a: =3/4-19/18-3/4

=-19/18

b: =24-24x47-24x54

=24(1-47-54)

=24x(-100)=-2400

a) \(=\dfrac{-19}{18}\)

b) \(=-2400\)

a:=-19/18

b: -2400

Bài 5:

a: x(x-4)=0

=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b: Đề thiếu vế phải rồi bạn

Bài 6:

a: \(\left(-5\right)\cdot\left(-6\right)\cdot\left(-4\right)\cdot2\)

\(=-\left(2\cdot5\right)\cdot\left(4\cdot6\right)\)

\(=-24\cdot10=-240\)

b: \(\left(-3\right)\cdot2\cdot\left(-8\right)\cdot5\)

\(=3\cdot2\cdot8\cdot5\)

\(=\left(3\cdot8\right)\cdot\left(2\cdot5\right)\)

\(=24\cdot10=240\)

Câu 5:

a: \(31\cdot\left(-18\right)+31\cdot\left(-81\right)-31\)

\(=31\left(-18-81-1\right)\)

\(=31\cdot\left(-100\right)=-3100\)

b: \(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)

\(=\left(-12\right)\left(47+52+1\right)\)

\(=-12\cdot100=-1200\)

c: \(13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\)

\(=13\cdot45-3\cdot45\)

\(=45\cdot10=450\)

d: \(-48+48\left(-78\right)+48\left(-21\right)\)

\(=48\left(-1-78-21\right)\)

\(=48\left(-100\right)=-4800\)

Câu 4:

a: \(\left(-6-2\right)\left(-6+2\right)=\left(-8\right)\cdot\left(-4\right)=32\)

b: \(\dfrac{\left(7\cdot3-3\right)}{-6}=\dfrac{21-3}{-6}=\dfrac{18}{-6}=-3\)

c: \(\left(-5+9\right)\cdot\left(-4\right)=4\cdot\left(-4\right)=-16\)

d: \(\dfrac{72}{-6\cdot2+4}=\dfrac{72}{-12+4}=\dfrac{72}{-8}=-9\)

a: \(=\dfrac{48-45}{54}\cdot\dfrac{9}{4}=\dfrac{3}{54}\cdot\dfrac{9}{4}=\dfrac{27}{216}=\dfrac{1}{8}\)

b: \(=\dfrac{6}{5}+1=\dfrac{11}{5}\)

c: \(=\dfrac{7\cdot28\cdot15}{9\cdot25\cdot14}=\dfrac{1}{2}\cdot\dfrac{28}{25}\cdot\dfrac{5}{3}=\dfrac{140}{150}=\dfrac{14}{15}\)

a, -10

b, 20

c, -20

d, -30

a) 10 – 12 – 8 = 10 – (12 + 8) = 10  - 20 = - 10

b) 4 – (- 15) – 5 + 6 = (4 + 6) – [(-15) + 5)] = 10 – (- 10) = 10 + 10 = 20

c) 2 – 12 – 4 – 6 = (2 – 12) - (4 + 6) = -10 + (- 10) = - 20

d) – 45 – 5 – (- 12) + 8 = - (45 + 5) + (12 + 8) = (- 50) + 20 = - 30

a) 10-12-8=2-12=-10

b) 4-(-15)-5+6=4+15-5+6=10+10=20

c) 2-12-4-6=-10-10=-20

d) -45-5-(-12)+8=-50+12+8=-50+20=-30

a, 10 -12 - 8

= 10 - (12 + 8 )

= 10 - 20 = -10

b, 4-(-15)-5+6

= 4+15-5+6

= (4+6) + (15-5)

= 10+10=20

c, 2-12-4-6

= (2-12)-(4+6)

= (-10) - 10

=-20

d, -45-5-(-12)+8

= -45-5+12+8

= [(-45)-5]+(12+8)

= (-50)+20

= -30

\(a.10-12-8\\ =10+\left(-12-8\right)\\ =10+\left(-20\right)=-10\\ b.4-\left(-15\right)-5+6\\ =4+5-5+6\\ =\left(4+6\right)+\left(15-5\right)\\ =10+10=20\\ c.2-12-4-6\\ =\left(2-12\right)+\left(-4-6\right)\\ =-10+\left(-10\right)=-20\\ d.-45-5-\left(-12\right)+8\\ =-45-5+12+8\\ =\left(-45-5\right)+\left(12+8\right)\\ =-50+20=-30\)

Giá trị của biểu thức C tại x=25 là C(25).

Theo định lý Bezout, C(25) = số dư khi chia C(x) cho x-25.

Ta dùng sơ đồ Hooc-ne để tìm số dư này:

Vậy: C(25)=1 

\(\dfrac{4^{10}.9^6+3^{12}.8^5}{6^{13}.4-2^{16}.3^{12}}\)

\(=\dfrac{\left(2^2\right)^{10}.\left(3^2\right)^6+3^{12}.\left(2^3\right)^5}{\left(2.3\right)^{13}.2^2-2^{16}.3^{12}}\)

\(=\dfrac{2^{20}.3^{12}+3^{12}.2^{15}}{2^{13}.3^{13}.2^2-2^{16}.3^{12}}\)

\(=\dfrac{2^{20}.3^{12}+3^{12}.2^{15}}{2^{15}.3^{13}-2^{16}.3^{12}}\)

\(=\dfrac{2^{15}.3^{12}.\left(2^5+1\right)}{2^{15}.3^{13}.\left(3-2\right)}\)

\(=\dfrac{2^5+1}{3-2}\)

\(=\dfrac{32+1}{1}=33\)