\(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{16}\)

Ta có: \(\frac{1}{a^2+b^2}+\frac{2}{ab}+ab\)

 \(=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{3}{2ab}+384ab-383ab\)

\(\ge\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{3}{2ab}.384ab}-383.\frac{1}{16}\)

\(=\frac{4}{\left(a+b\right)^2}+2.24-\frac{383}{16}=\frac{641}{16}\)

Dấu "=" xảy ra <=> a = b = 1/4

\(x^3-4x^2-8x+8=\left(x^2-6x+4\right)\left(x+2\right)\)

Hc tốt 

Trả lời:

\(x^3-4x^2-8x+8\)

\(=x^3+2x^2-6x^2-12x+4x+8\)

\(=x^2.\left(x+2\right)-6x.\left(x+2\right)+4.\left(x+2\right)\)

\(=\left(x+2\right).\left(x^2-6x+4\right)\)

Trả lời:

\(\left(x+2y-3\right)-4.\left(x+2y-3\right)+4\)

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

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

Bài làm:

Ta có: \(\frac{x-1}{3}+\frac{x-3}{4}=2\)

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

\(\Leftrightarrow\frac{7}{12}x=\frac{37}{12}\)

\(\Leftrightarrow x=\frac{37}{12}\div\frac{7}{12}\)

\(\Rightarrow x=\frac{37}{7}\)

\(\frac{x-1}{3}+\frac{x-3}{4}=2\)

\(\Leftrightarrow\frac{4\left(x-1\right)}{12}+\frac{3\left(x-3\right)}{12}=\frac{24}{12}\)

\(\Leftrightarrow4\left(x-1\right)+3\left(x-3\right)=24\)

\(\Leftrightarrow4x-4+3x-9=24\)

\(\Leftrightarrow7x-13=24\)

\(\Leftrightarrow7x=37\)

\(\Leftrightarrow x=\frac{37}{7}\)

Bài làm:

Ta có: \(x=-9\Leftrightarrow-10=x-1\Rightarrow10=1-x\)nên thay vào ta tính:

\(P\left(-9\right)=1+\left(1-x\right)x+\left(1-x\right)x^2+\left(1-x\right)x^3+...+\left(1-x\right)x^{19}+\left(1-x\right)x^{20}\)

\(P\left(-9\right)=1+x-x^2+x^2-x^3+x^3-x^4+...+x^{20}-x^{21}\)

\(P\left(-9\right)=1+x-x^{21}\)

\(P\left(-9\right)=1-9+9^{21}\)

\(P\left(-9\right)=9^{21}-8\)

Vậy khi \(x=-9\)thì \(P\left(x\right)=9^{21}-8\)

Học tốt!!!!

Xem lại đề bài đi. Đó có phải là bài toán không?

thieu de ban oi 

a, de phuong trinh tren co nghia thi \(3x-9\ge0\)

\(3x\ge9< =>x\ge3\)

b, de phuong trinh tren co nghia thi \(5-10x\ge0\)

\(< =>10x\le5\)\(< =>x\le\frac{1}{2}\)

c, de phuong trinh tren co nghia thi \(\frac{3}{2x+1}\ge0\)(DK: x khac -1/2)

\(< =>2x+1\ge0\)\(< =>x>-\frac{1}{2}\)

d, de phuong trinh tren co nghia thi \(\frac{2x-4}{3}\ge0\)

\(< =>2x-4\ge0\)\(< =>x\ge2\)

e, de phuong trinh tren co nghia thi \(\frac{x^2}{2x-3}\)

do \(x^2\ge\)suy ra \(2x-3\ge0\)

\(< =>2x\ge3\)\(< =>x\ge\frac{3}{2}\)