b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)

\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)

\(a,\Leftrightarrow2x^3-x^2+ax+b=\left(x-1\right)\left(x+1\right)\cdot a\left(x\right)\)

Thay \(x=1\Leftrightarrow2-1+a+b=0\Leftrightarrow a+b=-1\)

Thay \(x=-1\Leftrightarrow-2-1-a+b=0\Leftrightarrow b-a=3\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)

\(b,\Leftrightarrow ax^3+bx^2+2x-1=\left(x-1\right)\left(x+6\right)\cdot b\left(x\right)\)

Thay \(x=1\Leftrightarrow a+b+2-1=0\Leftrightarrow a+b=-1\)

Thay \(x=-6\Leftrightarrow-216a+36b+12-1=0\Leftrightarrow216a-36b=11\)

Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\216a-36b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{25}{252}\\b=-\dfrac{227}{252}\end{matrix}\right.\)

\(c,\Leftrightarrow ax^4+bx^3+1=\left(x+1\right)^2\cdot c\left(x\right)\)

Thay \(x=-1\Leftrightarrow a-b+1=0\Leftrightarrow b=a+1\)

\(\Leftrightarrow ax^4+\left(a+1\right)x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^4+ax^3+x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^3\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(ax^3+x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow ax^3+x^2-x+1⋮\left(x+1\right)\)

Thay \(x=-1\Leftrightarrow-a+1+1+1=0\Leftrightarrow a=3\Leftrightarrow b=4\)

a: \(\dfrac{2x^3-x^2+ax+b}{x^2-1}\)

\(=\dfrac{2x^3-2x-x^2+1+\left(a+2\right)x+b-1}{x^2-1}\)

\(=2x-1+\dfrac{\left(a+2\right)x+b-1}{x^2-1}\)

Để đây là phép chia hết thì a+2=0 và b-1=0

=>a=-2; b=1

b: \(\Leftrightarrow x^4-1+ax^2-a+bx+a⋮x^2-1\)

=>bx+a=0

=>a=b=0

â) viết lại biểu thức bên trái = (x²+5x-3)(x²-2x-4)+(14+a)x+b-12

Để là phép chia hết thì số dư =0

Số dư chính là (14+a)x+b-12=0 => a+14=0 và b-12=0 <=>a=-14 và b=12

b) làm tương tự phân tích vế trái thành (x³-2x²+4)(x²+9x+18)+(a+32)x²+(b-36)x

số dư là (a+32)x²+(b-36)x=0 =>a=-32 và b=36

c) Tương tự (x²-1)4x+(a+4)x+b

số dư là (a+4)x+b =2x-3 =>a+4=2 và b=-3 <=>a=-2 và b=-3

1)

<=> \(x^2-3x=0\)

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

x= 0 

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

\(2,x^2-3x=4\)

\(\Leftrightarrow x^2-3x-4=0\)

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

\(4,x^3=9x\)

\(\Leftrightarrow x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

Vậy \(S=\left\{0;\pm3\right\}\)

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

\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)

\(\Leftrightarrow-x-2=0\)

\(\Leftrightarrow-x=2\)

\(\Leftrightarrow x=-2\)

Vậy \(S=\left\{-2\right\}\)