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\(\hept{\begin{cases}\\\end{cases}\hept{\begin{cases}\\\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^{ }^2_{ }\widebat{ }\varnothing\alpha}\)ok
chua

a ) \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}\)
\(=\frac{3^{28}\left(11\cdot3-3^2\right)}{2^2\cdot3^{28}}=\frac{33-9}{4}=\frac{24}{4}=6\)
b ) \(27^{16}\div9^{10}\)
\(=\left(3^3\right)^{16}\div\left(3^2\right)^{10}\)
\(=3^{48}\div3^{20}\)
\(=3^{48-20}\)
\(=3^{28}\)
Bg
a) \(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
= \(\frac{11.3^{22}.3^7-3^{30}}{2^2.3^{28}}\)
= \(\frac{11.3^{29}-3.3^{29}}{4.3^{28}}\)
= \(\frac{8.3^{29}}{4.3^{28}}\)
= \(\frac{2.3}{1.1}\)
= 6
b) 2716 ÷ 910
= 33.16 ÷ 32.10
= 348 ÷ 320
= 348 - 20
= 328

Bài 35 :
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)
\(A=\frac{2^8.2^2.98}{2^8.104}\)
\(A=\frac{2^8.4.98}{2^8.4.26}\)
\(A=\frac{49}{13}\)
Vậy \(A=\frac{49}{13}\)
\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.\left(3^{14}\right)^2}\)
\(B=\frac{11.3^{29}-9^{15}}{2^2.3^{28}}\)
\(B=\frac{11.3^{29}-\left(3^2\right)^{15}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)
\(B=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)
\(B=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)
\(B=\frac{3^{29}.8}{4.3^{28}}\)
\(B=\frac{3^{28}.3.4.2}{4.3^{28}}\)
\(B=3.2\)
\(B=6\)
Vậy B = 6
A = 2^10 . 13 + 2^10 . 65 / 2^8 . 104
= 2^10 ( 13 + 65 ) / 2^8 . 104 = 2^10 . 78 / 2^8 . 104 = 2^8 . 2^2 . 78 / 2^8 . 104 = 2^8 . 4 . 78 / 2^8 . 104 = 2^8 . 312 / 2^8 . 104
= 312/104
= 3
B = 11 . 3^22 . 3^7 - 9^15 / ( 2.3^14)^2
= 11 . 3^29 - (3^2)^15 / ( 3.2^14)^2
= 11 . 3^29 - 3^30 / ( 3. 2 )^28
= ( 8 + 3 ) . 3^29 - 3^30 / ( 3. 2)^28
= 8 . 3^29 + 3.3^29 - 3^30 / ( 3.2)^28
= 8 . 3^29 + 3^30 - 3^30 / ( 3 . 2)^28
= 8 . 3^29 / 3^28 . 2^28
= 2^3 . 3 / 2^28
= 3/ 2^25

a) 2x - 15 = 17
2x = 25
b) ( 7x - 11 )3 = 25. 52 + 200
(7x - 11)3 = 32 . 25 + 200
(7x -11)3 = 1000
(7x-11)3 = 103
7x - 11 = 10
7x = 10+11
7x = 21
x = 21 : 7
x = 3
like nha

\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)
\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)
\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)
\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)

Gợi ý
bn thực hiện phép tính tử mẫu bình thường , khi ra nhưng số trùng nhau bn gạch ra nháp cho đến nhưng số tối giản là ra nha
chúc bn
học tốt
A = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
= \(\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
= \(\frac{3^{10}.16}{3^9.2^4}\)
= \(\frac{3^{10}.2^4}{3^9.2^4}=3\)
B = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
= \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)
= \(\frac{2^{10}.78}{2^8.104}\)
= \(\frac{2^{10}.13.2.3}{2^8.2^3.13}\)
= \(\frac{2^{11}.13.3}{2^{11}.13}=3\)

Bài 2
\(a,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)
\(\Rightarrow x\in\left\{4;5;6\right\}\)
\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)
\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)
Mà \(\frac{15}{2}\notin n\)
\(\Rightarrow x\in\left\{7;8\right\}\)
#)Giải :
Bài 1 :
a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)
Bài 2 :
a) \(\left(x-5\right)^2=\left(x-5\right)^6\)
\(\Leftrightarrow x^4-625=x^6-15625\)
\(\Leftrightarrow x^6-x^4=15000\)
\(\Leftrightarrow x^6-x^4=5^6-5^4\)
\(\Leftrightarrow x=5\)
b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow2x-15=1\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
lam nhanh ho voi a