B1 : -(-256)+(-156)-324+32

=256+(-156)-324+32

=[256+(-156)]+(-324+32)

=(-100)+(-292)

=-192

B2 : a) (-76).11=(-76)(10+1)=(-76)+(-760)=-836

b) 65(-101)=65[(-1)+(-100)]=(-65)+(-6500)=-6565

Theo mình thì những bài như vầy bạn nên tự làm.

~~~Học tốt~~~

a,(-87) + (-34) + (-13)

=[(-87) + (-13)] + (-34)

= -100 + (-34)

= -134

b,(-26) + 117 + 1026 + (-17)

=[(-26) + 1026] + [117 + (-17)]

=1000 + 100

=1100

c,142 . 23 + 142 . 17 - 40 . 42

= 142 . (23 + 17) - 40 . 42

= 142 . 40 - 40 . 42

= (142 - 42) . 40

= 100 . 40

= 4000

#họctốt 

&YOUTUBER&

a=-134;b=1100;c=4000 bạn nhé (sai thì bỏ qua nhé)

\(\left(x-3\right)\left(21+5\right)=5\)

\(\Leftrightarrow\left(x-3\right).26=5\)

\(\Leftrightarrow x-3=\frac{5}{26}\)

\(\Leftrightarrow x=\frac{83}{26}\)

Vậy \(x=\frac{83}{26}\)

83/26 nha ban

\(27\cdot\left(-3\right)^5\cdot\left(-3\right)\cdot81=3^3\cdot\left(-3\right)^5\cdot\left(-3\right)\cdot3^4=3^7\cdot\left(-3\right)^6=3^7\cdot3^6=3^{13}\)

Với mọi x,y luôn có \(\left(3x-6\right)^{100}\ge0,\left(4y-8\right)^{200}\ge0\)

Mà \(\left(3x-6\right)^{100}+\left(4y-8\right)^{200}=0\)Nên \(\hept{\begin{cases}4y-8=0\\3x-6=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=2\end{cases}}\)

Ta thấy : \(\left(3x-6\right)^{100}\ge0\forall x,\left(4y-8\right)^{200}\ge0\forall y\)

\(\Rightarrow\left(3x-6\right)^{100}+\left(4y-8\right)^{200}\ge0\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-6=0\\4y-8=0\end{cases}}\)  \(\Leftrightarrow\hept{\begin{cases}x=2\\y=2\end{cases}}\)

Vậy : (x,y) = (2,2)

\(\left(-\right)\frac{1989\cdot1990+3970}{1992\cdot1991+3984}=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991+2\right)}=\frac{1989}{1993}\)

\(\left(-\right)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{3^{10}\cdot\left(-5\right)^{20}\cdot\left(-5\right)}{3^{10}\cdot\left(-5\right)^{20}\cdot3^2}=-\frac{5}{9}\)

\(\left(-\right)\frac{\left(-11\right)^5\cdot13^7}{11^5\cdot13^8}=\frac{11^5\cdot13^7\cdot\left(-1\right)}{11^5\cdot13^7\cdot13}=-\frac{1}{13}\)

\(\left(-\right)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{11}\cdot3^9}{2^9\cdot3^{10}}=\frac{4}{3}\)

Phân tích ra \(6^9\cdot2^{10}+12^{10}=\left(2\cdot3\right)^9\cdot2^{10}+\left(3\cdot4\right)^{10}=2^9\cdot3^9\cdot2^{10}+3^{10}\cdot\left(2^2\right)^{10}=2^{19}\cdot3^9+\left(-3\right)^{10}\cdot2^{20}\)