1. Tính
a) $\frac{-1}{16}$+ $\frac{-2}{5}$
b) $\frac{-1}{8}$ – $\frac{3}{15}$
c) $\frac{-18}{10}$ + $\text{0,4}$
d) $\text{6,5}$ – $\frac{-1}{5}$
e) $\frac{-2}{26}$+ $\text{4}$
f) $\frac{12}{17}$ – $\text{(-2)}$
g) ($\frac{3}{2}$ + $\frac{5}{3}$) + $\frac{-1}{3}$
h) $\text{-}$ ($\frac{12}{7}$ – $\frac{1}{3}$) + $\frac{12}{7}$
2. Tìm x:
a) x + $\frac{1}{3}$= $\frac{6}{7}$
b) $\frac{5}{15}$ – x =$\text{0,25}$
c) -x + $\frac{12}{5}$= $\text{-0,4}$
1. Tính
a) $\frac{-1}{16}$+$\frac{-2}{5}$=$\frac{-5+(-32)}{80}$= $\frac{-37}{80}$
b) $\frac{-1}{8}$-$\frac{3}{15}$=$\frac{-15-24}{120}$=$\frac{-39}{120}$=$\frac{-13}{40}$
c) $\frac{-18}{10}$+0,4=-1,8+0,4=-1,4
d) 6,5-$\frac{-1}{5}$=6,5+0,2=6,7
e) $\frac{-2}{26}$+4=$\frac{-2}{26}$+$\frac{4}{1}$ =$\frac{-1+52}{13}$=$\frac{51}{13}$
f) $\frac{12}{17}$-(-2)=$\frac{12}{17}$+$\frac{2}{1}$=$\frac{12+34}{17}$=$\frac{46}{17}$
g) ($\frac{3}{2}$+$\frac{5}{3}$)+$\frac{-1}{3}$=$\frac{3}{2}$+($\frac{5}{3}$+$\frac{-1}{3}$)=($\frac{3}{2}$+$\frac{4}{3}$=$\frac{9+8}{6}$=$\frac{17}{6}$
h) -($\frac{12}{7}$- $\frac{1}{3}$)+$\frac{12}{7}$=-$\frac{12}{7}$+$\frac{1}{3}$+$\frac{12}{7}$= (-$\frac{12}{7}$+$\frac{12}{7}$)+$\frac{1}{3}$=0+$\frac{1}{3}$=$\frac{1}{3}$
2. Tìm x:
a) x+$\frac{1}{3}$=$\frac{6}{7}$
⇒ x=$\frac{6}{7}$-$\frac{1}{3}$
⇒ x=$\frac{11}{21}$
b) $\frac{5}{15}$-x=0,25
⇒ x=$\frac{1}{3}$-0,25
⇒ x=$\frac{1}{3}$-$\frac{1}{4}$=$\frac{1}{12}$
c) -x+$\frac{12}{5}$=-0,4
⇒ -x=-0,4-$\frac{12}{5}$
⇒ -x=-$\frac{2}{5}$-$\frac{12}{5}$
⇒ -x=-$\frac{14}{5}$
⇒ x=$\frac{14}{5}$
Đáp án:
1.
a) `\frac{-1}{16}` + `\frac{-2}{5}`
= `\frac{-5}{80}` + `\frac{-32}{5}`
= `\frac{-37}{80}`
b) `\frac{-1}{8}` – `\frac{3}{15}`
= `\frac{-1}{8}` – `\frac{1}{5}`
= `\frac{-5}{40}` – `\frac{8}{40}`
= `\frac{-13}{40}`
c) `\frac{-18}{10}` + `0,4`
= `\frac{-18}{10}` + `\frac{4}{10}`
= `\frac{-14}{10}`
d) `6,5` – `\frac{-1}{5}`
= `\frac{65}{10}` – `\frac{-1}{5}`
= `\frac{65}{10}` – `\frac{-2}{10}`
= `\frac{63}{10}`
e) `\frac{-2}{26}` + `4`
= `\frac{-2}{26}` + `\frac{4}{1}`
= `\frac{-2}{26}` + `\frac{104}{26}`
= `\frac{51}{13}`
f) `\frac{12}{17}` – `(“-2“)`
= `\frac{12}{17}` + `2`
= `\frac{12}{17}` + `\frac{2}{1}`
= `\frac{12}{17}` + `\frac{34}{17}`
= `\frac{46}{17}`
g) `(“\frac{3}{2}` + `\frac{5}{3}` `)` +`\frac{-1}{3}`
= `\frac{19}{6}` + `\frac{-1}{3}`
= `\frac{19}{6}` + `\frac{-2}{6}`
= `\frac{17}{6}`
h) -`(“\frac{12}{7}` – `\frac{1}{3}` `)` + `\frac{12}{7}`
= -`\frac{12}{7}` – `\frac{1}{3}` + `\frac{12}{7}`
= `(`-`\frac{12}{7}` + `(“\frac{12}{7}“)` – `\frac{1}{3}`
= `0` – `\frac{1}{3}`
= -`\frac{1}{3}`
2.
a) x + `\frac{1}{3}` = `\frac{6}{7}`
x = `\frac{6}{7}` – `\frac{1}{3}`
x = `\frac{11}{21}`
Vậy x = `\frac{11}{21}`
b) `\frac{5}{15}` – `x` = `0,25`
`\frac{5}{15}` – `x` = `\frac{1}{4}`
`x` =`\frac{5}{15}` – `\frac{1}{4}`
`x` =`\frac{1}{12}`
Vậy `x` =`\frac{1}{12}`
c) `-x` + `\frac{12}{5}` = `0,4`
`-x` + `\frac{12}{5}` =`\frac{-2}{5}`
`-x` =`\frac{-2}{5}`- `\frac{12}{5}`
`-x` =`\frac{-14}{5}`
⇒ `x` =`\frac{14}{5}`
Vậy `x` =`\frac{14}{5}`