a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)

\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)

\(=a^{12}\cdot b^3\)

b) \(b^6\cdot b\cdot c^7\cdot c^8\)

\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)

\(=b^7\cdot c^{15}\)

c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)

\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)

\(=a^{18}\cdot c^{21}\)

d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)

\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)

\(=a^5\cdot b^4\cdot c^4\)

a) Kiểm tra lại nhé

b) \(b^6.b^7.c^8\)

\(=b^{6+7}.c^8=b^{13}.c^8\)

c) \(a^8.a^9.a.c.c^{20}\)

\(=a^{8+9+1}.c^{1+20}\)

\(=a^{18}.c^{21}\)

d) \(a^2.a^3.b^4.c.c^3\)

\(=a^{2+3}.b^4.c^{1+3}\)

\(=a^5.b^4.c^4\)

\(#WendyDang\)

a, \(\left(x+\frac{4}{3}y^2\right)^2\)

\(=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)

b, \(\left(2x-3y\right)^2\)

\(=4x^2-12xy+9y^2\)

c, \(\left(x^2+2x\right)\left(2x-x^2\right)\)

\(=\left(2x+x^2\right)\left(2x-x^2\right)\)

\(=4x^2-x^4\)

d, \(\left(x+\frac{1}{2}\right)^3\)

\(=x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

e, \(\left(2x-\frac{6}{5}y\right)^3\)

\(=8x^3-\frac{72}{5}x^2y+\frac{216}{25}xy^2-\frac{216}{125}y^3\)

a) a mũ 3 x a mũ 9 là : a³ x b⁹

b) ( a mũ 5 ) mũ 7 là : (a⁵)⁷

c) ( 2 mũ 3 ) mũ 5 x ( 2 mũ 3 ) mũ 3 là : (2³)⁵ x (2³)³

Hok tốt !

Đầu bài là thế ạ còn viết thì em ko biết anh giải giùm em ạ

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

ỏ cảm mơn nhaaaa ! có j giúp típ nha thank kiuuu 

giúp e đg cần gấp ạa

\(a^2.a^3.a^7.b^2.b\)

\(=a^{2+3+7}.b^{2+1}\)

\(=a^{12}.b^3\)

\(P\left(x\right)=B\left(x\right)-A\left(x\right)\)

\(=(2+5x^3-8x+x^2)-\left(2x^2+2x+5x^3+2\right)\)

\(=2+5x^3-8x+x^2-2x^2-2x-5x^3-2\)

\(=-x^2-10x\)

Vậy \(P\left(x\right)=-x^2-10x\)

b: Ta có: \(2^{x+3}+2^x=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

hay x=4

a) (x ^ 54)^2 = x                                         

         x^108  = x

Để: x^108  = x 

=> x=0 hoặc x=1

mình cũng hỏi về cái này gấp nha ai biết trả lời câu nào thì trả