a) \(0,\left(3\right)+3\dfrac{1}{2}+0,4\left(2\right)\)

\(\)\(Ta\) đưa \(0,\left(3\right)\) và \(0,4\left(2\right)\) về phân số như sau:

+ Đặt \(x=0,\left(3\right)\) thì \(10x=3,\left(3\right)=3+0,\left(3\right)=3+x\)

Suy ra \(9x=3\) hay \(x=\dfrac{1}{3}=0,\left(3\right)\)

+ Ta có \(0,4\left(2\right)=0,4+0,0\left(2\right)\)

Đặt \(y=0,0\left(2\right)\) thì \(100y=2,\left(2\right)=2+10y\)

Suy ra \(90y=2\) hay \(y=0,0\left(2\right)=\dfrac{1}{45}\)

Do đó, \(0,4\left(2\right)=0,4+0,0\left(2\right)=\dfrac{19}{45}\)

quay trở lại bài toán \(0,\left(3\right)+3\dfrac{1}{2}+0,4\left(2\right)=\dfrac{1}{3}+\dfrac{7}{2}+\dfrac{19}{45}=\dfrac{383}{90}\)

b) \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)\)

Ta đưa \(1,2\left(31\right)\) và \(0,\left(13\right)\) về phân số nhứ sau:

Đặt \(x=0,\left(01\right)\) thì \(100x=1,\left(01\right)=1+x\)

Suy ra \(99x=1\) hay \(x=\dfrac{1}{99}=0,\left(01\right)\)

+ Tính \(1,2\left(31\right)\)

Xét \(0,\left(31\right)=0,\left(01\right).31=31.\dfrac{1}{899}=\dfrac{31}{99}\)
Vậy \(1,2\left(31\right)=1+0,2+0,0\left(31\right)=1+\dfrac{1}{5}+\dfrac{31}{990}=\dfrac{1219}{990}\)

+ Tính \(0,\left(13\right)=13.0,\left(01\right)=13.\dfrac{1}{99}=\dfrac{13}{99}\)

Quay trở lại bài toán: \(\dfrac{4}{9}+1,2\left(31\right)-0,\left(13\right)=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}=\dfrac{139}{90}\)

a) \(A=x^2-2x+3\) khi \(|x|=0,5\)

Ta có \(|x|=0,5\) thì x=0,5 hoặc x=-0,5

+ Với x=0,5 ta có \(A=0,5^2-2.0,5+3=2,25\)

+ Với x=-0,5 ta có \(A=\left(-0,5\right)^2-2.\left(-0,5\right)+3=4,25\)

b) \(b=x-3+|1-3x|\) khi \(|x|=\dfrac{1}{3}\)

Ta có \(|x|=\dfrac{1}{3}\) thì \(x=\dfrac{1}{3}\) hoặc \(x=-\dfrac{1}{3}\)

+ Với \(x=\dfrac{1}{3}\) ta có \(B=\dfrac{1}{3}-3+|1-3.\dfrac{1}{3}|=-\dfrac{8}{3}\)

+ Với \(x=-\dfrac{1}{3}\) ta có \(B=-\dfrac{1}{3}-3+|1-3.\left(-\dfrac{1}{3}\right)|=\dfrac{-1}{3}-3+2=-\dfrac{4}{3}\)

a) m=\(\sqrt{25+9}\) và n=\(\sqrt{25}+\sqrt{9}\)

Ta có \(m=\sqrt{34}\) và \(n=5+3=8=\sqrt{64}\)

Mà \(34< 64\) nên \(m< n\)

b) \(y=\sqrt{49-16}\) và \(z=\sqrt{81}-\sqrt{9}\)

Ta có \(y=\sqrt{33}\) và \(z=9-3=6=\sqrt{36}\)

