\(7^4.7^2=7^{4+2}=7^6\)

=> Đáp số đúng là đáp số (D) \(7^6\)

đáp án (D)

a) 4x - 15 = - 75 - x

<=> 4x + x = -75 + 15

<=> 5x = -60

<=> x = -12

b) 72 - 3x = 5x + 8

<=> -3x -5x = 8 - 72

<=> -8x = -64

<=> x = 8

d) 3 . | x - 7 | = 21

<=> | x - 7 | = 7

\(\Leftrightarrow\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=0\end{cases}}\)

\(-7\left|x+3\right|=-49\) 

\(\Leftrightarrow\left|x+3\right|=7\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)

Bài làm

a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)

= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)

= \(-\frac{3}{7}+\frac{7}{2}\)

\(=-\frac{7}{14}+\frac{49}{14}\)

\(=\frac{42}{14}=3\)

b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)

\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)

\(=5-\frac{1}{3}+\frac{2}{9}\)

\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)

\(=\frac{44}{9}\)

c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)

\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)

\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)

\(=\frac{5}{12}+\frac{7}{12}\)

\(=\frac{12}{12}=1\)

d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)

\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)

\(=-\frac{5}{9}+\frac{1}{90}\)

\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)

a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)

(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)

\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)

\(\frac{44}{7}.x=\frac{-77}{35}\)

x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)

(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1

(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)

(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)

\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)

(44\(x\) + 3) : \(\frac{11}{5}\) = - 4

44\(x\) + 3 = - 4 x \(\frac{11}{5}\)

44\(x\) + 3 = - \(\frac{44}{5}\)

44\(x\) = - \(\frac{44}{5}\) - 3

44\(x\) = - \(\frac{59}{5}\)

\(x=-\frac{59}{5}:44\)

\(x\) = - \(\frac{59}{250}\)

Vậy \(x\) = - \(\frac{59}{250}\)

a=1135/23-((167/32+330/23)

a=1135/23-14401/736

a=953/32

Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)

a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)

\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)

b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)

\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)

\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)

\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)

c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)

\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)

d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)

=\(\frac{7}{9}\)

làm thế nào

a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)

=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)

=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)

=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)

=\(1+0+14+1-\frac{19}{5}\)

=\(15+1-\frac{19}{5}\)

=\(16-\frac{19}{5}\)

=\(\frac{80}{5}-\frac{19}{5}\)

=\(\frac{61}{5}\)

a) Cách 1:

\(3\dfrac{4}{9}+5\dfrac{1}{6}=\left(3+5\right)+\left(\dfrac{4}{9}+\dfrac{1}{6}\right)=8+\dfrac{11}{18}=8\dfrac{11}{18}\)

Cách 2:

\(3\dfrac{4}{9}+5\dfrac{1}{6}=\dfrac{31}{9}+\dfrac{31}{6}=\dfrac{155}{18}=8\dfrac{11}{18}\)

b) Cách 1:

\(8\dfrac{1}{14}-6\dfrac{3}{7}=\left(7-6\right)+\left(\dfrac{15}{14}-\dfrac{3}{7}\right)=1+\dfrac{9}{14}=1\dfrac{9}{14}\)

Cách 2:

\(8\dfrac{1}{14}-6\dfrac{3}{7}=\dfrac{113}{14}-\dfrac{45}{7}=\dfrac{23}{14}=1\dfrac{9}{14}\)

c) Cách 1:

\(7-3\dfrac{6}{7}=\left(6-3\right)+\left(1-\dfrac{6}{7}\right)=3+\dfrac{1}{7}=3\dfrac{1}{7}\)

Cách 2:

\(7-3\dfrac{6}{7}=\dfrac{49}{7}-\dfrac{27}{7}=\dfrac{22}{7}=3\dfrac{1}{7}\)