-5 - x = -11

=> x = -5 - (-11)

=> x = -5 + 11

=> x = 11 - 5 = 6

=> x = 6

-5 - x = -11

=> x = -5 - (-11)

=> x = -5 + 11

=> x = 11 - 5 = 6

=> x = 6

j: \(\dfrac{7}{23}+\dfrac{-10}{18}+\dfrac{-4}{9}+\dfrac{16}{23}+\dfrac{-5}{8}\)

\(=\left(\dfrac{7}{23}+\dfrac{16}{23}\right)+\left(-\dfrac{10}{18}-\dfrac{4}{9}\right)-\dfrac{5}{8}\)

\(=1-1-\dfrac{5}{8}=-\dfrac{5}{8}\)

k: \(\dfrac{1}{3}+\dfrac{-3}{4}+\dfrac{3}{5}+\dfrac{-1}{36}+\dfrac{1}{15}+\dfrac{-2}{9}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(-\dfrac{3}{4}-\dfrac{1}{36}-\dfrac{2}{9}\right)\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}\)

\(=\dfrac{15}{15}-\dfrac{36}{36}=1-1=0\)

l: \(-\dfrac{2}{15}+\dfrac{12}{13}+\dfrac{-13}{15}+\dfrac{-1}{13}\)

\(=\left(-\dfrac{2}{15}-\dfrac{13}{15}\right)+\left(\dfrac{12}{13}+\dfrac{-1}{13}\right)\)

\(=\dfrac{-15}{15}+\dfrac{11}{13}=-1+\dfrac{11}{13}=-\dfrac{2}{13}\)

m: \(-\dfrac{5}{7}+\dfrac{3}{4}+\dfrac{-1}{5}+\dfrac{-2}{7}+\dfrac{1}{4}\)

\(=\left(-\dfrac{5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{-1}{5}\)

\(=\dfrac{-7}{7}+\dfrac{4}{4}+\dfrac{-1}{5}=-1+1+\dfrac{-1}{5}=-\dfrac{1}{5}\)

n: \(\dfrac{4}{7}+\dfrac{-1}{9}+\dfrac{3}{7}+\dfrac{7}{10}+\dfrac{-8}{9}\)

\(=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\left(-\dfrac{1}{9}-\dfrac{8}{9}\right)+\dfrac{7}{10}\)

\(=\dfrac{7}{7}-\dfrac{9}{9}+\dfrac{7}{10}=1-1+\dfrac{7}{10}=\dfrac{7}{10}\)

o: \(\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}\)

\(=\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)\)

\(=\dfrac{7}{7}+\dfrac{8}{4}\)

=1+2

=3

p: \(\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{3}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)

\(=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{3}{8}+\dfrac{6}{8}\right)\)

\(=\dfrac{12}{2}+\dfrac{11}{11}+\dfrac{9}{8}\)

\(=6+1+\dfrac{9}{8}=7+\dfrac{9}{8}=\dfrac{65}{8}\)

q: \(\dfrac{4}{3}+\dfrac{3}{5}+\dfrac{7}{3}+\dfrac{2}{5}+\dfrac{1}{3}\)

\(=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\)

\(=\dfrac{12}{3}+\dfrac{5}{5}=4+1=5\)

r: \(\dfrac{5}{17}+\dfrac{-9}{15}+\dfrac{-2}{17}+\dfrac{2}{-5}\)

\(=\left(\dfrac{5}{17}-\dfrac{2}{17}\right)+\left(-\dfrac{3}{5}-\dfrac{2}{5}\right)\)

\(=\dfrac{3}{17}-1=-\dfrac{14}{17}\)

j: 

k: 

l: 

m: 

n: 

o: 

=1+2

=3

p: 

q: 

r: 

Ta có : 90 = 2 . 3² . 5

 135 = 3³ . 5

=> ƯCLN(90;135) = 3² . 5 = 45 => a = 45

=> Chọn C

Số tự nhiên a lớn nhất thỏa mãn 90M và 135M là:

