Cộng, trừ số hữu tỉ

Cộng, trừ số hữu tỉ

6th Grade

16 Qs

quiz-placeholder

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Cộng, trừ số hữu tỉ

Cộng, trừ số hữu tỉ

Assessment

Quiz

Mathematics

6th Grade

Practice Problem

Medium

CCSS
6.EE.B.6, 5.NF.A.1, 6.NS.A.1

+10

Standards-aligned

Created by

Nhung Trương

Used 223+ times

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16 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Chọn câu đúng. Với x =  am\frac{a}{m}  , y =  bm\frac{b}{m}   

(a, b, m ∈ Z; m > 0) ta có:

 x+y = am+bm=abmx+y\ =\ \frac{a}{m}+\frac{b}{m}=\frac{a-b}{m} 

 x+y =am+bm=a(+b)mx+y\ =\frac{a}{m}+\frac{b}{m}=\frac{a-\left(+b\right)}{m} 

 x+y =am+bm=a+bmx+y\ =\frac{a}{m}+\frac{b}{m}=\frac{a+b}{m} 

 x+y =am+bm=a+(b)mx+y\ =\frac{a}{m}+\frac{b}{m}=\frac{a+\left(-b\right)}{m} 

Tags

CCSS.6.EE.A.2

CCSS.6.EE.A.3

CCSS.6.EE.A.4

CCSS.6.EE.B.6

CCSS.6.NS.A.1

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Tính  34720\frac{3}{4}-\frac{7}{20}  

 110\frac{1}{10}  

 3040\frac{30}{40}  

 25\frac{2}{5}  

 45\frac{4}{5}  

Tags

CCSS.4.NF.A.2

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Tính  1718\frac{-1}{7}-\frac{1}{8}  

 156\frac{-1}{56}  

 1556\frac{-15}{56}  

 2156\frac{21}{56}  

 1556\frac{15}{56}  

Tags

CCSS.5.NF.A.1

CCSS.5.NF.A.2

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Tìm x, biết:   x+17=1114x+\frac{1}{7}=\frac{11}{14}  

 914\frac{9}{14}  

 114\frac{1}{14}  

 1114\frac{11}{14}  

 12\frac{1}{2}  

Tags

CCSS.6.EE.B.5

CCSS.6.EE.B.6

CCSS.6.EE.B.7

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Gía trị của x thỏa mãn  1520x=716\frac{15}{20}-x=\frac{7}{16}  là

 516\frac{-5}{16}  

 516\frac{5}{16}  

 1916\frac{19}{16}  

 1916\frac{-19}{16}  

Tags

CCSS.6.EE.B.5

CCSS.6.EE.B.6

CCSS.6.EE.B.7

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Tính  47+54328\frac{4}{7}+\frac{-5}{4}-\frac{3}{28}  ta được:

 3528\frac{-35}{28}  

 1114\frac{11}{14}  

 2228\frac{22}{28}  

 1114\frac{-11}{14}  

Tags

CCSS.5.NF.A.1

CCSS.5.NF.A.2

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Chọn câu đúng. Với x =  am,\frac{a}{m},   y =   bm\frac{b}{m}  
(a, b, m ∈ Z; m > 0) ta có:

 xy=ambm=a+bmx-y=\frac{a}{m}-\frac{b}{m}=\frac{a+b}{m}  

 xy=ambm=abmx-y=\frac{a}{m}-\frac{b}{m}=\frac{a-b}{m}  

 xy=ambm=a(b)mx-y=\frac{a}{m}-\frac{b}{m}=\frac{a-\left(-b\right)}{m}  

Tags

CCSS.6.EE.A.2

CCSS.6.EE.A.3

CCSS.6.EE.A.4

CCSS.6.EE.B.6

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