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

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

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

\(=10^3+10^2+10+1=1111\)

a) Rút gọn Q = 54 x 3 , thay x = 10 vào tính được Q = 54000;

b) Gợi ý x 4 + y 2 = x + 2 y 4 = 8 4 = 2 .  Kết quả P = 2.

1) 3x = 45 + 15 = 60

x = 60 : 3 = 20

2) 5x = 50 - 35 = 15

x = 15 : 5 = 3

3) (2x - 5) + 17 = 6

2x - 5 = 6 - 17

2x - 5 = -11

2x = -11 + 5

2x = -6

x = -6 : 2 = -3

4) 10 - 2(4 -3x) = -4

2(4 - 3x) = -4 - 10 = -14

4 - 3x = -14 : 2 = -7

3x = -7 - 4 = -18

x = -18 : 3 = -6

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)

1: 3x-15=45

=>\(3x=45+15=60\)

=>\(x=\frac{60}{3}=20\)

2: 35-5x=50

=>5x=35-50=-15

=>\(x=-\frac{15}{5}=-3\)

3: (2x-5)+17=6

=>2x-5+17=6

=>2x+12=6

=>2x=6-12=-6

=>x=-3

4: 10-2(4-3x)=-4

=>10-8+6x=-4

=>6x+2=-4

=>6x=-6

=>x=-1

5: -12+3(-x+7)=-18

=>3(-x+7)=-18+12=-6

=>-x+7=-2

=>-x=-2-7=-9

=>x=9

6: 24:(3x-2)=-3

=>3x-2=-24:3=-8

=>3x=-6

=>x=-2

7: \(-45:5\cdot\left(-3-2x\right)=3\)

=>-9(-3-2x)=3

=>9(2x+3)=3

=>\(2x+3=\frac13\)

=>\(2x=\frac13-3=-\frac83\)

=>\(x=-\frac83:2=-\frac43\)

a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)

\(=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

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

b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)

\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)

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