a.

\(A\left(x\right)+B\left(x\right)+C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)+\left(8x^6+7x^4-x^2+11\right)+\left(x^6+7x^4-x^2+6\right)\)

\(=12x^6+9x^4+10\)

b.

\(A\left(x\right)-B\left(x\right)+C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)-\left(8x^6+7x^4-x^2+11\right)+\left(x^6+7x^4-x^2+6\right)\)

\(=-4x^6-5x^4+2x^2-12\)

c.

\(A\left(x\right)-B\left(x\right)-C\left(x\right)=\left(3x^6-5x^4+2x^2-7\right)-\left(8x^6+7x^4-x^2+11\right)-\left(x^6+7x^4-x^2+6\right)\)

\(=-6x^6-19x^4+4x^2-24\)

Do \(f\left(1\right)=g\left(2\right)\) nên:

\(2.1^2+a.1+4=2^3-5.2-b\)

\(\Rightarrow a+6=-2-b\)

\(\Rightarrow a=-b-8\) (1)

Do \(f\left(-1\right)=g\left(5\right)\) nên:

\(2.\left(-1\right)^2+a.\left(-1\right)+4=5^3-5.5-b\)

\(\Rightarrow-a+6=100-b\)

\(\Rightarrow a=b-94\) (2)

(1);(2) \(\Rightarrow-b-8=b-94\)

\(\Rightarrow2b=94-8\)

\(\Rightarrow b=43\)

\(\Rightarrow a=43-94=-51\)

9a.

Do \(P\left(-2\right)=7\)

\(\Rightarrow a.\left(-2\right)+5=7\)

\(\Rightarrow-2a=2\)

\(\Rightarrow a=-1\)

b.

Do \(P\left(0\right)=-2\)

\(\Rightarrow a.0+b=-2\)

\(\Rightarrow b=-2\)

Do \(P\left(-3\right)=-1\)

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

\(\Rightarrow-3a-2=-1\)

\(\Rightarrow-3a=1\)

\(\Rightarrow a=-\dfrac{1}{3}\)

- Kẻ EF,CH lần lượt vuông góc với AB (F,H thuộc AB).

- Ta có: EF,CH lần lượt vuông góc với AB (gt)

=>EF//CH.

- Xét tam giác ACH có:

EF//CH (cmt)

=>\(\dfrac{EF}{CH}=\dfrac{AE}{AC}=\dfrac{1}{4}\) (định lí Ta-let)

- \(\dfrac{S_{ABC}}{S_{ADE}}=\dfrac{CH}{EF}.\dfrac{AB}{AD}=4.3=12\)(cm²)

=>S_ADE=\(\dfrac{1}{12}S_{ABC}=\dfrac{1}{12}.60\)=5 (cm²)

* S_DBEC=S_ABC-S_ADE=60-5=55(cm²)

- Hic hic mình xin lỗi nhưng mình đang bận :). Đợi tối mình làm :)