a, A = \(\dfrac{3^{10}\times10+3^{10}\times6}{3^9\times2^4}\)

    A =  \(\dfrac{3^{10}\times\left(10+6\right)}{3^9\times2^4}\)

    A = \(\dfrac{3^{10}\times16}{3^9\times16}\)

    A = 3 

c, C = \(\dfrac{36^{10}\times25^{15}}{30^8}\)

    C = \(\dfrac{\left(6^2\right)^{10}.\left(5^2\right)^{15}}{30^8}\)

    C = \(\dfrac{6^{20}.5^{30}}{6^8.5^8}\)

   C  =  6¹².5²²

e, E = \(\dfrac{11\times3^{22}\times3^7-9^{15}}{2\times3^{14}}\)

   E  =  \(\dfrac{11\times3^{29}-3^{30}}{2\times3^{14}}\)

  E = \(\dfrac{3^{29}\times\left(11-3\right)}{2\times3^{14}}\)

 E = \(\dfrac{3^{15}\times8}{2}\)

E = 4 \(\times\) 3¹⁵

a, Ta có: \(A=\left|x+2\right|+\left|9-x\right|\ge\left|X+2+9-x\right|=11\)

Dấu "=' xảy ra khi \(\left(x+2\right)\left(9-x\right)\ge0\Leftrightarrow-2\le x\le9\)

Vậy Min_A = 11 khi -2 =< x =< 9

b, Vì \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow B=\frac{3}{4}-\left(x-1\right)^2\le\frac{3}{4}\)

Dấu "=" xảy ra khi x = 1

Vậy Max_B = 3/4 khi x=1

Ta có :\(A=\left|x+2\right|+\left|9-x\right|\ge\left|x+2+9-x\right|=11\)

Vậy \(A_{min}=11\) khi \(2\le x\le9\)

\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)

\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)

\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)

(x+2)(4x^2 -2x+1)+(3-2x)(9+6x+4x^2) =  -4x^3+6x^2-3x+29

nha bạn chúc bạn học tốt nha 

Trả lời:

( x + 2 )( 4x² - 2x + 1 ) + ( 3 - 2x )( 9 + 6x + 4x² )

= 4x³ - 2x² + x + 8x² - 4x + 2 + 27 + 18x + 12x² - 18x - 12x² - 8x³ 

= - 4x³ + 6x² - 3x + 29

-3/11.(-22)/66.121/15

=(-3).(-22).121

 11.66.15

=11

15

3/7.2/5.7/3.20.19/72

=3.2.7.20.19

7.5.3.72

=76

16

6/7.8/13+6/13.9/7-3/13.6/7

=6/7.8/13+6/7.9/13-3/13.6/7

=6/7.(8/13+9/13-3/13)

=6/7.14/13

=12/13

-1/4.152/11+68/4.(-1)/11

=152/4.(-1)/11+68/4.(-1)/11

=(-1)/11.(152/4+68/4)

=(-1)/11.220/4

=-110/22

-5/7.2/11+(-5)/7.9/11+12/7

=-5/7.2/11+-5/7.9/11+12/7

=-5/7.(2/11+9/11)+12/7

=-5/7.1+12/7

=(-5)/7+12/7

=7/7

=1

146/13-(18/7+68/13)

=146/13-18/7-68/13

=(146/13-68/13)-18/7

=78/13-18/7

=6-18/7

=42/7-18/7

=24/7

\(A=100+98+96+...+2-97-95-...-1\)

\(A=100+\left(98-98\right)+\left(96-95\right)+...+\left(2-1\right)\)

\(A=100+1+1+...+1\)

\(A=100+1\cdot49\)

\(A=100\cdot49\)

\(A=4900\)

\(B=1+2-3-4+5+6-7-8+9+10-11-12+...-299-300+301+302\)

\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(298-299-300+301\right)+302\)

\(B=1+0+0+...+302\)

\(B=1+302\)

\(B=303\)

�=100+(98−98)+(96−

b) Để \(B=\frac{x+2}{x+1}\)có giá trị nguyên thì \(x+2⋮x+1\)

Ta có : \(x+2⋮x+1\)

\(\Rightarrow x+1+1⋮x+1\)

Mà \(x+1⋮x+1\)

\(\Rightarrow1⋮x+1\)

\(\Rightarrow x+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

+) x+1=-1\(\Rightarrow\)x=-2  (thỏa mãn)

+) x+1=1\(\Rightarrow\)x=0  (thỏa mãn)

Vậy \(x\in\left\{-2;0\right\}\)

Các phần sau bạn làm tương tự nhé!

Học tốt!

#Huyền#

\(A=\frac{15\times3^{11}+4\times27^4}{9^7}\)

\(A=\frac{15\times177147+4\times531441}{4782969}\)

\(A=\frac{2657205+2125764}{4782969}\)

\(A=\frac{47829969}{47829969}=1\)

Tự nhiên phần Σ của tớ bị lỗi nên tớ tính tử số rồi tính mẫu nhé :

Tử số của A :

15 x 3¹¹ + 4 x 27⁴

= 5 x 3¹² + 4 x 3¹²

= 3¹² x ( 5 + 4 ) 

= 3¹² x 9 

= 3¹² x 3²

= 3¹⁴

Mẫu số của A :

97 = (3²)⁷ = 3^{2 x 7} = 3¹⁴

Vậy A = 3¹⁴ : 3¹⁴ = 1

\(B=\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

\(B=\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

\(B=\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(B=\frac{1}{11}\)