\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

Ok

Bài 2:

a. 8x(5x - 2) - 12(5x - 2) = 0

<=> (8x - 12)(5x - 2) = 0

<=> \(\left[{}\begin{matrix}8x-12=0\\5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1 

a)2x(5x-2)+(2x+1)(8-5x)
=10x^2-4x+16x-10x^2+8-5x

=7x+8

b)(x+7)(x-7)-(x-3)^2

=x^2-49-(x^2-6x+9)

=6x-40

a) 3x(2x-y)+(x-y)(x+y)-7x²+y²

<=>6x²-3xy +x²-y²-7x²+y²

<=> -3xy

b) Thay x=-2/3 và y=2 vaò A ta có:

-3.-2/3.2=4

Lời giải:

a.

$x(x-7)+2x^2=(x+1)(3x-2)$

$\Leftrightarrow x^2-7x+2x^2=3x^2+x-2$

$\Leftrightarrow 3x^2-7x=3x^2+x-2$

$\Leftrightarrow 2=8x$

$\Leftrightarrow x=\frac{1}{4}$

b.

$(5x-3)(2x+1)-(2x-1)^2+4=0$

$\Leftrightarrow (10x^2-x-3)-(4x^2-4x+1)+4=0$

$\Leftrightarrow 6x^2+3x=0$

$\Leftrightarrow 3x(2x+1)=0$

$\Rightarrow 3x=0$ hoặc $2x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\frac{-1}{2}$

c.

$4x^2-9=(3-2x)(x+2)$

$\Leftrightarrow (2x-3)(2x+3)+(2x-3)(x+2)=0$

$\Leftrightarrow (2x-3)(2x+3+x+2)=0$

$\Leftrightarrow (2x-3)(3x+5)=0$

$\Rightarrow 2x-3=0$ hoặc $3x+5=0$

$\Leftrightarrow x=\frac{3}{2}$ hoặc $x=-\frac{5}{3}$

\(2,\\ a,=x^3-8-6x^2+12x=\left(x-2\right)^3\\ b,=25-4x^2+5x+4x^2-25-20x=-15x\\ c,=\left(2x+1\right)^2-\left(2x-1\right)^2\\ =\left(2x+1+2x-1\right)\left(2x+1-2x+1\right)=8x\\ d,=x^3+5x^2+25x+5x^2+25x+125-10x^2-50x-x^3\\ =125\)

Bài 2: 

a: Ta có: \(x\left(x+3\right)=6x+18\)

\(\Leftrightarrow\left(x+3\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)

b: Ta có: \(x^2\left(2x+1\right)-x\left(2x+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)\left(x-1\right)=0\)

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

\(Ba\) và \(CO_3\)

\(\xrightarrow[]{}\)\(BaCO_3\)

\(Ca\) và \(CO_3\)

\(\xrightarrow[]{}\)\(CaCO_3\)

\(Na\) và \(CO_3\)

\(\xrightarrow[]{}Na_2CO_3\)

Câu 14: B

Câu 15: Không có câu nào đúng

Câu 15 đáp án C nhé.