\(\left(xy+1\right)^2-\left(x+y\right)^2\)

\(\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)

\(x-\frac{2}{4}-\frac{2}{3}\ge5x-\frac{9}{12}\)

\(\Leftrightarrow x-\frac{7}{6}\ge5x-\frac{3}{4}\)

\(\Leftrightarrow-4x\ge\frac{5}{12}\)

\(\Leftrightarrow-\frac{5}{56}\ge x\)

\(2\sqrt{b+c-4}\le\frac{4+b+c-4}{2}=\frac{b+c}{2}\Rightarrow\frac{a}{\sqrt{b+c-4}}\ge\frac{4a}{b+c}\)

Tương tự ta cũng có: \(\frac{b}{\sqrt{a+c-4}}\ge\frac{4b}{a+c},\frac{c}{\sqrt{a+b-4}}\ge\frac{4c}{a+b}\).

Bất đẳng thức cần chứng minh sẽ đúng nếu ta chứng minh được: 

\(\frac{4a}{b+c}+\frac{4b}{a+c}+\frac{4c}{a+b}\ge6\)

\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)

(đúng, theo bất đẳng thức Nesbitt) 

Do đó ta có: \(\frac{a}{\sqrt{b+c-4}}+\frac{b}{\sqrt{a+c-4}}+\frac{c}{\sqrt{a+b-4}}\ge6\)

Dấu \(=\)khi \(a=b=c=2\).

Nhân phá ra thôi bn 

\(\left(x+1\right)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\) 

\(=x^7+x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(=x^7+1\)

\(\left(x+1\right)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)\)

\(=\)\(x^7-x^6+x^5-x^4+x^3-x^2+x+x^6-x^5+x^4-x^3+x^2-x+1\)

\(=\)\(x^7+1\)

\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)\)

Ta có x⁴ + x³ - x² + ax + b = (x² + x - 2)(x² + cx + d)

<=>  x⁴ + x³ - x² + ax + b = (x - 2)(x + 1)(x² + cx + d)

=> x = 2 là nghiệm phương trình 

=> 2⁴ + 2³ - 2² + 2a + b = 0

<=> 2a + b = -20 (1)

x = -1 là nghiệm phương trình 

(-1)⁴ + (-1)³ - (-1)² -a + b = 0  

<=> -a + b = 1 (2)

Từ (1) và (2) => a = -7 ; b = -6

Khi đó x⁴ + x³ - x² - 7x - 6 = (x² + x - 2)(x² + cx + d) 

<=> x³(x + 1) - (x + 1)(x + 6) = (x + 1)(x - 2)(x² + cx + d)

<=> (x + 1)(x³ - x - 6) = (x + 1)(x - 2)(x² + cx + d) 

<=> (x + 1)(x - 2)(x² + 2x + 3) = (x + 1)(x - 2)(x² + cx + d) 

<=> x² + 2x + 3 = x² + cx + d

=> c = 2 ; d = 3

Vậy a = -7 ; b = -6 ; c = 2 ; d = 3

\(\sqrt{8-4\sqrt{3}}+\sqrt{8+4\sqrt{3}}\)   

\(=\sqrt{6-4\sqrt{3}+2}+\sqrt{6+4\sqrt{3}+2}\)   

\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{6}\right)^2+2\cdot\sqrt{6}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)   

\(=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)   

\(=|\sqrt{6}-\sqrt{2}|+|\sqrt{6}+\sqrt{2}|\)   

\(=\sqrt{6}-\sqrt{2}+\sqrt{6}+\sqrt{2}\)   

\(=2\sqrt{6}\)

\(39,\sqrt{6-4\sqrt{2}}-\sqrt{7-4\sqrt{3}}\)

\(\sqrt{2^2-4\sqrt{2}+\sqrt{2}^2}-\sqrt{2^2-4\sqrt{3}+\sqrt{3}^2}\)

\(\sqrt{\left(2-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(2-\sqrt{2}-2-\sqrt{3}\)

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

\(40,\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)

(x -1)x² - 4x(x - 1) + 4(x - 1) 

= (x - 1)x - 4(x - 1)² 

= (x - 1)[(x - 4(x - 1)]

= (x - 1)(-3x + 4)

Thay x = 3 vào biểu thức : 

(3 - 1)(-3.3 + 4) = 2.(-5) = -10

Bằng 2 nha bạn

Ta có: \(\frac{x^2\left(y+z\right)}{yz}+\frac{y^2\left(z+x\right)}{zx}+\frac{z^2\left(x+y\right)}{xy}\)

\(=\frac{x^2}{z}+\frac{x^2}{y}+\frac{y^2}{z}+\frac{y^2}{x}+\frac{z^2}{x}+\frac{z^2}{y}\)

\(\ge\frac{\left(2x+2y+2z\right)^2}{2\left(x+y+z\right)}=\frac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}=2\)

Dấu "=" xảy ra khi: x = y = z = 1/3

Ta có: \(\frac{1}{x}+\frac{2}{y}=\frac{1}{x}+\frac{1}{y}+\frac{1}{y}\)

\(\ge\frac{\left(1+1+1\right)^2}{x+y+y}=\frac{3^2}{x+2y}=\frac{9}{3}=3\)

Dấu "='' xảy ra khi: x = y = 1