A. 15       B. 30      C. 45      D. 60

a: \(-\dfrac{5}{7}+\dfrac{11}{13}+\dfrac{5}{7}=\left(-\dfrac{5}{7}+\dfrac{5}{7}\right)+\dfrac{11}{13}=0+\dfrac{11}{13}=\dfrac{11}{13}\)

b: \(2+\left(\dfrac{8}{11}+\dfrac{3}{11}\right)=2+\dfrac{11}{11}=2+1=3\)

c: \(\dfrac{-5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}\)

\(=\dfrac{-5-2}{21}+\dfrac{1}{3}\)

\(=-\dfrac{7}{21}+\dfrac{1}{3}=\dfrac{1}{3}-\dfrac{1}{3}=0\)

d: \(\dfrac{1}{3}+\dfrac{-2}{6}+\dfrac{-3}{9}=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{1}{3}\)

e: \(\dfrac{5}{12}+\dfrac{5}{3}+\dfrac{2}{12}+\dfrac{2}{3}\)

\(=\left(\dfrac{5}{12}+\dfrac{2}{12}\right)+\left(\dfrac{5}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{12}+\dfrac{7}{3}\)

\(=\dfrac{7+28}{12}=\dfrac{35}{12}\)

f: \(\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}\)

\(=\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)\)

\(=\dfrac{7}{7}+\dfrac{8}{4}=2+1=3\)

g: \(\dfrac{1}{28}+\dfrac{-1}{14}+\dfrac{3}{28}+\dfrac{-1}{7}+\dfrac{3}{14}\)

\(=\left(\dfrac{1}{28}+\dfrac{3}{28}\right)+\left(-\dfrac{1}{14}+\dfrac{3}{14}\right)+\dfrac{-1}{7}\)

\(=\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{2}{14}=\dfrac{1}{7}\)

h: \(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)

\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(-\dfrac{20}{41}-\dfrac{21}{41}\right)+\dfrac{-5}{7}\)

\(=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)

i: \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{24}{30}\)

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

\(=\dfrac{6}{5}-1=\dfrac{1}{5}\)

j: \(\dfrac{5}{-4}+\dfrac{-18}{4}=\dfrac{-5}{4}+\dfrac{-18}{4}=\dfrac{-5-18}{4}=-\dfrac{23}{4}\)

k: \(\dfrac{5}{7}-\dfrac{6}{7}=\dfrac{5-6}{7}=\dfrac{-1}{7}\)

l: \(-\dfrac{3}{5}-\dfrac{12}{5}=\dfrac{-3-12}{5}=-\dfrac{15}{5}=-3\)

m: \(\dfrac{3}{8}-\dfrac{5}{-8}=\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)

n: \(\dfrac{11}{-6}-\dfrac{7}{6}=\dfrac{-11}{6}-\dfrac{7}{6}=\dfrac{-11-7}{6}=-\dfrac{18}{6}=-3\)

o: \(\dfrac{4}{9}-\dfrac{-5}{-9}=\dfrac{4}{9}-\dfrac{5}{9}=\dfrac{4-5}{9}=-\dfrac{1}{9}\)

p: \(\dfrac{13}{-7}-\dfrac{1}{-7}=\dfrac{13-1}{-7}=\dfrac{14}{-7}=-2\)

q: \(\dfrac{13}{11}-\dfrac{2}{-11}=\dfrac{13}{11}+\dfrac{2}{11}=\dfrac{13+2}{11}=\dfrac{15}{11}\)

r: \(\dfrac{12}{5}-\dfrac{7}{-5}=\dfrac{12}{5}+\dfrac{7}{5}=\dfrac{12+7}{5}=\dfrac{19}{5}\)

s: \(-\dfrac{11}{6}-\dfrac{1}{-6}=\dfrac{-11}{6}+\dfrac{1}{6}=\dfrac{-11+1}{6}=-\dfrac{10}{6}=-\dfrac{5}{3}\)

t: \(-\dfrac{1}{11}-\dfrac{10}{11}=\dfrac{-1-10}{11}=\dfrac{-11}{11}=-1\)

