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) \(a^2.a^3.b^4.c.c^3\)

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

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

\(#WendyDang\)

* là gì đó bạn

@Hưng Què Dấu nhân nhé

a: Thay x=-3 vào B, ta được:

\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)

b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)

1)

ĐKXĐ: x>4

Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)

\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)

\(\Leftrightarrow8x+6x=8-15\)

\(\Leftrightarrow14x=-7\)

hay \(x=-\dfrac{1}{2}\)(loại)

2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)

\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

93873498347

\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)

C1: $\sqrt{28}=\sqrt{4.7}=2\sqrt 7$

Ta có: $3>2$

$\Leftrightarrow 3\sqrt 7>3\sqrt 7$ hay $3\sqrt 7>\sqrt{28}$

C2: $3\sqrt{7}=\sqrt{63}$

Ta có: $63>28$

$\Leftrightarrow\sqrt{63}>\sqrt{28}$ hay $3\sqrt 7>\sqrt{28}$

`4/7 x -x=-9/14`

`=> 4/7 x - x . 1 =-9/14`

`=> (4/7-1) x=-9/14`

`=> -3/7 x=-9/14`

`=>x=-9/14 : (-3/7)`

`=>x=-9/14 xx (-7/3)`

`=>x=3/2`