Đáp án cần chọn là: A

Thay a=25 vào biểu thức ta được : 25+(−45)=−(45−25)=−(20)

Đáp án là A

Thay a = -25 vào biểu thức ta được: (-25) + (-45) = -(25 + 45) = -70

C

C. 260

C

Đặt \(f\left(x\right)=10x\)

Khi đó ta có \(f\left(1\right)=10=P\left(1\right)\), \(f\left(2\right)=20=P\left(2\right)\), \(f\left(3\right)=30=P\left(3\right)\)

Do đó \(P\left(x\right)-f\left(x\right)=g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow P\left(x\right)=10+g\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

Vì \(P\left(x\right)\)là đa thức bậc 4 mà \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\)là đa thức bậc 3 nên \(g\left(x\right)\)là đa thức bậc 1 hay \(g\left(x\right)=x+n\)

Vậy \(P\left(x\right)=\left(x+n\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)+10\)

\(\Rightarrow P\left(12\right)=\left(12+n\right)\left(12-1\right)\left(12-2\right)\left(12-3\right)=\left(n+12\right).11.10.9=990\left(n+12\right)\)

\(=990n+11880\)

Và \(P\left(-8\right)=\left(-8+n\right)\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)=\left(n-8\right)\left(-9\right)\left(-10\right)\left(-11\right)\)\(=-990\left(n-8\right)=-990n+7920\)

Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}+25=\frac{990n+11880-990n+7920}{10}+25=\frac{19800}{10}+25=2005\)

a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)

    = \(-\dfrac{73}{10}\) : - 5

    = \(\dfrac{73}{50}\)

b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)

    = \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)

    = \(\dfrac{2}{41}\)

a.

\(\frac{45^{10}\times5^{20}}{75^{15}}=\frac{\left(3^2\times5\right)^{10}\times5^{20}}{\left(3\times5^2\right)^{15}}=\frac{3^{20}\times5^{10}\times5^{20}}{3^{15}\times5^{30}}=3^5=243\)

b.

\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,8\right)^5}{\left(0,4\right)^5}\times\frac{1}{\left(0,4\right)}=\left(\frac{0,8}{0,4}\right)^5\times\frac{1}{\frac{4}{10}}=2^5\times\frac{5}{2}=2^4\times5=16\times5=80\)

c.

\(\frac{2^{15}\times9^4}{6^6\times8^3}=\frac{2^{15}\times\left(3^2\right)^4}{\left(2\times3\right)^6\times\left(2^3\right)^3}=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=3^2=9\)

Chúc bạn học tốt ^^

\(log_216=log_22^4=4\)

\(log_32187=log_33^7=7\)

\(log_{10}\dfrac{1}{100}=log_{10}10^{-2}=-2\)

\(log10000=log10^4=4\)

\(9^{log_312}=3^{2log_312}=3^{log_3144}=144\)

\(8^{log_25}=2^{3log_25}=2^{log_2125}=125\)

\(\left(\dfrac{1}{25}\right)^{log_5\dfrac{1}{3}}=5^{-2log_5\dfrac{1}{3}}=5^{log_59}=9\)

\(\left(\dfrac{1}{4}\right)^{log_23}=2^{-2log_23}=2^{log_2\dfrac{1}{9}}=\dfrac{1}{9}\)

\(\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{10}\cdot2\cdot5\cdot5^6+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot5^2}\\ =\dfrac{2^8\cdot5^7\left(2^3+5^3\right)}{2^5\cdot5^4\left(2^3+5^3\right)}\\ =\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\\ =2^3\cdot5^3\\ =8\cdot125\\ =1000\)

( -27 - 15 ) : ( -6 ) + (-4 )

=      -42    :  6   + 4 

=   -7    +     4 

= -3

( -3 + 9 ) . ( -7 ) : ( -14 ) + ( -12 )

=   6        .    7    :     ( -14 ) + ( -12 )

= 42 : ( -14 ) + ( -12 )

=    -3      +  ( -12 )

= -15

Hai câu cuối tự làm  nha ! Mik mới làm mẫu hai cái cho đó

d . 15 . ( 25 - 35 ) + 35 . ( 15 - 25 ) 

= 15. 25 - 15. 35 + 35. 15 - 35.25

= 15. 25 - 35.25

= 25. ( 35 - 25 )

= 25 . 10

= 250

 15.(25 - 35) + 35.(15 - 25) 

= 15.25 - 15.35 + 35.15 - 35.25

= 15.25 - 35.25

= 25.(35 - 25)

= 25.10

= 250

25 – (20 - 10) = 25 - 10 = 15

80 – (30 + 25) = 80 - 55