a: =>6x^2+2xb-15x-5b=ax^2+x+c

=>6x^2+x(2b-15)-5b=ax^2+x+c

=>a=6; 2b-15=1; -5b=c

=>a=6; b=8; c=-40

b: =>ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1

=>x^2(-a+b)+x(-a-b)-b=cx^2-1

=>-b=-1; -a+b=c; -a-b=0

=>b=1; c=b-a; a=-b=-1

=>c=b-a=1-(-1)=2; b=1; a=-1

f(0) = 1

\(\Rightarrow\) a.0² + b.0 + c = 1 

\(\Rightarrow\) c = 1

Vậy hệ số a = 0; b = 0; c = 1

f(1) = 2

\(\Rightarrow\) a.1² + b.1 + c = 2

\(\Rightarrow\) a + b + c = 2

Vậy hệ số a = 1; b = 1; c = 1

f(2) = 4

\(\Rightarrow\) a.2² + b.2 + c = 4

\(\Rightarrow\) 4a + 2b + c = 4

Vậy hệ số a = 4; b = 2; c = 1

Chúc bn học tốt! (chắc vậy :D)

Khai triển VT, ta có: \(VT=ax^3+\left(b+ac\right)x^2+\left(bc+2a\right)x+2b=x^3-x^2+2\)

Đồng nhất hệ số ta có hệ điều kiện:

\(\left\{{}\begin{matrix}a=1\\b+ac=-1\\bc+2a=0\\2b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=-2\end{matrix}\right.\)

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

1 ) Ta có :

\(ax+2x+ay+2y+4\)

\(=x\left(a+2\right)+y\left(a+2\right)+4\)

\(=\left(x+y\right)\left(a+2\right)+4\)

\(=\left(a-2\right)\left(a+2\right)+4\) ( do \(x+y=a-2\) )

\(=a^2-4+4\)

\(=a^2\left(đpcm\right)\)

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

\(\Leftrightarrow ax^3+bx^2-ax^2-bx-ax-b=ax^3+cx^2-1\)

\(\Leftrightarrow ax^3+x^2\left(b-a\right)-\left(b+a\right)x-b=ax^3+x^2c-0.x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}b-a=c\\b+a=0\\b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-a=c\\1+a=0\\b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-a=c\\a=-1\\b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=2\\a=-1\\b=1\end{matrix}\right.\)

Vậy \(a=-1;b=1;c=2\)

Ta có:

\(ax+2x+ay+2y+4\)

\(=\left(ax+ay\right)+\left(2x+2y\right)+4\)

\(=a\left(x+y\right)+2\left(x+y\right)+4\)

\(=\left(x+y\right)\left(a+2\right)+4\)

Thay \(x+y=a-2\), ta được

\(=\left(a-2\right)\left(a+2\right)+4\)

\(=a^2-4+4\)

\(=a^2\)

\(\left(x^2+cx+2\right)\left(ax+b\right)=x^3+x^2-2\)

\(\Leftrightarrow ax^{3\:}+\left(ac+b\right)x^2+\left(2a+bc\right)x+2b=x^3+x^2-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\ac+b=1\\2a+bc=0\\2b=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-1\\c=2\end{matrix}\right.\) ( TM )

Ta có: f(0)=1

<=> ax² +bx+c=1

<=> c=1

          f(1)=0

<=>ax² +bx+c=0

<=> a+b+c=0

mà c=1

=>a+b=-1(1)

      f(-1)=10

<=> ax² +bx +c=10

<=>a-b+c=10

mà c=1

=>a-b=9(2)

Lấy (1) trừ (2) ta được (a+b)-(a-b)=-1-9

                           <=> 2b=-10

                           <=> b=-5

                           =>a=4

Vậy a=4,b=-5,c=1

Nhớ k đúng cho mik