chọn sai môn học rồi nha

1. B
2. A

Bài 1:

a. $=2x(x-3)$

b. $=x^3(x+3)+(x+3)=(x^3+1)(x+3)=(x+1)(x^2-x+1)(x+3)$

c. $=64-(x^2-2xy+y^2)=8^2-(x-y)^2$

$=(8-x+y)(8+x-y)$

Bài 2:

$(x+5)(x+1)+(x-2)(x^2+2x+4)-x(x^2+x-2)$

$=x^2+6x+5+(x^3-2^3)-(x^3+x^2-2x)$

$=x^2+6x+5+x^3-8-x^3-x^2+2x$

$=8x-3$

Ta có đpcm.

Bài 4:

$P=a^3+a^2c-abc+b^2c+b^3$

$=a^2(a+c)-abc+b^2(b+c)$

$=a^2(-b)-abc+b^2(-a)=-a^2b-abc-b^2a$

$=-ab(a+c+b)=-ab.0=0$

\(\left(2x+5\right)^2=\left(x+2\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=x+2\\2x+5=-x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=2-5\\2x+x=-2-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

easy thôi

\(\left(2x+5\right)^2=\left(x+2\right)^2\)

\(\Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\)

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

Vậy \(S=\left\{-3;-\dfrac{7}{3}\right\}\)

\(\left(2x+5\right)^2=\left(x+2\right)^2\)

<=> \(\left(2x+5\right)^2-\left(x+2\right)^2=0\)

<=> \(\left[{}\begin{matrix}2x+5-x-2=0\\2x+5+x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\3x+7=0\end{matrix}\right.\)

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

Câu 6: B

Câu 7: A

Nào gg thẳng tiến 

308:5,5=56

18,5:7,4=2,5

8,568:36=0,238

8,16:3=2,72

85:14=7(dư 1)