a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)

1)x²-8x-9

= x^2 - 9x +x -9

= x(x+1) - 9 (x+1)

= (x-9) (x+1)

2)x²+3x-18

3)x³-5x²+4x

=x^3 - 4x^2 - x^2 + 4x 

= x^2 (x-1) - 4x(x-1)

= (x^2 - 4x) (x-1)

= x(x-4)(x-1)

4)x³-11x²+30x

5)x³-7x-6

6)x¹⁶-64

\(=\left(x^8\right)^2-8^2\)

\(=\left(x^8-8\right)\left(x^8+8\right)\)

7)x³-5x²+8x-4

8)x²-3x+2

= x^2 - 2x - x +2

= x(x-1) -2(x-1)

= (x-2)(x-1)

1)   \(\left(x-9\right)\left(x+1\right)\)             2)   \(\left(x-3\right)\left(x+6\right)\)                                           3)   \(x\left(x-4\right)\left(x-1\right)\)

4)    \(x\left(x-6\right)\left(x-5\right)\)         5)\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\)                               6)   ........

7)  \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\)          8)   \(\left(x-2\right)\left(x-1\right)\)

a: ĐKXĐ: \(3x^2+6x\ne0\)

=>\(x^2+2x\ne0\)

=>\(x\cdot\left(x+2\right)\ne0\)

=>\(x\notin\left\{0;-2\right\}\)

b: ĐKXĐ: \(x^3+64\ne0\)

=>\(x^3\ne-64\)

=>\(x\ne-4\)

c: ĐKXĐ: \(x^2-1\ne0\)

=>\(x^2\ne1\)

=>\(x\notin\left\{1;-1\right\}\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)

4: \(\Leftrightarrow3^{x+4}\cdot\dfrac{1}{3}-4\cdot3^x=3^{16}\left(1-4\cdot3^3\right)\)
=>\(3^x\cdot27-4\cdot3^x=3^{16}\cdot\left(-107\right)\)

=>3^x*23=3^16*(-107)

=>\(x\in\varnothing\)

2: \(\Leftrightarrow2^x\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)=2^{10}\left(\dfrac{3}{5}+\dfrac{7}{5}\cdot2^3\right)\)

=>2^x=2^10

=>x=10

3: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>8^x=8^9

=>x=9

1: \(\Leftrightarrow3^x\cdot\left(4\cdot\dfrac{1}{9}+2\cdot3\right)=3^4\left(4+2\cdot3^3\right)\)

=>3^x=3^4*3^2

=>x=4+2=6

647 - [6. (107 . 8 - 7 . 2³) + 5 . 2⁵ : 2⁴ ]

= 647 - [6. (856 - 7 .8) + 5 . 2 ]

= 647 - [6. (856 - 56) + 5 . 2]

= 647 - [6. 800 + 5 . 2]

= 647 - [ 4800 + 10]

= 647 - 4810 

= -4163

-4153 ạ

đảm bảo đúng luôn

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$