Đúng như đã hứa nhé:

\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}+1}}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}+1\right)}{\sqrt{3+1}-1}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}-1\right)}{\sqrt{3+1}-1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3+1}}+1-\sqrt{\sqrt{3+1}}+1\right)}{2-1}=2\sqrt{3}\)

đề đúng hông?? Nhìn hơi lạ.....

\(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4\)

Kl: Min_A = 4

\(B=x^2-x+1=\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

kl:.......

\(C=5x^2+5x+1=5\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+1-\dfrac{5}{4}=5\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

kl:.......

\(D=3x^2+4x+2=3\left(x^2+2\cdot\dfrac{2}{3}x+\dfrac{4}{9}\right)+2-\dfrac{4}{3}=3\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)

kl:......

\(E=\dfrac{1}{2}\cdot x^2+x-1=\dfrac{1}{2}\left(x^2+2x+1\right)-1-\dfrac{1}{2}=\dfrac{1}{2}\left(x+1\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}\)

kl:............

\(F=\dfrac{1}{9}x^2+3x+2=\dfrac{1}{3}\left(x^2+2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+2-\dfrac{1}{12}=\dfrac{1}{3}\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\)

kl:..........

a) đa thức

\(=\left(x-3\right)\left(x+3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+6x+9-x^2-3x-9\right)=3x\left(x-3\right)=3x^2-9x\)

b) đa thức \(=\left(x+6\right)^2-2x\left(x+6\right)+x^2-36=\left(x+6-x\right)^2-36=0\)

áp dụng tính chất của dãy tỉ số bằng nhau,ta có:

a/b = b/c = c/d = d/a => a+b+c+d/b+c+d+a = 1

khi đó : a/b =1 =>a=b ;b/c =1 => b=c ;c/d =1 =>c=d ;d/a =1 =>d=a

=>a=b=c=d

Câu hỏi tương tự nha bạn !