A=x³+y³=(x+y)(x²-xy+y²)

=(x+y)²\(\ge\)0

Dấu "=" xảy ra khi x=-y

\(x^2-x-4=2\sqrt{\left(x-1\right)\left(1-x\right)}\)

\(\Leftrightarrow\left(x^2-x-4\right)^2=\left[2\sqrt{\left(x-1\right)\left(1-x\right)}\right]^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+16=8x-4x^2-4\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+16+4=8x-4x^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+20=8x-4x^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+20+4x^2=8x\)

\(\Leftrightarrow x^4-2x^3-3x^2+8x+20=8x\)

\(\Leftrightarrow x^4-2x^3-3x^2+8x+20-8x=0\)

\(\Leftrightarrow x^4-2x^3-3x^2+20=0\)

Vậy: phương trình vô nghiệm

Có gì không hiểu ib

#)Giải :

Áp dụng BĐT Cauchy :

\(\left(ab+c\right)\left(bc+a\right)\le\left(\frac{ab+c+bc+a}{2}\right)^2=\frac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)

Tương tự với các cặp còn lại, ta được :

\(\left(bc+a\right)\left(ca+b\right)\le\frac{\left(c+1\right)^2\left(a+b\right)^2}{4}\)

\(\left(ab+c\right)\left(ca+b\right)\le\frac{\left(a+1\right)^2\left(b+c\right)^2}{4}\)

Nhân theo vế :

\(\left[\left(ab+c\right)\left(ca+b\right)\left(bc+a\right)\right]^2\le\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\frac{\left[\left(a+1\right)\left(b+1\right)\left(c+1\right)\right]^2}{64}\)

Mà : \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\frac{a+1+b+1+c+1}{3}\right)^3=8\)

Do đó \(\left[\left(ab+c\right)\left(ac+b\right)\left(bc+a\right)\right]^2\le\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2.\frac{8^2}{64}\)

Từ đó suy ra \(\left(ab+c\right)\left(ca+b\right)\left(bc+a\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\Rightarrowđpcm\)

ai giúp mk với