u: \(\dfrac{7}{4}-\dfrac{-3}{4}=\dfrac{7}{4}+\dfrac{3}{4}=\dfrac{10}{4}=\dfrac{5}{2}\)

a: \(\dfrac{9}{25}+\dfrac{6}{25}=\dfrac{9+6}{25}=\dfrac{15}{25}=\dfrac{3}{5}\)

b: \(-\dfrac{4}{7}+\dfrac{11}{7}=\dfrac{-4+11}{7}=\dfrac{7}{7}=1\)

c: \(\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-1-5}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

d: \(\dfrac{8}{24}+\dfrac{-12}{24}=\dfrac{8-12}{24}=\dfrac{-4}{24}=-\dfrac{1}{6}\)

e: \(\dfrac{1}{6}+\dfrac{5}{6}=\dfrac{1+5}{6}=\dfrac{6}{6}=1\)

f: \(\dfrac{5}{12}+\dfrac{7}{-12}=\dfrac{5-7}{12}=\dfrac{-2}{12}=-\dfrac{1}{6}\)

g: \(\dfrac{3}{8}+\dfrac{5}{-8}=\dfrac{3}{8}-\dfrac{5}{8}=\dfrac{3-5}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)

h: \(-\dfrac{3}{7}+\dfrac{5}{-7}=\dfrac{-3-5}{7}=-\dfrac{8}{7}\)

i: \(\dfrac{1}{-5}+\dfrac{9}{-5}=\dfrac{1+9}{-5}=\dfrac{10}{-5}=-2\)

Vì số học sinh 6A xếp hàng 3; 4; 8 thì vừa đủ, nên số học sing của lớp chia hết cho 3; 4; 8

Gọi số học sinh là \(x\); \(x\in\) N

Theo bài ra ta có: \(x\) ⋮ 3;4;8

                             \(x\) \(\in\) B(3;4;8)

                            3; 3; 4 = 2²; 8 = 2³; 

BCNN(3;4;8) = 2³.4  = 24

\(x\) \(\in\) BC(24) = {0; 24; 48; 72;...;}

vì số học sinh nằm trong khoảng từ 40 đến 50 nên số học sinh của lớp 6A là 48 học sinh

cảm ơn nhé

Lời giải:

Vì $ƯCLN(a,b)=21$ và $a>b$ nên đặt $a=21x,b=21y$ với $x,y$ là số tự nhiên nguyên tố cùng nhau, $x>y$

Ta có:

$a+b=21(x+y)=336$

$\Rightarrow x+y=16$

Vì $x>y$ và $x,y$ nguyên tố cùng nhau nên:

$(x,y)=(15,1), (13,3), (11,5), (9,7)$

Với mỗi cặp $(x,y)$ thỏa mãn ta có 1 cặp $(a,b)$ thỏa mãn. Có 4 cặp $(x,y)$ thỏa mãn nên có $4$ cặp $(a,b)$ thỏa mãn.

Lời giải:

Vì $ƯCLN(a,b)=21$ và $a>b$ nên đặt $a=21x,b=21y$ với $x,y$ là số tự nhiên nguyên tố cùng nhau, $x>y$

Ta có:

$a+b=21(x+y)=336$

$\Rightarrow x+y=16$

Vì $x>y$ và $x,y$ nguyên tố cùng nhau nên:

$(x,y)=(15,1), (13,3), (11,5), (9,7)$

Với mỗi cặp $(x,y)$ thỏa mãn ta có 1 cặp $(a,b)$ thỏa mãn. Có 4 cặp $(x,y)$ thỏa mãn nên có $4$ cặp $(a,b)$ thỏa mãn.

     xy+x-3y=0

  => (xy+x)-3y-3=0-3

=>x(y+1)-(3y+3)=(-3)

=>x(y+1)-3(y+1)=(-3)

=>(y+1).(x-3)=(-3)

     Mà (-3)=1.(-3)=(-1).3

Lập bảng giá trị:

Vậy các cặp số nguyên (x;y) là (4;-4);(2;2);(6;-2);(0;0)