`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)

`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`

`C(x)= -3x^4-2x^3+x^2+x+5`

`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)

`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`

`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`

`D(x)=-x^4-4x^3-4x+7`

`P(x)=C(x)+D(x)`

`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`

`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`

`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`

`P(x)=-4x^4-6x^3+x^2-3x+12`

`Q(x)=C(x)-D(x)`

`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`

`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`

`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`

`Q(x)=-2x^4+2x^3+x^2+5x-2`

`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`

`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`

`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`

`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`

`F(x)=5x+10`

Đặt `5x+10=0`

`\Leftrightarrow 5x=0-10`

`\Leftrightarrow 5x=-10`

`\Leftrightarrow x=-10 \div 5`

`\Leftrightarrow x=-2`

Vậy, nghiệm của đa thức là `x=-2.`

Câu 1: 

a) Ta có: x-3 là ước của 13

\(\Leftrightarrow x-3\inƯ\left(13\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)

Vậy: \(x\in\left\{4;2;16;-10\right\}\)

b) Ta có: \(x^2-7\) là ước của \(x^2+2\)

\(\Leftrightarrow x^2+2⋮x^2-7\)

\(\Leftrightarrow x^2-7+9⋮x^2-7\)

mà \(x^2-7⋮x^2-7\)

nên \(9⋮x^2-7\)

\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)

\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(x^2-7\ge-7\forall x\)

nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)

\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)

\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)

mà \(x\in Z\)

nên \(x\in\left\{2;-2;4;-4\right\}\)

Vậy: \(x\in\left\{2;-2;4;-4\right\}\)

Câu 2: 

a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-6-3x+15=12-4x-18\)

\(\Leftrightarrow-x+9+4x+6=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

hay x=-5

Vậy: x=-5

Câu 1 : \(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)

Ta có : \(VT=-a.\left(c-d\right)-d\left(a+c\right)\)

                 \(=-ac+ad-da-dc\)

                 \(=-ac-dc\)

                 \(=-c\left(a+d\right)=VP\)

\(\Rightarrow-a\left(c-d\right)-d\left(a+c\right)=-c\left(a+d\right)\left(đpcm\right)\)

Câu 2 : 

1,  \(x.\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)

2, \(\left(x+12\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)

3, \(\left(-x+5\right)\left(3-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)

4, \(x\left(2+x\right)\left(7-x\right)=0\)

\(\Rightarrow x=0;2+x=0\)hoặc \(7-x=0\)

\(\Rightarrow x=0;x=-2\)hoặc \(x=7\)

Thanks Bạn!!

\(\begin{array}{l} - 2{x^3}{y^4}:D = x{y^2}\\ \Rightarrow D =  - 2{x^3}{y^4}:x{y^2} =  - 2{x^2}{y^2}\end{array}\)

\(\begin{array}{l}\left( {10{x^5}{y^2} - 6{x^3}{y^4} + 8{x^2}{y^5}} \right):\left( { - 2{x^2}{y^2}} \right)\\ = \left( {10{x^5}{y^2}} \right):\left( { - 2{x^2}{y^2}} \right) - \left( {6{x^3}{y^4}} \right):\left( { - 2{x^2}{y^2}} \right) + \left( {8{x^2}{y^5}} \right):\left( { - 2{x^2}{y^2}} \right)\\ =  - 5{x^3} + 3x{y^2} - 4{y^3}\end{array}\)

\(B=8x^2+2x-8x^3-8x^2+8x^3-2x+3=3\)

\(C=x^3-3x^2+3x-1+x^3+3x^2+3x+1+2x^3-8x=4x^3-2x\)

\(D=\left(x+y-5\right)^2-2\left(x+y-5\right)\left(x+3\right)+\left(x+3\right)^2=\left(x+y-5-x-3\right)^2=\left(y-8\right)^2\)

câu 2. ta có 

a.\(\left(x-y\right)^2=\left(x+y\right)^2-4xy=7^2-4\times12=1\)

b.\(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)=3\left(x+y\right)^2-6xy-2\left(x+y\right)^3+6xy\left(x+y\right)=3-6xy-2+6xy=1\)

Giúp mk vs Akai HarumaLovers