\(C=9\cdot\left(\frac{\frac{26299}{2023}-\frac{3757}{2023}-\frac{91}{2023}}{\frac{8092}{2023}-\frac{1156}{2023}-\frac{28}{2023}}\right):\left(\frac{\frac{840255}{21545}+\frac{168051}{21545}+\frac{6045}{21545}+\frac{27105}{21545}}{\frac{344720}{21545}+\frac{68944}{21545}+\frac{2480}{21545}+\frac{11120}{21545}}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\left(\frac{\frac{22451}{2023}}{\frac{6908}{2023}}\div\frac{\frac{1041456}{21545}}{\frac{427264}{21545}}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\left(\frac{13}{4}\div\frac{39}{16}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\frac{4}{3}\cdot\frac{2\cdot79\cdot2\cdot79}{2\cdot82\cdot2\cdot82}=9\cdot\frac{4}{3}\cdot\frac{79}{82}\)

\(C=\frac{474}{41}\)

\(\dfrac{4}{13}+\dfrac{-\left(-12\right)}{39}\)

\(=\dfrac{4}{13}+\dfrac{12}{39}\)

\(=\dfrac{12}{39}+\dfrac{12}{39}\)

\(=\dfrac{24}{39}\)

\(=\dfrac{8}{13}\)

= 

Ủa bạn ??

Zậy là tính cả gì vậy

Bài 1:

a: \(\frac{13}{20}<1\)

\(\frac{19}{20}<1\)

\(\frac{17}{23}<1\)

Do đó: \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<1+1+1=3\)

hay \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<3\)

b: \(\frac75>1;\frac54>1;\frac98>1;\frac32>1\)

Do đó; \(\frac75+\frac54+\frac98+\frac32>1+1+1+1\)

=>\(\frac75+\frac54+\frac98+\frac32>4\)

Bài 2:

\(\frac87+\frac{11}{13}+\frac{23}{30}+\frac{39}{45}<4\)

a) <     b) >     c) <

Bài 1:

a: \(\frac{13}{20}<1\)

\(\frac{19}{20}<1\)

\(\frac{17}{23}<1\)

Do đó: \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<1+1+1=3\)

hay \(\frac{13}{20}+\frac{19}{20}+\frac{17}{23}<3\)

b: \(\frac75>1;\frac54>1;\frac98>1;\frac32>1\)

Do đó; \(\frac75+\frac54+\frac98+\frac32>1+1+1+1\)

=>\(\frac75+\frac54+\frac98+\frac32>4\)

Bài 2:

\(\frac87+\frac{11}{13}+\frac{23}{30}+\frac{39}{45}<4\)

\(\left(\frac{3}{5}\right)^{10}.\left(\frac{5}{3}\right)^{10}-\frac{13^4}{39^4}+2017^0\\ =\left(\frac{3}{5}.\frac{5}{3}\right)^{10}-\left(\frac{13}{39}\right)^4+1\\ =1^{10}-\left(\frac{1}{3}\right)^4+1\)

\(=1-\frac{1}{81}+1\\ =2+\frac{-1}{81}\\ =\frac{161}{81}\)

(3/5)^10.(5/3)^10-13^4/39^4+2017^0

=1-1/81+1

=161/81

Đề khó nhìn quá bạn ơi!

\(\left(-22\right)\cdot\left(-5\right)>0\)

\(\left(-7\right)\cdot20< -7\)

(-22).(-5)và 0

do 2 số nguyên âm nhân với nhau ra số nguyên dương nên ta có thể rút gọn biểu thức thành 22.5 và 0 từ đó => 22.5>0

(-7).20 < -7

(-39).12 = 39.(-12)

(35-15).(-4)+24(-13-17)=30.(-4)+24(-13-17)=-120+24.30=-120+720=600

(-13)(57-34)+57(13-45)=-13.57-(-13).34+57.13-57.45=13.(-57)-13.(-34)+57.13-57.45=13(-57-(-34)+57)-57.45=13.34-57.45=442-2565=-2123

1: \(=-145\cdot13+145\cdot57+57\cdot10-57\cdot145=-1315\)

2: \(=17\cdot15-17\cdot16+16\cdot17-16\cdot20=255-320=-65\)

3: \(=-38\cdot25+38\cdot4-25\cdot4+25\cdot38=13\cdot4=52\)

4: \(=23\cdot145-23\cdot17-145\cdot23+145\cdot6=479\)

5: \(=24\cdot15-24\cdot4+4\cdot24-4\cdot15=360-60=300\)

6: \(=199\left(15-17+17-5\right)=199\cdot10=1990\)

7: \(=-39\cdot5+39\cdot99+99\cdot10-99\cdot39=795\)

bạn ơi viết rõ cách làm ra

a: \(B=\frac12-\left\lbrack\frac38+\left(-\frac74\right)\right\rbrack\)

\(=\frac12-\frac38+\frac74\)

\(=\frac48-\frac38+\frac{14}{8}=\frac{15}{8}\)

b: \(-\frac{4}{12}-\left(-\frac{13}{39}-0,25\right)+0,75\)

\(=-\frac13+\frac13+0,25+0,75\)

=0,25+0,75

=1

c: \(\frac12-\frac13+\frac{1}{23}+\frac16\)

\(=\left(\frac12-\frac13+\frac16\right)+\frac{1}{23}\)

\(=\left(\frac16+\frac16\right)+\frac{1}{23}=\frac13+\frac{1}{23}=\frac{26}{69}\)

d: \(\left(-\frac{13}{7}-\frac49\right)-\left(-\frac{10}{7}-\frac49\right)\)

\(=-\frac{13}{7}-\frac49+\frac{10}{7}+\frac49\)

\(=-\frac{13}{7}+\frac{10}{7}=-\frac37\)

e: \(\left(\frac78-\frac52+\frac47\right)-\left(-\frac37+1-\frac{13}{8}\right)\)

\(=\frac78-\frac52+\frac47+\frac37-1+\frac{13}{8}\)

\(=\frac{20}{8}-\frac52=0\)

f: \(-\frac37+\left(3-\frac34\right)-\left(2,25-\frac{10}{7}\right)\)

\(=-\frac37+3-\frac34-2,25+\frac{10}{7}\)

\(=\frac77+0,75-\frac34\)

=1
g: \(\left(\frac53-\frac37+9\right)-\left(2+\frac57-\frac23\right)+\left(\frac87-\frac43-10\right)\)

\(=\frac53-\frac37+9-2-\frac57+\frac23+\frac87-\frac43-10\)

\(=\left(\frac53+\frac23-\frac43\right)+\left(-\frac37-\frac57+\frac87\right)+\left(9-2-10\right)\)

\(=\frac33+7-10=1-3=-2\)