Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot0^3+b\cdot0+c=2\\a+b+c=0\\-a-b+c=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=2\\a+b=-2\\-a-b=4\end{matrix}\right.\Leftrightarrow\left(a,b,c\right)\in\varnothing\)

Bn oi bn bỏ phần bôi đen đc ko? Chữ khó nhìn wá

a) Ta có: \(f\left(0\right)=5\Rightarrow a.0^2+b.0+c=5\)

\(\Rightarrow c=5\)

\(f\left(1\right)=0\Rightarrow a.1^2+b.1+c=0\)

\(\Rightarrow a+b+c=0\left(1\right)\)

Thay \(c=5\) vào (1) được:

\(a+b+5=0\Rightarrow a+b=-5\left(2\right)\)

\(f\left(5\right)=0\Rightarrow a.5^2+5b+c=0\)

\(\Rightarrow25a+5b+c=0\)

\(\Rightarrow5\left(5a+b+1\right)=0\)

\(\Rightarrow5a+b+1=0\)

\(\Rightarrow5a+b=-1\)

\(\Rightarrow b=-1-5a\left(3\right)\)

Thay \(\left(3\right)\rightarrow\left(2\right):a+\left(-1-5a\right)=-5\)

\(\Rightarrow a-1-5a=-5\)

\(\Rightarrow-1-4a=-5\)

\(\Rightarrow4a=4\)

\(\Rightarrow a=1\)

Khi đó: \(1+b=-5\Rightarrow b=-6\)

Vậy \(\left\{{}\begin{matrix}a=1\\b=-6\\c=5\end{matrix}\right.\).

b) Kết hợp \(y=-3\) với câu a) ta có:

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

\(\Rightarrow x^2-3x-3x+5=-3\)

\(\Rightarrow x^2-3x-3x+ 9-4=-3\)

\(\Rightarrow x\left(x-3\right)-3\left(x-3\right)-4=-3\)

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

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\).

a) thay f(0) = 5 vào hàm số ta có : \(5=a0^2+b0+c\) \(\Leftrightarrow\) \(c=5\)

thay f(1) = 0 và f(5) = 0 vào hàm số ta có hệ phương trình

\(\left\{{}\begin{matrix}a+b+5=0\\25a+5b+5=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=-5\\25a+5b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}5a+5b=-25\\25a+5b=-5\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}20a=20\\a+b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\1+b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\b=-6\end{matrix}\right.\)

vậy \(a=1;b=-6;c=5\)

câu b thay a;b;c vào tìm x là được

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot0+b\cdot0+c=5\\4a+2b+c=0\\25a+5b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\4a+2b=-5\\25a+5b=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\a=\dfrac{1}{2}\\b=-\dfrac{7}{2}\end{matrix}\right.\)

\(f\left(0\right)=5=>c=5;f\left(2\right)=4.a+2.b+5=0;f\left(5\right)=25a+5b+5=0\Leftrightarrow5a+b+1=0\)

\(\hept{\begin{cases}4a+2b+5=0\\5a+b+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\10a+2b+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\6a-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=-\frac{7}{2}\\a=\frac{1}{2}\end{cases}}\)

\(f\left(x\right)=\frac{1}{2}x^2-\frac{7}{2}x+5\)

b)

\(f\left(-1\right)=\frac{1}{2}+\frac{7}{2}+5=9=>P\left(-1;3\right)kothuocHS\)

\(f\left(\frac{1}{2}\right)=\frac{1}{2}.\frac{1}{4}-\frac{7}{2}.\frac{1}{2}+5=\frac{\left(1-14+5.8\right)}{8}=\frac{27}{8}=>Qkothuoc\)

c)

\(\frac{1}{2}x^2-\frac{7}{2}x+5=-3\Rightarrow\frac{1}{2}x^2-\frac{7}{2}x+8=0\)

\(x^2-7x+16=0\Leftrightarrow\left(x^2-2.\frac{7}{2}x+\frac{49}{4}\right)+\frac{15}{4}\)vo nghiem

no hiểu gì hết THIS IS HOW I DO NOT KNOW HOW TO APOLOGIZE OFFLINE

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