= 1 - 4x² - ( x² - 4)

= 1 - 4x² - x² + 4

= -5x² + 5

= - ( 5x² - 5 )

= - 5 (x² - 1)

a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)

\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)

\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)

b) \(3x\left(5-x\right)+4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

c) \(x\left(x-3\right)+4x-12=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

d) \(x^2-36=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

e) \(x^2+3x+1=2\)

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

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

\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)

Còn lại ........... Tự lm nất nha 

cách dễ nhất để chứng minh nesbitt

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ab}+\frac{b^2}{ba+bc}+\frac{c^2}{ca+cb}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}}=\frac{3}{2}\)

Dấu = xảy ra \(< =>a=b=c\)

Có thể dùng AM-GM ^^ 

Bổ sung điều kiện a,b,c dương

Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(=\left(\frac{a}{b+c}+\frac{b+c}{b+c}\right)+\left(\frac{b}{c+a}+\frac{c+a}{c+a}\right)+\left(\frac{c}{a+b}+\frac{a+b}{a+b}\right)-3\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(\ge\frac{1}{2}\cdot3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot\frac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}-3\)( AM-GM )

\(=\frac{1}{2}\cdot9-3=\frac{3}{2}\)

Đẳng thức xảy ra <=> a = b = c

=> đpcm

(x+3)² =x²+6x+9

(x-1/2)(x+1/2)=x²-1/2

Câu 2 phải là thế này nhé: (x-1/2)(x+1/2)=x²-1/4

PTĐTTNT :

a) 3x² - 6x = 3x( x - 2 )

b) x² - 2x + 1 - y² = ( x² - 2x + 1 ) - y² = ( x - 1 )² - y² = ( x - y - 1 )( x + y - 1 )

c) 3x³ - 9x²y - 4x + 4y ( đề sai )

d) x³ - 2x² - 3x = x( x² - 2x - 3 ) = x( x² - 3x + x - 3 ) = x[ x( x - 3 ) + ( x - 3 ) ] = x( x - 3 )( x + 1 )

Tìm x

a) chưa rõ đề

b) 4x² - 36x = 0

⇔ 4x( x² - 9 ) = 0

⇔ 4x( x - 3 )( x + 3 ) = 0

⇔ x = 0 hoặc x = ±3

c) 2x² - 2x = ( x - 1 )²

⇔ 2x( x - 1 ) - ( x - 1 )² = 0

⇔ ( x - 1 )( 2x - x + 1 ) = 0

⇔ ( x - 1 )( x + 1 ) = 0

⇔ x = ±1 

d) ( x - 7 )( x² - 9 )( x + 1 ) = 2 ( đề sai ) --

x³ - 9x - 5x² + 45 = 0

⇔ ( x³ - 5x² ) - ( 9x - 45 ) = 0

⇔ x²( x - 5 ) - 9( x - 5 ) = 0

⇔ ( x - 5 )( x² - 9 ) = 0

⇔ ( x - 5 )( x - 3 )( x + 3 ) = 0

⇔ x - 5 = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

⇔ x = 5 hoặc x = ±3

\(x^3-9x-5x^2+45=0\)   

\(x^3-5x^2-9x+45=0\)   

\(x^2\left(x-5\right)-9\left(x-5\right)=0\)   

\(\left(x-5\right)\left(x^2-9\right)=0\)   

\(\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)   

\(\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)

a) x² - 8x + 19 = ( x² - 8x + 16 ) + 3 = ( x - 4 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) x² + y² - 4x + 2 = ( x² - 4x + 4 ) + y² - 2 = ( x - 2 )² + y² - 2 ≥ -2 ∀ x, y ( chưa cm được -- )

c) 4x² + 4x + 3 = ( 4x² + 4x + 1 ) + 2 = ( 2x + 1 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

d) x² - 2xy + 2y² + 2y + 5 = ( x² - 2xy + y² ) + ( y² + 2y + 1 ) + 4 = ( x - y )² + ( y + 1 )² + 4 ≥ 4 > 0 ∀ x, y ( đpcm )

câu a sai đề ko bạn

8x³ - 12x² + 8x - 1 =0
=> ( 2x )³ + 3.( 2x )² . 1 + 3.2x.1² + 1³
=> ( 2x + 1 )³ = 0 
=> 2x + 1 = 0
=> 2x = - 1 => x = 1/2