Câu a:

(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)

= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5

= 1 + 15 - 3 4/5

= 16 - 3 - 4/5

= 13 - 4/5

= 61/5

Câu b:

(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))

= (17/6 + 13/9) :(121/12 - 19/2)

= (51/18 + 26/18) :(121/12 - 114/2)

= 77/18 : 7/12

= 77/18 x 12/7

= 22/3

B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)

=\(11\frac34-6\frac56+4\frac12+1\frac23\)

=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)

=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=10+\(\frac{13}{12}\)

=\(\frac{120}{12}+\frac{13}{12}\)

=\(\frac{133}{12}\)

b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)

= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)

=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)

=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)

=\(\frac{108}{60}\)

=\(\frac95\)

\(a,-13-\left(-15\right)=-13+15=2\)

\(b,43-\left(56-43\right)+\left(56-12\right)=43-13+44=30+44=74\)

\(c,4^4:2^8+\left[13-\left(-5\right)\right]\cdot\left(-3\right)=256:256+\left(13+5\right)\cdot\left(-3\right)\)\(=1+18\cdot\left(-3\right)=1+\left(-54\right)=-53\)

\(d,135\cdot\left(13-37\right)+37\cdot\left(135-13\right)=135\cdot\left(-24\right)+37\cdot122\)\(=-3240+4514=1274\)

a, -13-(-15)=-13+15=2

b, 43-(56-43)+(56-12)=43-13+44=30+44=74

c, 4^4:2^8+[13-(-5)].(-3)=256:256+(13+5).(-3)=1+18.(-3)=1+(-54)=-53

d, 135.(13-37)+37.(135-13)=135.(-24)+37.122=-3240+4514=1274

a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)

\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)

\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)

b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)

c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)

a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)

\(=\frac{8+9+10}{12}\)

\(=\frac{27}{12}=\frac{9}{4}\)

b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)

\(=\frac{45-14+20}{24}\)

\(=\frac{51}{24}=\frac{17}{8}\)

2)

a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)

\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)

\(=1+\frac{7}{13}+\frac{1}{7}\)

\(=\frac{20}{13}+\frac{1}{7}\)

\(=\frac{153}{91}\)

Tí tớ trả lời tiếp

b)\(5\frac14+3\frac25-4\frac14\)

=\(\left(5\frac14-4\frac14\right)+3\frac25\)

=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)

=\(1+0+\frac{17}{5}\)

=\(\frac55+\frac{17}{5}\)

=\(\frac{22}{5}\)

dễ mà bạn đây là bài cơ bản lớp 6 dấy

 câu a nhé bạn bạn nếu ko làm kiểu khó thì đổi về phaan số bình thường nà sau đó tính trong ngoặc trước rồi tính xoong bỏ dấu ngoặc nhưng ko đổi dấu né thế lad đc tương tự như các câu dưới
 

a)\(8\frac{2}{3}:2\frac{1}{6}-2\frac{27}{51}=\frac{26}{3}.\frac{6}{13}-\frac{43}{17}=4-\frac{43}{17}=\frac{25}{17}\)

b)\(\frac{27}{20}.\frac{15}{4}+\frac{19}{8}=\frac{119}{16}\)

c)\(\left(\frac{1}{12}+\frac{5}{6}\right)+\left(\frac{13}{35}+\frac{23}{35}\right)=\frac{11}{12}+\frac{36}{35}=\frac{817}{420}\)

d)\(\frac{24}{37}.\left(\frac{13}{18}+\frac{2}{9}+\frac{1}{18}\right)=\frac{24}{37}.1=\frac{24}{37}\)

a, \(4\dfrac{5}{37}\)-\(3\dfrac{4}{5}\)+ \(8\dfrac{15}{29}\)- \(3\dfrac{5}{37}\)+ \(6\dfrac{14}{29}\)

=(\(4\dfrac{5}{37}\)-\(3\dfrac{5}{37}\))+(\(8\dfrac{15}{29}\)+\(6\dfrac{14}{29}\))-\(3\dfrac{4}{5}\)

=(4-3)+(\(\dfrac{5}{37}\)-\(\dfrac{5}{37}\))+(8+6)+(\(\dfrac{15}{29}\)+\(\dfrac{14}{29}\))-3\(\dfrac{4}{5}\)

=1+ 15-\(3\dfrac{4}{5}\)=13-\(\dfrac{4}{5}\)=\(\dfrac{61}{5}\)

b, 60\(\dfrac{7}{13}\)+ 50\(\dfrac{8}{13}\)-11\(\dfrac{2}{13}\)

=(60+50-11)+(\(\dfrac{7}{13}\)+ \(\dfrac{8}{13}\)-\(\dfrac{2}{13}\))

=99+1=100

c, đáp án bằng \(\dfrac{-2}{3}\). bạn tự tính nha

Bài 2:

- \(\frac75\) + \(x\) = - \(\frac34\)

\(x\) = - \(\frac34+\frac75\)

\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)

\(x\) = \(\frac{13}{20}\)

Vậy \(x=\frac{13}{20}\)

Bài 3:

- \(\frac74\) + \(\frac14:x\) = \(\frac32\)

\(\frac14:x\) = \(\frac32+\frac74\)

\(\frac14:x\) = \(\frac64+\frac74\)

\(\frac14:x=\frac{13}{4}\)

\(x=\frac14:\frac{13}{4}\)

\(x=\frac14\times\frac{4}{13}\)

\(x=\frac{1}{13}\)

Vậy \(x=\frac{1}{13}\)

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)