Gọi I là giao của AC và BD

Ta sẽ chứng minh MN cũng đi qua I

Ta có: AB // CD => \(\frac{AI}{IC}=\frac{BI}{ID}=\frac{AB}{DC}=\frac{\frac{2}{3}AB}{\frac{2}{3}DC}=\frac{AM}{NC}\)

Xét 2 tam giác: AMI và CNI có:

\(\hept{\begin{cases}\frac{AM}{NC}=\frac{AI}{IC}\left(cmt\right)\\\widehat{MAI}=\widehat{NCI}\left(soletrong\right)\end{cases}}\)

\(\Rightarrow\widehat{AIM}=\widehat{NIC}\Rightarrow\overline{M,I,N}\) => đpcm

Giải thích các bước giải:

 a) Ta có:

{ˆBchungˆADB=ˆCFB=900⇒ΔADB∼ΔCFB(g.g)⇒BABD=BCBF⇒BF.BA=BD.BC{B^chungADB^=CFB^=900⇒ΔADB∼ΔCFB(g.g)⇒BABD=BCBF⇒BF.BA=BD.BC

Lại có:

⎧⎪⎨⎪⎩BABD=BCBFˆBchung⇒ΔBFD∼ΔBCA(c.g.c)⇒ˆBFD=ˆBCA{BABD=BCBFB^chung⇒ΔBFD∼ΔBCA(c.g.c)⇒BFD^=BCA^

b) Ta có:

{ˆHFB=ˆHEC=900ˆFHB=ˆEHC(dd)⇒ΔFHB∼ΔEHC(g.g)⇒HFHE=HBHC⇒HF.HC=HE.HB{HFB^=HEC^=900FHB^=EHC^(dd)⇒ΔFHB∼ΔEHC(g.g)⇒HFHE=HBHC⇒HF.HC=HE.HB

Lại có:

HFHE=HBHC⇒HFHB=HEHCHFHE=HBHC⇒HFHB=HEHC

Khi đó:

⎧⎪⎨⎪⎩ˆEHF=ˆCHB(dd)HFHB=HEHC⇒ΔEHF∼ΔCHB(g.g)⇒ˆFEH=ˆBCH⇒ˆFEB=ˆFCB{EHF^=CHB^(dd)HFHB=HEHC⇒ΔEHF∼ΔCHB(g.g)⇒FEH^=BCH^⇒FEB^=FCB^

c) Ta có:

{ˆCchungˆCDH=ˆCFB=900⇒ΔCDH∼ΔCFB(g.g)⇒CDCF=CHCB⇒CF.CH=CD.CB⇒CF.CH+BF.BA=CD.CB+BD.BC=BC.(CD+BD)⇒CF.CH+BF.BA=BC2{C^chungCDH^=CFB^=900⇒ΔCDH∼ΔCFB(g.g)⇒CDCF=CHCB⇒CF.CH=CD.CB⇒CF.CH+BF.BA=CD.CB+BD.BC=BC.(CD+BD)⇒CF.CH+BF.BA=BC2

d) Ta có:

{ˆAchungˆAEB=ˆAFC=900⇒ΔAEB∼ΔAFC(g.g)⇒AEAF=ABAC⇒AEAB=AFAC{A^chungAEB^=AFC^=900⇒ΔAEB∼ΔAFC(g.g)⇒AEAF=ABAC⇒AEAB=AFAC

Khi đó:

⎧⎪⎨⎪⎩ˆAchungAEAB=AFAC⇒ΔAEF∼ΔABC(g.g)⇒ˆAFE=ˆACB⇒ˆIFB=ˆACB⇒ˆIFB=ˆICE{A^chungAEAB=AFAC⇒ΔAEF∼ΔABC(g.g)⇒AFE^=ACB^⇒IFB^=ACB^⇒IFB^=ICE^

Như vậy:

