Áp dụng định lý PYTAGO vào tam giác ABC có

BC^2=AB^2+AC^2= 9^2+12^2=225

=>BC= 15

Sabc= 1/2.AB.AC = 54 mà Sabc = 1/2.AH.BC 

=>1/2.AH = Sabc: BC = 3.6=> AH =7,2

mình ghi thiếu đề ạ:

Cho P=...  

\(P=\left(\frac{x^2+x-4}{x^2-2x-3}\right):\left(1-\frac{x-3}{x-2}\right)\)

\(=\frac{x^2+x-4}{\left(x+1\right)\left(x-3\right)}:\left(\frac{x-2-x+3}{x-2}\right)\)

\(=\frac{x^2+x-4}{\left(x+1\right)\left(x-3\right)}.\frac{x-2}{1}=\frac{\left(x^2+x-4\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}\)

Tạm thời mình làm câu bất trước :)) Các câu Đại còn lại để tối mình làm nhé ( chiều mình bận với mình không giỏi Hình lắm )

Câu 6. Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

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

Lại có \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{1}=4\)(2)

Từ (1) và (2) => \(A\ge\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)=> \(A\ge18\)

Đẳng thức xảy ra <=> x = y = 1/2

Vậy MinA = 18

Cau 2

1. TA CO :\(\frac{1}{A}\)+\(\frac{1}{B}\)+\(\frac{1}{c}\)=\(\frac{1}{2019}\)
2. <=>\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=\(\frac{1}{a+b+c}\)
3. <=>(\(\frac{1}{a}\)+\(\frac{1}{b}\))+(\(\frac{1}{c}\)-\(\frac{1}{a+b+c}\)=0
4. <=>\(\frac{a+b}{ab}\)+\(\frac{a+b}{c\left(a+b+c\right)}\)=0
5. <=>(a+b)(\(\frac{1}{ab}\)+\(\frac{1}{c\left(a+b+c\right)}\))=0
6. <=>\(\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}\)=0
7. <=>\(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
8.  
9. gia su\(\left(a+b\right)=0\)=>c=2016 khi do
10. \(\frac{a^{2019}+b^{2019}}{\left(ab\right)^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019+c^{2019}}}\)
11. cac truong hop kia tuong tu
12.  

1. \(\left(a+b+c\right)^2=a^2+b^2+c^2\)
2. <=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
3. ,=>\(2\left(ab+bc+ac\right)=0\)
4. <=>\(ab+bc+ac=0\)
5. <=>\(ab=-\left(bc+ac\right),bc=-\left(ab+ac\right),ac=-\left(ab+bc\right)\)
6. voi \(ab=-\left(bc+ac\right)\)=>\(c^2+2ab=c^2+ab-bc-ac=\left(a-c\right)\left(b-c\right)\)
7. tuong tu \(a^2+2bc=\left(a-c\right)\left(a-b\right)\)
8.               \(b^2+2ac=-\left(b-c\right)\left(a-b\right)\)
9. =>\(P=\frac{a^2}{\left(a-c\right)\left(a-b\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)\(=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Cosi 2 số dương \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}\frac{y}{x}}=2\)

Dấu ''='' xảy ra <=> x = y 

Ta có: \(x^2-3x+5=\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{11}{4}\)

                                    \(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}>0\forall x\)

mà \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy.......

Ta có : \(x^2-3x+5=0\)

\(\Leftrightarrow\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-\frac{11}{4}\)(Vô lí , do \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\))

Vậy PT vô nghiệm