10}\right)=36:\left(3+1\right)=36:4=9\)

d) \(\left(\dfrac{3}{5}\right)^2-\left(4,5-6\dfrac{1}{2}\right)=\dfrac{9}{25}-\left(4,5-6,5\right)\\ =\dfrac{9}{25}-\left(-2\right)=\dfrac{9}{25}+2=\dfrac{59}{25}\)

e) \(\dfrac{9}{25}-2^2:\left(-0,2\right)=\dfrac{9}{25}-4:\dfrac{-2}{10}\\ =\dfrac{9}{25}-4\times\left(-5\right)=\dfrac{9}{25}+20=20\dfrac{9}{25}\)

a) \(\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:\dfrac{-9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\times\dfrac{2}{-9}\\ =\dfrac{1}{4}-\left(-\dfrac{1}{12}\right)=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)

b) \(\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{3}{10}\\ =\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

c) \(\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}\times\dfrac{9}{5}-1=\dfrac{1}{5}-1\\ =-\dfrac{4}{5}\)

Nửa chu vi HCN:

  630 : 2 = 315 (m)

Chiều dài gấp rưỡi chiều rộng hay chiều dài bằng \(\dfrac{3}{2}\) chiều rộng

Coi chiều dài 3 phần và chiều rộng 2 phần

Tổng số phần bằng nhau:

  3 + 2 = 5 (phần)

Chiều rộng HCN:

  315 : 5 x 2 = 126 (m)

Chiều dài HCN:

  315 - 126 = 189 (m)

         Đáp số: 189m và 126m

Nửa chu vi HCN là:

  92 : 2 = 46 (m)

Nếu tăng chiều rộng 5m và giảm chiều dài 5m thì mảnh vườn trở thành hình vuông hay lúc này chiều dài và rộng bằng nhau

Ban đầu chiều dài hơn chiều rộng:

  5 + 5 = 10(m)

Chiều rộng HCN:

  (46 - 10) : 2 = 18 (m)

Chiều dài HCN:

  18 + 10 = 28 (m)

Diện tích HCN:

  28 x 18 = 504 \(\left(m^2\right)\)

    Đáp số: 504\(m^2\)

Mai còn số quả cam là:

  5 - 4 = 1 (quả cam)

   Đáp số: 1 quả cam

a) \(\dfrac{2}{3}+\dfrac{-2}{18}=\dfrac{2}{3}-\dfrac{1}{9}\\ =\dfrac{6}{9}-\dfrac{1}{9}=\dfrac{5}{9}\)

b) \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0\)

c) \(\dfrac{-6}{11}:\dfrac{7}{4}+\dfrac{-6}{11}:\dfrac{7}{3}+1\dfrac{1}{11}\\ =-\dfrac{6}{11}\times\dfrac{4}{7}+\dfrac{-6}{11}\times\dfrac{3}{7}+\dfrac{12}{11}\\ =-\dfrac{6}{11}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{12}{11}\\ =-\dfrac{6}{11}\times1+\dfrac{12}{11}\\ =-\dfrac{6}{11}+\dfrac{12}{11}=\dfrac{6}{11}\)