34758-45+45-34567+21(100x2334)=
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a) -(12+21-23)-(23-21+10)
=-12-21+23-23+21-12=-12-12=-24
b)(57-725)-(605-53)
=57-725-605+53=-1220
c)(55+45+45)-(15-55+45)
=55+45+45-15+55-45=110+45-15=140
-(12+21-23)-(23-21+10)
=-12-21+23-23+21-10
=(21-21)+(23-23)-(12+10)
=0+0-22=-22
(57-725)-(605-53)
=57-725-605+53
=(57+53)-(725+605)
=110-1330
=-1220
(55+45+45)- (15-55+45)
=55+45+45-15+55-45
=(45+55)+(55-15)+(45-45)
=100+40+0=140
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65 . (- 45 ) – 35. 45 + 24 .( -21 )+ 76.( -21 )
=65.(-1).45-35.45+24.(-21)+76.(-21)
=45.(-65-35)+(-21).(24+76)
=45.(-100)+(-21).100
=-4500-2100
=-6600
65 . (- 45 ) – 35. 45 + 24 .( -21 )+ 76.( -21 )
= ( 65 + 35 ) . ( 76 + 24 ) . ( -21) - (-21) . ( 45 + (-45)
= 100 . 100 . 0 .0
= cuối cùng thì đều bằng 0
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\(\frac{3}{67}\left(17\frac{21}{45}-13\frac{21}{45}\right)\)
\(=\frac{3}{67}.4\)
\(=\frac{12}{67}\)
\(\frac{3}{67}\left(17\frac{21}{45}-13\frac{21}{45}\right)\)
\(=\frac{3}{67}\times4\)
\(=\frac{12}{67}\)
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\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}-1,5\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)-1,5\)
\(=\frac{47}{47}+\frac{45}{45}\)
\(=1+1\)
=2
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a) ( 12 + 21 - 23 ) - ( 23 - 21 + 10 )
= 12 + 21 - 23 - 23 - 21 + 10
= ( 12 + 10 ) - ( 21 - 21 ) - ( 23 - 23 )
= 22 - 0 - 0
= 22
b) ( 55 + 45 + 15 ) - ( 15 - 55 + 45 )
= 55 + 45 + 15 - 15 - 55 + 45
= ( 55 + 45 ) + ( 55 + 45 ) + ( 15 - 15)
= 100 + 100 + 0
= 200
\(a)\)\(-\left(12+21-23\right)-\left(23-21+10\right)\)
\(=-12-21+23-23+21-10\)
\(=\left(-21+21\right)+23-23-12-10\)
\(=0+0-12-10\)
\(=-22\)
\(b)\)\(\left(55+45+15\right)-\left(15-55+45\right)\)
\(=55+45+15-15+55-45\)
\(=\left(15-15\right)+\left(45-45\right)+55+55\)
\(=0+0+55+55\)
\(=110\)
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\(1,=38-42+14-25+27+15=27\)
\(2,=12+21-23+21-10=21\)
\(3,=57-725-605+53=-1220\)
\(4,=55+45+15-15+55-45=110\)
34758-45+45-34567+21(100.2334)
=34758-45+45-34567+21.233400
=34758-45+45-34567+4901400
=4901591
kết quả là : 4901591 nhớ k nha