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\(A=\frac{17}{23}\cdot\frac{8}{16}\cdot\frac{23}{17}\cdot\left(-80\right)\cdot\frac{3}{4}\)\(=\frac{17\cdot4\cdot2\cdot23\cdot16\cdot\left(-5\right)\cdot3}{23\cdot16\cdot17\cdot4}\)
=> \(A=\frac{2\cdot\left(-5\right)\cdot3}{1}=-30\)
\(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\left(\frac{7}{12}-\frac{7}{12}\right)\)
=> \(B=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right)\cdot0=0\)
a)\(A=\frac{17}{23}.\frac{8}{16}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)
\(A=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{8}{16}.\frac{3}{4}\right).\left(-80\right)\)
\(A=\frac{3}{8}.\left(-80\right)\)
\(A=-30\)
b)\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
\(C=\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)
\(C=0\)
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé! 
a)\(\left(\dfrac{1}{9}:\dfrac{8}{27}\right):\left(\dfrac{16}{48}:\dfrac{8}{21}\right)\)
=\(\dfrac{3}{8}:\dfrac{7}{8}\)
=\(\dfrac{3}{7}\)
b)\(\dfrac{19}{23}+\dfrac{1}{9}\cdot\dfrac{36}{23}\)
=\(\dfrac{19}{23}+\left(\dfrac{1}{9}\cdot\dfrac{36}{23}\right)\)
=\(\dfrac{19}{23}+\left(\dfrac{1}{1}\cdot\dfrac{4}{23}\right)\)
=\(\dfrac{19}{23}+\dfrac{4}{23}\)
=\(\dfrac{23}{23}=1\)
c)\(\left(\dfrac{13}{23}+\dfrac{1313}{2323}-\dfrac{131313}{232323}\right)\cdot\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)
=\(\left(\dfrac{13}{23}+\dfrac{1313}{2323}-\dfrac{131313}{232323}\right)\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{7}{12}\right)\)
=\(\left(\dfrac{13}{23}+\dfrac{1313}{2323}-\dfrac{131313}{232323}\right)\cdot0\)
=0
chúc bn học tốt!
câu 1 bỏ dấu ngoặc rồi tính
( 36 + 79 ) + ( 145 _ 79 _ 36 )
\(=36+79+145-79-36\)
\(=\left(36-36\right)+\left(79-79\right)+145\)\
\(=0+0+145=145\)
10 _ [ 12 _ ( -9 _ 1 ) ]
\(=10-12-10\)
\(=10-10-12\)
\(=0-12=-12\)
( 38 _ 29 + 43) _ ( 43 + 38 )
\(=38-29+43-43-38\)
\(=\left(38-38\right)+\left(43-43\right)-29\)
\(=0+0-29=-29\)
271 _ [ ( -43 ) + 271 _ ( -17 ) ]
\(=271+43-271-17\)
\(=\left(271-271\right)+\left(43-17\right)\)
\(=0+26=26\)
- 144 _ [ 29 _ ( + 144 ) _ ( + 144 )]
\(=-144-19+144+144\)
\(=\left(-144+144+144\right)-19\)
\(=144-19=125\)
đợi mk lm tiếp câu 2 nha .
bài 2 tính tổng các số nguyên
- 18 < hoặc bằng x < hoặc bằng 17
\(\Rightarrow x\in\left\{-18;-17;-16;....;17\right\}\)
tổng \(x=-18+\left(-17\right)+\left(-16\right)+...+17=-18\)
- 27 < hoặc bằng x < hoặc bằng 27
\(\Rightarrow x\in\left\{-27;-26;-25;..;27\right\}\)
Tổng \(x=-27+\left(-26\right)+\left(-25\right)+...+27=0\)
\(a,157.17-157.7=157\left(17-7\right)=1570\)
\(b,5.\left(-3+2\right)-7.\left(5-4\right)=-5-7=-12\)
\(c,17.\left(-84\right)+17\left(-16\right)=17\left[\left(-84\right)+\left(-16\right)\right]=-1700\)
\(d,-145.\left(13-57\right)+57.\left(10-145\right)=6380-7695=-1315\)
\(e,199.\left(15-17\right)-199.\left(-17+5\right)=199.10=1990\)
a) \(\frac{17}{23}.\frac{18}{16}.\frac{23}{17}-\left(-80\right)-\frac{3}{4}\)
= \(\frac{17.18.23}{23.16.17}+80-\frac{3}{4}\)
= \(\frac{9}{8}+80-\frac{3}{4}\)
= \(\frac{3}{8}+80=\frac{643}{8}\)
b) \(\frac{5}{11}.\frac{18}{29}-\frac{5}{11}.\frac{8}{29}+\frac{5}{11}.\frac{19}{29}\)
= \(\frac{5}{11}.\left(\frac{18}{29}-\frac{8}{29}+\frac{19}{29}\right)\)
= \(\frac{5}{11}.1=\frac{5}{11}\)
c) \(\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).\left(\frac{1}{3}+\frac{1}{4}-\frac{7}{12}\right)\)
= \(\left(\frac{13}{23}+\frac{1313}{2323}-\frac{131313}{232323}\right).0\)
= \(0\)