{ˆIchungˆIFB=ˆICE⇒ΔIFB∼ΔICE(g.g)⇒IFIC=IBIE⇒IB.IC=IF.IE(∗){I^chungIFB^=ICE^⇒ΔIFB∼ΔICE(g.g)⇒IFIC=IBIE⇒IB.IC=IF.IE(∗)

Lại có:

ˆIFD=ˆIFB+ˆBFD=ˆICE+ˆBCA=2ˆBCA(1)IFD^=IFB^+BFD^=ICE^+BCA^=2BCA^(1)

Mà ΔBEC;ˆE=900;OΔBEC;E^=900;O là trung điểm của BCBC

⇒OB=OC=OE=12BC⇒OB=OC=OE=12BC

⇒ΔCOE⇒ΔCOE cân ở OO

⇒ˆBOE=2ˆOCE⇒ˆIOE=2ˆBCA(2)⇒BOE^=2OCE^⇒IOE^=2BCA^(2)

Từ (1),(2)⇒ˆIFD=ˆIOE(1),(2)⇒IFD^=IOE^

Khi đó:

{ˆIchungˆIFD=ˆIOE⇒ΔIFD∼ΔIOE(g.g)⇒IFIO=IDIE⇒IO.ID=IF.IE(∗∗){I^chungIFD^=IOE^⇒ΔIFD∼ΔIOE(g.g)⇒IFIO=IDIE⇒IO.ID=IF.IE(∗∗)

Từ (∗),(∗∗)⇒IO.ID=IB.IC(∗),(∗∗)⇒IO.ID=IB.IC

học tốt

a) Ta có: \(a^2+b^2+c^2=\frac{1}{2}\left(2a^2+2b^2+2c^2\right)\)

\(=\frac{1}{2}\left[\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\right]\ge\frac{1}{2}\left(2ab+2bc+2ca\right)\)

\(=\frac{1}{2}\cdot2\left(ab+bc+ca\right)=ab+bc+ca\)

\(\Rightarrow ab+bc+ca\le a^2+b^2+c^2\le3\)

Dấu "=" xảy ra khi: a = b = c = 1

b) \(P=\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\ge\frac{\left(1+1+1\right)^2}{1+ab+1+bc+1+ca}\) (Cauchy cộng mẫu)

\(=\frac{9}{3+\left(ab+bc+ca\right)}\ge\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

Ta có: \(A=\frac{1}{x}+\frac{1}{y}+x^2+y^2\)

\(\ge\frac{\left(1+1\right)^2}{x+y}+\frac{\left(x+y\right)^2}{2}=\frac{2^2}{1}+\frac{1^2}{2}=\frac{9}{2}\)

Dấu "=" xảy khi: \(x=y=\frac{1}{2}\)

Ta có: \(a^4+b^4+c^4=\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{\left(a+b+c\right)^4}{27}=\frac{3^4}{3^3}=3\)

Dấu "=" xảy ra khi: a = b = c = 1

tại s lại là \(\frac{\left(a+b+c\right)^4}{27}\)

\(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\left(x,y,z\ne0\right)\).

Ta có:

\(a+b+c=0\).

Ta phải chứng minh rằng nếu \(a+b+c=0\)thì \(a^3+b^3+c^3=3abc\).

Thật vậy, xét hiệu  \(A=a^3+b^3+c^3-3abc\)với \(a+b+c=0\).

\(A=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\).

\(A=\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]\).

\(A=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\).

\(A=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2-3ab\right)\).

\(A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

\(A=0\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(vì \(a+b+c=0\)).

Do đó \(a^3+b^3+c^3-3abc=0\).

\(\Rightarrow a^3+b^3+c^3=3abc\)với \(a+b+c=0\)

\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)với \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)(điều phải chứng minh).

Đặt \(A=\sqrt{8}+3\)   

\(B=6+\sqrt{2}\)

\(A-B=\sqrt{8}+3-\left(6+\sqrt{2}\right)\)   

\(=2\sqrt{2}+3-6-\sqrt{2}\)   

\(=\sqrt{2}-3\)   

\(=\sqrt{2}-\sqrt{9}< 0\)

Vậy A < B