a) 126+ (-29)+ 2004+ (-106)

= [126+ (-106)] +2004+ (-29)

= 20+ 2004- 29

= 2024-29

= 1995

b) (-199)+ (-200)+ (-201)

= -(199+200+201)

= -600

a) 126 + (-20) + 2004 + (-106)

= [126 + (-20) + (-106)] + 2004

= 0 + 2004

= 2004

b) (-199) + (-200) + (-201)

= [(-199) + (-201)] + (-200)

= (-400) + (-200)

= -600

\(\left(\frac{1}{216}-\frac{1}{1^3}\right)\) Ko phải là \(\left(\frac{1}{216}-\frac{1}{13}\right)\) nha

Trong tích trên có thừa số \(\frac{1}{216}-\frac{1}{6^3}=\frac{1}{216}-\frac{1}{216}=0\)

Vậy biểu thức trên bằng 0. Chúc bạn học tốt.

a) \(126+\left(-20\right)+2004+\left(-106\right)\)

\(=106+2004+\left(-106\right)\)

\(=0+2004=2004\)

b) \(70+69+68+...+1+0+\left(-1\right)+\left(-2\right)+...+\left(-70\right)\)

\(=\left[70+\left(-70\right)\right]+\left[69+\left(-69\right)\right]\)\(+...+\left[2+\left(-2\right)\right]+\left[1+\left(-1\right)\right]+0\)

\(=0+0+...+0+0+0=0\)

a, 4590:(x-24+14)=30

    x-24+14=4590 : 30

    x-24+14=153

    x-24=153-14

    x-24=139

    x=139+24

    x=163

phần b mik ko bt

Sửa đề: 8x-1

=>2(8x^2-x)(8x^2-x+2)-126=0

=>2[(8x^2-x)^2+2(8x^2-x)]-126=0

=>(8x^2-x)^2+2(8x^2-x)-63=0

=>(8x^2-x+9)(8x^2-x-7)=0

=>8x^2-x-7=0

=>x=1 hoặc x=-7/8

m )\(-\left|-5\right|+\left(-19\right)+18+\left|11-4\right|-57\)

\(=-5-19+18+7-57\)

\(=-56\)

n ) \(126+\left(-20\right)+\left|124\right|-\left(-320\right)-\left|-150\right|\)

\(=126-20+124+320-150\)

\(=400\)

m)

n)

=400

Chúc bạn học tốt

\(m) -|-5|+(-19)+18+|11-4|-57\)

\(=(-5)+(-19)+18+7-57\)

\(=(-24)+18+(-50)\)

\(=(-56)\)

\(n) 126+(-20)+|124|-(-320)-|-150|\)

\(=126+(-20)+124+320+150\)

\(=(126+124+150)+(320+-20)\)

\(=400+300\)

\(=700\)

n=6

a,\(\left(x-3\right).\left(2y+1\right)=7\)

Vì \(x;y\inℤ=>x-3;2y+1\inℤ\)

\(=>x-3;2y+1\inƯ\left(7\right)\)

Nên ta có bảng sau

Vậy ...

b,\(A=-126-\left(4^2-5\right)^2+870:29\)

\(=-126-\left(16-5\right)^2+30\)

\(=-126-11^2+30\)

\(=-247+30=-217\)

\(a,\left(2432515-315\right)+\left(51-126\right)-74\)

\(=2432200+\left(-75\right)-74\)

\(=2432125-74\)

\(=2432051\)

\(b,\dfrac{\left(12.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\dfrac{12^2.2^{16.2}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\dfrac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\dfrac{3^2.2^{4+32}}{11.2^{13+22}-2^{36}}\)

\(=\dfrac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

\(=\dfrac{3^2.2}{11.1-2^{36}}\)

\(=\dfrac{18}{11-2^{36}}\)

a) (2432515 - 315 ) + (51 - 126) - 74

= 2432515 - 315 + 51 - 126 - 74
= ( 2432515 - 315 ) + 51 - ( 126 + 74 )
= 2432200 + 51 - 200
= 2432200 - 200 + 51
= 2432000 + 51 = 2432051
b, \(\dfrac{\left(12.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\dfrac{12^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)=
\(\dfrac{\left(2.2.3\right)^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\dfrac{2^2.2^{2.}.3^2.2^{32}}{11.2^{35}-2^{35}.2}\)= \(\dfrac{2^{36}.3^2}{2^{35}.\left(11-2\right)}=\dfrac{2^{35}.2.3^2}{2^{35}.9}=\dfrac{2.3^2}{9}=\dfrac{18}{9}=2\)