c)

\(x^2+7x+12=0\\ x^2+3x+4x+12=0\\ x\left(x+3\right)+4\left(x+3\right)=0\\ \left(x+3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

d)

\(x^2-5x+6=0\\ x^2-2x-3x+6=0\\ x\left(x-2\right)-3\left(x-2\right)=0\\ \left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

a)

\(\left(x-1\right)^2=\left(2x+5\right)^2\\ \left(x-1\right)^2-\left(2x+5\right)^2=0\\ \left(x-1+2x+5\right)\left(x-1-2x-5\right)=0\\ \left(3x+4\right)\left(-x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+4=0\\-x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=-6\end{matrix}\right.\)

b)

\(x^2+8x+15=0\\ x^2+3x+5x+15=0\\ x\left(x+3\right)+5\left(x+3\right)=0\\ \left(x+3\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)

\(n_{MgO}=\dfrac{6}{40}=0,15mol\)

\(PTHH\left(1\right):Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ ­\: ­\: ­\: ­\: ­\: ­\: ­­\: ­\: ­\: ­\: ­­­\: ­\: ­\: ­\left(2\right):MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{0,3.3}{1}+\dfrac{0,15.1}{1}=1,05mol\)

\(m_{H_2SO_4}=1,05.98=102,9g\)

a)

\(x-2\left|x+1\right|=3\\ -2\left|x+1\right|=3-x\)

\(\left[{}\begin{matrix}nếu\:x\ge-1\:thì\left|x+1\right|=x+1\\nếu\:x< -1\:thì\:\left|x+1\right|=-x-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-2\left(x+1\right)=3-x\left(với\: x\ge-1\: \right)\\-2\left(-x-1\right)=3-x\left(với\: x< -1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x-2=3-x\\2x+2=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=-\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

vậy phương trình đã cho vô nghiệm.

b)

\(6-\left|3x-1\right|=5\\ -\left|3x-1\right|=-1\\ \left|3x-1\right|=1\\ \Rightarrow\left[{}\begin{matrix}3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)

vậy phương trình đã cho có tập nghiệm là S={0;2/3}

c)

\(\left|2x-1\right|=x+2\\ \Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\\ \left(2x-1\right)^2-\left(x+2\right)^2=0\\ \left(2x-1+x+2\right)\left(2x-1-x-2\right)=0\\ \left(3x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=3\end{matrix}\right.\)

vậy phương trình đã cho có tập nghiệm là S={-1/3;3}

d)

\(\left|2x-7\right|-x-3=0\\ \left|2x-7\right|=x+3\\ \Rightarrow\left(2x-7\right)^2=\left(x+3\right)^2\\ \left(2x-7\right)^2-\left(x+3\right)^2=0\\ \left(2x-7+x+3\right)\left(2x-7-x-3\right)=0\\ \left(3x-4\right)\left(x-10\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\x-10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=10\end{matrix}\right.\)

vậy phương trình đã cho có tập nghiệm là S={4/3;10}

42.

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\\ =ab^2-a^2b-b^2c+bc^2-ac\left(c-a\right)\\ =b^2\left(a-c\right)+b\left(c^2-a^2\right)-ac\left(c-a\right)\\ =\left(a-c\right)\left(b^2-ac+ba+bc\right)\)

33.

\(x^{10}+x^5+1\\ =x^{10}+x^9+x^8-x^9-x^8-x^7+x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\\ =x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ \left(x^2+x+1\right)\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

34.

đặt: \(t=x^2+x+1,5\)

khi đó:

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\\ =\left(t-0,5\right)\left(t+0,5\right)-12\\ =t^2-0,25-12\\ =t^2-12,25\\ =\left(t-3,5\right)\left(t+3,5\right)\\ =\left(x^2+x-2\right)\left(x^2+x+5\right)\)

35.

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

36.

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+15\\ =\left(x^2-10x+16\right)\left(x^2-10x+24\right)+15\\ =\left(x^2-10x+20-4\right)\left(x^2-10x+20+4\right)+15\\ =\left(x^2-10x+20\right)^2-4^2+15\\ =\left(x^2-10x+20\right)^2-1\\ =\left(x^2-10x+19\right)\left(x^2-10x+21\right)\)

37.

\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\\ =\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\\ =\left(x^2-10x+20-4\right)\left(x^2-10x+20+4\right)+16\\ =\left(x^2-10x+20\right)^2-4^2+16\\ =\left(x^2-10x+20\right)^2\)

38.

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)-24\\ =\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\\ =\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24\\ =\left(x^2+5x+5\right)^2-1-24\\ =\left(x^2+5x+5\right)^2-5^2\\ =\left(x^2+5x+10\right)\left(x^2+5x\right)\\ =x\left(x+5\right)\left(x^2+5x+10\right)\)

39.

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\\ =\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\\ =\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\\ =\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+1\\ =\left(x^2+5x+5\right)^2-1+1\\ =\left(x^2+5x+5\right)^2\)

40.

\(a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\\ =a^3b^2-a^2b^3-c^3b^2+c^2b^3+a^2c^2\left(c-a\right)\\ =b^2\left(a^3-c^3\right)+b^3\left(c^2-a^2\right)+a^2c^2\left(c-a\right)\\ =b^2\left(a-c\right)\left(a^2+ac+c^2\right)+b^3\left(c-a\right)\left(c+a\right)+a^2c^2\left(c-a\right)\\ =-b^2\left(c-a\right)\left(a^2+ac+c^2\right)+\left(c-a\right)\left(cb^3+ab^3+a^2c^2\right)\\ =\left(c-a\right)\left(cb^3+ab^3+a^2c^2-a^2b^2-acb^2-b^2c^2\right)\)

A = 3 + 3³ + 3⁵ + ... + 3¹⁹⁹¹

   = ( 3 + 3³ + 3⁵ ) + ( 3⁷ + 3⁹ + 3¹¹ ) + ... + ( 3¹⁹⁸⁷ + 3¹⁹⁸⁹ + 3¹⁹⁹¹ )

   = 3(1+3²+3⁴) + 3⁷(1+3²+3⁴) + ... + 3¹⁹⁸⁷(1+3²+3⁴)

   = 3.91 + 3⁷.91 + ... + 3¹⁹⁸⁷.91

   = 91.(3+3⁷+...+3¹⁹⁸⁷) chia hết cho 91

Mà 91 = 13.7 nên A cũng chia hết cho 13

\(x^3+x^2+4=x^3+2x^2-x^2-2x+2x+4\\ =x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\\ =\left(x+2\right)\left(x^2-x+2\right)\)

\(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)=n^2+5n-n^2+n+6\\ =6n+6=6\left(n+1\right)⋮6\)

vậy \(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)⋮6\:\forall x\in Z\)

gọi 3 số tự nhiên chẵn liên tiếp là 2a-2;2a;2a+2(a là số tự nhiên lớn hơn 0)

theo đề bài, ta có:

\(2a.\left(2a+2\right)-2a.\left(2a-2\right)=192\\ 4a^2+4a-4a^2+4a=192\\ 8a=192\Rightarrow a=\dfrac{192}{8}=24\left(thõa\:mãn\right)\)

vậy 3 số cần tìm là :46;48;50