Mà \(33< 36\) nên \(y< z\)

a) \(A=\sqrt{36}.\left(3\sqrt{4}-\sqrt{\dfrac{1}{9}}\right)+2=6.\left(3.2-\dfrac{1}{3}\right)+2=36-2+2=36\\ \)

b) \(B=\sqrt{\dfrac{1}{9}+\dfrac{1}{16}}=\sqrt{\dfrac{9+6}{9.16}}=\sqrt{\dfrac{5^2}{3^2.4^2}}=\dfrac{5}{12}\)

c) \(C=\left(\sqrt{\dfrac{1}{9}}+\sqrt{\dfrac{25}{36}}-\sqrt{\dfrac{49}{81}}\right):\sqrt{\dfrac{441}{324}}=\left(\dfrac{1}{3}+\dfrac{5}{6}-\dfrac{7}{9}\right):\sqrt{\dfrac{21^2}{18^2}}=\dfrac{7}{18}:\dfrac{7}{6}=\dfrac{1}{3}\)

d)\(D=\sqrt{\left(\dfrac{-2}{5}\right)^2}+\sqrt{1,44}-\sqrt{256}=\dfrac{2}{5}+1,2-16=\dfrac{-72}{5}\)

Dọc theo chiều dài, ta trồng được:

\(5,5:\dfrac{1}{4}=22\)(khóm hoa)

Dọc theo chiều rộng, ta trồng được:

\(3,75:\dfrac{1}{4}=15\\\)(khóm hoa)

Như vậy, số khóm hoa trồng được dọc theo hai cạnh của mảnh vườn là:

[(22+15).2]-4=70 (khóm hoa)  

               Đáp số: 70 khóm hoa

a) \(\dfrac{1}{5}+\dfrac{4}{5}:x=0,75\\ \dfrac{1}{5}+\dfrac{4}{5}:x=\dfrac{3}{4}\\ \dfrac{4}{5}:x=\dfrac{3}{4}-\dfrac{1}{5}\\ \dfrac{4}{5}:x=\dfrac{11}{20}\\ x=\dfrac{16}{11}\)

b) \(x+\dfrac{1}{2}=1-x\\ 2x=1-\dfrac{1}{2}\\ 2x=\dfrac{1}{2}\\ x=\dfrac{1}{4}\)

a) \(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}=\dfrac{2}{3}\left(\dfrac{5}{4}-\dfrac{3}{4}\right)=\dfrac{2}{3}.\dfrac{1}{2}=\dfrac{1}{3}\)

b) \(2.\left(\dfrac{-3}{2}\right)^2-\dfrac{7}{2}=2.\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)

c) \(\dfrac{-3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}=\dfrac{-3}{4}.5\dfrac{3}{13}-\dfrac{3}{4}-\dfrac{36}{13}=\dfrac{-3}{4}\left(5\dfrac{3}{13}+\dfrac{36}{13}\right)=\dfrac{-3}{4}.8=-6\)

a) \(\left(\dfrac{1}{2}+1,5\right).x=\dfrac{1}{5}\\ 2x=\dfrac{1}{5}\\ x=\dfrac{1}{5}:2\\ x=\dfrac{1}{10}\)

b) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\\ -1\dfrac{3}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\\ x=2+1\dfrac{3}{5}\\ x=3\dfrac{3}{5}\)

c) \(\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=\dfrac{-3}{8}\\ x.\dfrac{3}{7}.\dfrac{1}{7}=\dfrac{-3}{8}\\ x=\dfrac{-3}{8}:\dfrac{3}{49}\\ x=\dfrac{-49}{8}=-6\dfrac{1}{8}\)

d) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{2}\right)\\ \dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}.\dfrac{-3}{5}\\ \dfrac{-4}{7}x=\dfrac{-3}{40}-\dfrac{7}{5}\\ x=\dfrac{-59}{40}:\dfrac{-4}{7}=\dfrac{413}{=160}=2\dfrac{93}{160}\)

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

\(x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{1}{5}-x=\dfrac{1}{5}\)

\(x=1-\dfrac{1}{5}-\dfrac{1}{5}\)

\(x=\dfrac{3}{5}\)

a) \(x=7-\dfrac{2}{5}+1,62=8,22\)

b)\(x=4\dfrac{3}{5}+\dfrac{1}{5}-\dfrac{1}{2}=4\dfrac{3}{10}\)

c)\(2x-x=\dfrac{3}{5}+\dfrac{4}{7}\)

\(x=\dfrac{41}{35}\)

d) \(x=3\dfrac{1}{2}-\dfrac{5}{7}+\dfrac{1}{13}-0,25=2\dfrac{223}{364}\)