a) \(M\left(x\right)+N\left(x\right)=3x^3-3x+x^2+2x^2-x+3x^3+9\)

\(M\left(x\right)+N\left(x\right)=\left(3x^3+3x^3\right)+\left(x^2+2x^2\right)+\left(-3x-x\right)+9\)

\(M\left(x\right)+N\left(x\right)=6x^3+3x^2-4x+9\)

b) \(M\left(x\right)+N\left(x\right)-\left(6x^3+3x^2+2x\right)=P\left(x\right)\)

\(6x^3+3x^2-4x+9-6x^3-3x^2-2x=P\left(x\right)\)

Ta có:\(Q\left(x\right)=0\)

\(\Leftrightarrow2x^2-3x=0\)\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x=0,x=\dfrac{3}{2}là\) nghiệm của Q(x)

d)Ta có \(\widehat{BAC}=\widehat{DCA}=90\)

Góc BAM=DCM vì tam giác MAB=MDC

=>góc AC=góc MCA hay góc NAK=góc ICK

Ta có tgiac ABK=tgiac CDK

=>góc BKA=góc DKC hay góc NKA=góc IKC

Xét tgiac ANK và tgiac CIK,có

góc NAK=góc ICK (cmt)

AK=CK

góc NKA=góc IKC(cmt)

=> tgiac ANK = tgiac CIK(g-c-g)

=>NK=IK(2 cạnh t/ứng)

Xét \(\Delta KNI,có\)

NK=IK

\(\Delta KNI\) cân

Vì x=14 nên 15=x+1

\(A\left(x\right)=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-15\)

\(A\left(x\right)=x^{15}-x^{15}-x^{14}+x^{14}-x^{13}+...+x^4+x^3-x^3+x^2+x^2+x-15\)

\(A\left(x\right)=x^2+x^2+x-15\)

\(\Leftrightarrow A\left(x\right)=14^2+14^2+14-15=196+196-1\)

\(A\left(x\right)=391\)

a) Áp dụng định lý Py-Ta -Go vào \(\Delta\perp ABC\)

BC²=AC²+AB²

BC²=8²+6²

BC²=100

\(\Rightarrow BC=\sqrt{100}=10cm\)

b) Xét \(\Delta ABD\) và\(\Delta EBD\) ,có

^ ^

DAB=DEB=90 (gt)

DB chung

=> \(\Delta ABD\) =\(\Delta EBD\) (ch-gn)

a) Xét ABM và ACM,có

AB=AC (gt)

AM chung

BM=CM (gt)

=>$\Delta$ ABM=$\Delta$ ACM(c-c-c)

b)Ta có BM+CM=BC

Mà BC=10cm; BM=CM

=>BM+BM=BC

=>2BM=BC

=>BM=BC/2=10/2=5cm

Ta có ABM= ACM(cmt)

=>Góc BMA=góc CMA(2 góc t/ứng)

Mà \(\widehat{BMA}+\widehat{CMA}=180\left(kb\right)\)

=> \(\widehat{BMA}=\widehat{CMA}=90\)

AM²=AB²-BM²

AM²=13²-5²

AM²=144

=>\(AM=\sqrt{144}=12\)