giúp e vs

We subtitute \(ab+bc+ca=1\) into \(a^2+1\). We have: \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)\)\(=\left(a+b\right)\left(a+c\right)\)

Similarly, we have \(b^2+1=\left(a+b\right)\left(b+c\right)\) and \(c^2+1=\left(a+c\right)\left(b+c\right)\)

From these, we have \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)\(=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Thus, we must have \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}=\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)\(=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)

Because both \(a,b,c\) are rational numbers, \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) must be a rational number. Therefore, \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\) is also a rational number.

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)

\(=\sqrt{\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)}\)

\(=\sqrt{\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(b+c\right)+a\left(b+c\right)\right]\left[c\left(c+a\right)+b\left(c+a\right)\right]}\)

\(=\sqrt{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}\)

\(=\sqrt{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}=\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\)

Do \(a,b,c\) là các số hữu tỉ nên \(\left|\left(a+b\right)\left(b+c\right)\left(c+a\right)\right|\) là số hữu tỉ.

\(\Rightarrowđpcm\)

\(a,b\ne0\)

\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\dfrac{b^2\left(a+b\right)^2+a^2\left(a+b\right)^2+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\dfrac{\left(a^2b^2+2ab^3+b^4\right)+\left(a^4+2a^3b+a^2b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\dfrac{\left(a^4+2a^2b^2+b^4\right)+2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a+b\right)^2}}\)

\(=\sqrt{\dfrac{\left(a^2+b^2+ab\right)^2}{a^2b^2\left(a+b\right)^2}}=\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\)

Do \(a,b\) là số hữu tỉ nên \(\left|\dfrac{a^2+b^2+ab}{ab\left(a+b\right)}\right|\) cũng là số hữu tỉ.

\(\Rightarrowđpcm\)

234 x456 =

234 x456 = 106704

a/

Ta có

\(\widehat{BEC}=\widehat{BDC}=90^o\) => E và D cùng nhìn BC dưới 1 góc \(90^o\)

=> E và D cùng nằm trên đường tròn đường kính BC => BCDE là tứ giác nội tiếp

b/

Ta có H là trực tâm cuat tg ABC \(\Rightarrow AH\perp BC\)

Xét tg vuông ADH và tg vuông BCD có

\(\widehat{DAH}=\widehat{DBC}\) (cùng phụ với \(\widehat{ACB}\) )

=> tg ADH đồng dạng với tg BCD (g.g.g)

\(\Rightarrow\dfrac{DA}{DB}=\dfrac{DH}{DC}\Rightarrow DA.DC=DH.DB\)

c/

Gọi Q là giao của AO với (O)

Gọi I là giao của AO với ED

Gọi K là giao của AO với CE

Xét (H) có

\(HD\perp AN\) => DA=DN (trong đường tròn đường thẳng đi qua tâm và vuông góc với dây cung thì chia đôi dây cung )

\(HE\perp AM\) => EA=EM (lý do như trên)

Xét tg AMN có

\(\dfrac{DA}{DN}=\dfrac{EA}{EM}=1\) => ED//MN (talet đảo trong tam giác)

Xét tứ giác nt BCDE có

\(\widehat{DEC}=\widehat{DBC}\) (góc nội tiếp cùng chắn cung DC) (1)

Xét (O) có

\(\widehat{ACB}=\widehat{AQB}\) (góc nt cùng chắn cung AB) (2)

Ta có \(\widehat{ABQ}=90^o\) (góc nội tiếp chắn nửa đường tròn)

\(\Rightarrow QB\perp AB\) mà \(CE\perp AB\) => CE//QB \(\Rightarrow\widehat{AKE}=\widehat{AQB}\) (góc đồng vị) (3)

Từ (2) và (3) \(\Rightarrow\widehat{AKE}=\widehat{ACB}\) (4)

Xét tg vuông BCD có

\(\widehat{ACB}+\widehat{DBC}=90^o\) (5)

Từ (1) (4) (5) \(\Rightarrow\widehat{AKE}+\widehat{DEC}=90^o\)

Xét tg IKE có

\(\widehat{AKE}+\widehat{DEC}=90^o\Rightarrow\widehat{EIK}=90^o\Rightarrow AO\perp ED\)

Mà ED//MN (cmt)

\(\Rightarrow AO\perp MN\)

Draw your own figure. Now I'll give the solution:

a) Because BD and CE are the heights of the triangle ABC \(\left(D\in AC;E\in AB\right)\), we have \(\widehat{BEC}=\widehat{BDC}=90^o\)

Consider the quadrilateral BCDE, it has 2 adjacent vertices D, E which both look at the edge BC by a right angle. Thus, BCDE is an inscribed quadrilateral. And that's what we must prove!

b) We can easily have \(\widehat{EBD}=\widehat{ECD}\) due to the inscribed quadrilateral BCDE or \(\widehat{ABD}=\widehat{HCD}\)

Consider the 2 triangles DAB and DHC, which are both right at D, have \(\widehat{ABD}=\widehat{HCD}\). Therefore, \(\Delta DAB~\Delta DHC\left(a.a\right)\). This means \(\dfrac{DA}{DH}=\dfrac{DB}{DC}\) or \(DA.DC=DH.DB\) and again, that's what we must prove!

c) Draw the tangent Ax of (O). We have \(Ax\perp OA\) (at A)

Consider the circle (O), it has \(\widehat{BAx}\) is an angle that is formed by the tangent line Ax and the chord AB. Also, \(\widehat{ACB}\) is the inscribed angle intercept the arc AB. Therefore, \(\widehat{BAx}=\widehat{ACB}\)

On the other hand, \(\widehat{ACB}=\widehat{AED}\) due to the inscribed quadrilateral BCDE. Thus, we must have \(\widehat{BAx}=\widehat{AED}\), which means \(Ax//DE\) (because the 2 staggered angles are equal). We have already prove \(Ax\perp OA\), so, \(OA\perp DE\)

Consider the circle (H), we have \(HE\perp AM\) at E while AM is a chord of (H). Therefore, E is the midpoint of AM.

Similarly, D is the midpoint of AN.

Consider the triangle AMN, it has D, E, consecutively, are the midpoint of AN, AM. Thus, DE must be the average line of the triangle AMN. This means \(DE//MN\) 

Guess what? We've already had \(OA\perp DE\), so, \(OA\perp MN\), and that's what we must prove!

d) Sorry, I haven't had the solution for this yet.