Câu 15: (mãi mới nghĩ ra :v)

\(\dfrac{\left(a+b\right)^2}{ab}+\dfrac{\left(b+c\right)^2}{bc}+\dfrac{\left(c+a\right)^2}{ca}\ge9+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow\dfrac{a^2+2ab+b^2}{ab}+\dfrac{b^2+2bc+b^2}{bc}+\dfrac{c^2+2ca+a^2}{ca}\ge9+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow\dfrac{a}{b}+2+\dfrac{b}{a}+\dfrac{b}{c}+2+\dfrac{c}{b}+\dfrac{c}{a}+2+\dfrac{a}{c}\ge9+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{c}\ge3+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge3+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

-Áp dụng BĐT Caushy Schwarz ta có:

\(\left\{{}\begin{matrix}\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1\right)^2}{b+c}=\dfrac{4}{b+c}\\\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{\left(1+1\right)^2}{c+a}=\dfrac{4}{c+a}\\\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\end{matrix}\right.\)

-Từ đó suy ra: \(a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\)

-Ta c/m rằng: \(\dfrac{4a}{b+c}+\dfrac{4b}{c+a}+\dfrac{4c}{a+b}\ge3+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge3+2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)

\(\Leftrightarrow2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge3\)

\(\Leftrightarrow2\left(\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1-3\right)\ge3\)

\(\Leftrightarrow2\left(\dfrac{a+b+c}{b+c}+\dfrac{b+c+a}{c+a}+\dfrac{c+a+b}{a+b}\right)-6\ge3\)

\(\Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge9\left(1\right)\)

-Áp dụng BĐT Caushy Schwarz cho VT của BĐT ta được:

\(2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge2\left(a+b+c\right)\left(\dfrac{\left(1+1+1\right)^2}{a+b+b+c+c+a}\right)=2\left(a+b+c\right)\dfrac{9}{2\left(a+b+c\right)}=9\)

\(\Rightarrow\)BĐT (1) đúng.

\(\Rightarrowđpcm\)

-Dấu "=" xảy ra khi \(a=b=c\)

ủa đây là dạng toán lớp 6 mà

\(d\left(G;\left(ABCD\right)\right)=\dfrac{1}{3}d\left(S;\left(ABCD\right)\right)=\dfrac{1}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{6}\)

\(S_{\Delta ACD}=\dfrac{1}{2}S_{ABCD}=\dfrac{a^2}{2}\)

\(\Rightarrow V=\dfrac{1}{3}.\dfrac{a^2}{2}.\dfrac{a\sqrt{3}}{6}=\dfrac{a^3\sqrt{3}}{36}\)

Bạn gõ câu hỏi lên nhé, quy định là không được gửi câu hỏi dạng hình ảnh.

1: Ta có: \(A=25x^4-24x^2-1\)

\(=25x^4-25x^2+x^2-1\)

\(=\left(x^2-1\right)\left(25x^2+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(25x^2+1\right)\)

2: Ta có: \(A=64x^4+63x^2-1\)

\(=64x^4+64x^2-x^2-1\)

\(=\left(x^2+1\right)\left(64x^2-1\right)\)

\(=\left(x^2+1\right)\left(8x-1\right)\left(8x+1\right)\)

3: Ta có: \(A=x^4-15x^2+50\)

\(=x^4-5x^2-10x^2+50\)

\(=\left(x^2-5\right)\left(x^2-10\right)\)

4: Ta có: \(A=-10x^4+9x^2+1\)

\(=-10x^4+10x^2-x^2+1\)

\(=\left(x^2-1\right)\left(-10x^2-1\right)\)

\(=-\left(10x^2+1\right)\left(x-1\right)\left(x+1\right)\)

Bài 12: 

a: Xét ΔABM và ΔACN có

AB=AC

\(\widehat{ABM}=\widehat{ACN}\)

BM=CN

Do đó: ΔABM=ΔACN

\(\Delta'=4-\left(m-1\right)=5-m\)

để pt có nghiệm kép khi \(5-m=0\Leftrightarrow m=5\)

chọn B 

Phương trình có nghiệm kép khi:

\(\Delta'=4-\left(m-1\right)=0\Leftrightarrow5-m=0\)

\(\Rightarrow m=5\)

a Xét tứ giác BCEF có 

\(\widehat{BFC}=\widehat{BEC}=90^0\)

Do đó:BCEF là tứ giác nội tiếp

b: Xét ΔABE vuông tại E và ΔACF vuông tại F có

\(\widehat{BAE}\) chung

DO đó: ΔABE\(\sim\)ΔACF

Suy ra: AB/AC=AE/AF

hay \(AB\cdot AF=AE\cdot AC\)