Lời giải:

Gọi B(a,b)B(a,b) và C(c,d)C(c,d)

Ta có HA−→−=(0,4)⊥BC−→−=(c−a,d−b)⇒4(d−b)=0→b=dHA→=(0,4)⊥BC→=(c−a,d−b)⇒4(d−b)=0→b=d

Thay d=bd=b:

HB−→−=(a−1,b−2)⊥AC−→−=(c−1,b−6)HB→=(a−1,b−2)⊥AC→=(c−1,b−6)

⇒(a−1)(c−1)+(b−2)(b−6)=0⇒(a−1)(c−1)+(b−2)(b−6)=0

Lại có IA2=IB2=IC2↔{(a−2)2+(b−3)2=10(c−2)2+(b−3)2=10IA2=IB2=IC2↔{(a−2)2+(b−3)2=10(c−2)2+(b−3)2=10

⇒(a−2)2=(c−2)2→a+c=4⇒(a−2)2=(c−2)2→a+c=4 ( a≠ca≠c )

Ta thu được

{(a−2)2+(b−3)2=10(3−a)(a−1)+(b−2)(b−6)=0{(a−2)2+(b−3)2=10(3−a)(a−1)+(b−2)(b−6)=0

{a2+b2−4a−6b+3=0−a2+4a+b2−8b+9=0⇒2b2−14b+12=0→b=1{a2+b2−4a−6b+3=0−a2+4a+b2−8b+9=0⇒2b2−14b+12=0→b=1

hoặc b=6b=6

Thay vào PT suy ra [−a2+4a+2=0−a2+4a−3=0⇒[a=2+6–√a=1;a=3[−a2+4a+2=0−a2+4a−3=0⇒[a=2+6a=1;a=3

Vậy.....

Ta có

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

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy x^2+x+1 k có nghiệm

Ta có

\(x^2+x+1>0\)

\(\Rightarrow x^2+x+2>0\)

Vậy....

Bài 1:

a) Biến đổi \(f\left(x\right)\), ta có:

\(f\left(x\right)=x^2+x+2\)

\(=x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{7}{4}\)

\(=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{7}{4}\)

\(=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{7}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

\(\Rightarrow\forall x\) ta có \(f\left(x\right)\ne0\)

Vậy \(f\left(x\right)\) không có nghiệm

b) Tương tự

\(\frac{4x+1}{4x^2+2}=\frac{-\left(2x-1\right)^2}{4x^2+2}+1\le1\)

Dấu bằng xảy ra

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy................

đề là cân bằng pt hay j v

Ta có :

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

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

Mà \(a+b+c\ne0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

\(\Rightarrow\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

\(x^2+y^2+z^2+2x-4y-6z+14\)

\(=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2-6z+9\right)\)

\(=\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\)

Vì \(\left(x+1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\); \(\left(z-3\right)^2\ge0\forall z\)

\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

hay \(x^2+y^2+z^2+2x-4y-6z+14\ge0\)\(\forall x,y,z\)