\(3\left(x^2-x+1\right)^2-2\left(x+1\right)^2=5.\)\(\left(x^3+1\right)\)

\(\Leftrightarrow3\left(x^2-x+1\right)^2-2\left(x+1\right)^2=5\left(x+1\right)\left(x^2-x+1\right)\)

Đặt \(x+1=a,x^2-x+1=b\), phương trình trở thành:

\(3b^2-2a^2=5ab\) 

\(\Leftrightarrow3b^2-5ab-2a^2=0\)

\(\Leftrightarrow\)\(\left(3b+a\right)\left(b-2a\right)=0\)

\(\Leftrightarrow\left[3\left(x^2-x+1\right)+x+1\right]\left[x^2-x+1-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(3x^2-2x+4\right)\left(x^2-3x-1\right)=0\)

Vì \(3x^2-2x+4=\left(x-1\right)^2+2x^2+3>0\forall x\)nên:

\(x^2-3x-1=0:\left(3x^2-2x+4\right)\)

\(\Leftrightarrow x^2-3x-1=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{13}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{13}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{13}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}}}\)

Vậy phương trình có tập nghiệm: \(S=\left\{\frac{3\pm\sqrt{13}}{2}\right\}\)

\(2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\)\(\left(x^3-1\right)\)

\(\Leftrightarrow2\left(x^2+x+1\right)^2-7\left(x-1\right)^2=13\left(x-1\right)\left(x^2+x+1\right)\)

Đặt \(x-1=a,x^2+x+1=b\), phương trình trở thành:

\(2b^2-7a^2=13ab\)\(x=4\)

\(\Leftrightarrow2b^2-13ab-7a^2=0\)

\(\Leftrightarrow\left(b-7a\right)\left(a+2b\right)=0\)

\(\Leftrightarrow\left[x^2+x+1-7\left(x-1\right)\right]\left[x-1+2\left(x^2+x+1\right)\right]=0\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(2x^2+3x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(2x+1\right)\left(x+1\right)=0\)

-Xét các trường hợp sau:

+Với \(x-2=0\Leftrightarrow x=2\)

+Với \(x-4=0\Leftrightarrow x=4\)

+Với \(x+1=0\Leftrightarrow x=-1\)

+Với \(2x+1=0\Leftrightarrow x=-0,5\)

Vậy phương trình có tập nghiệm: \(S=\left\{-1;-0,5;2;4\right\}\)

\(x^4-9x^2+24x-16=\)\(0\)

\(\Leftrightarrow x^4-\left(9x^2-24x+16\right)=0\)

\(\Leftrightarrow x^4-\left(3x-4\right)^2=0\)

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

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]=0\)

Vì \(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)nên:

\(\left(x+4\right)\left(x-1\right)=0:\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

Vậy phương trình có tập nghiệm \(S=\left\{1;-4\right\}\)

\(x^4=6x^2+12x+\)\(8\)

\(\Leftrightarrow x^4-2x^2+1=4x^2+12x+9\)

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

\(\Leftrightarrow|x^2-1|=|2x+3|\)\(|\)

xét các trường hợp:

- Trường hợp 1:

\(x^2-1=2x+3\)

\(\Leftrightarrow x^2-1-2x-3=0\)

\(\Leftrightarrow x^2-2x-4=0\)

\(\Leftrightarrow\left(x-1\right)^2-5=0\Leftrightarrow\left(x-1\right)^2=5\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}}\)

-Trường hợp 2:

\(x^2-1=-2x-3\)

\(\Leftrightarrow x^2-1+2x+3=0\)

\(\Leftrightarrow x^2+2x+2=0\)

\(\Leftrightarrow\left(x+1\right)^2+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=-1\left(vn\right)\)(vô nghiệm)

Vậy phương trình có tập nghiệm: \(S=\left\{1\pm\sqrt{5}\right\}\)

* Vì \(a,b\ge1\)nên \(\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab+1\ge a+b\)

Một cách tương tự: \(bc+1\ge b+c;ca+1\ge c+a\)

Với mọi số thực \(a\ge1\) ta luôn có: \(\left(a-1\right)^2\ge0\Leftrightarrow a^2\ge2a-1\Leftrightarrow\frac{1}{2a-1}\ge\frac{1}{a^2}\)

Tương tự: \(\frac{1}{2b-1}\ge\frac{1}{b^2};\frac{1}{2c-1}\ge\frac{1}{c^2}\)

Từ đó suy ra \(VT\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{4ab}{ab+1}+\frac{4bc}{bc+1}+\frac{4ca}{ca+1}\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+4-\frac{4}{ab+1}+4-\frac{4}{bc+1}+4-\frac{4}{ca+1}\)\(\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{4}{ab+1}-\frac{4}{bc+1}-\frac{4}{ca+1}+12\)\(\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}-\frac{4}{a+b}-\frac{4}{b+c}-\frac{4}{c+a}+12\)\(=\left(\frac{2}{a+b}-1\right)^2+\left(\frac{2}{b+c}-1\right)^2+\left(\frac{2}{c+a}-1\right)^2+9\ge9\)

Đẳng thức xảy ra khi a = b = c = 1

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