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

=(y+1)(x-1)(x+1)

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

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

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

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

a) x/-2=-4/y=2/4

*x/-2=2/4=>4x=(-2)x2=>x=-1

*-4/y=2/4=>(-4)x4=2y=>y=-8

b)2/x=y/-3

=>xy=-6(câu này đề hơi lạ)

c) x+1/2=8/x+1

=>16=(x+1)(x+1)=>x^2+2x+1=16=>(x+1)^2=16=>(x+1)^2=4^2=>x+1=4=>x=3

Đề nghị viết lại đề

a) \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\) 

\(\Rightarrow\dfrac{x}{-2}=\dfrac{2}{4}\) 

\(\Rightarrow x=\dfrac{-2.2}{4}=-1\) 

\(\Rightarrow\dfrac{-4}{y}=\dfrac{2}{4}\) 

\(\Rightarrow x=\dfrac{-4.4}{2}=-8\) 

b) \(\dfrac{2}{x}=\dfrac{y}{-3}\) 

\(\Rightarrow x.y=2.-3\) 

\(\Rightarrow x.y=-6\) 

\(\Rightarrow x\) và \(y\inƯ\left(-6\right)=\left\{^+_-1;^+_-2;^+_-3;^+_-6\right\}\) 

\(\Rightarrow\left(x;y\right)=\left(-6;1\right);\left(1;-6\right);\left(6;-1\right);\left(-1;6\right);\left(-2;3\right);\left(3;-2\right);\left(2;-3\right);\left(-3;2\right)\) c) \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\) 
\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\) 

\(\Rightarrow\left(x+1\right)^2=16\) 

\(\Rightarrow\left(x+1\right)^2=4^2\) hoặc \(\left(x+1\right)^2=\left(-4\right)^2\) 

      \(x+1=4\) hoặc \(x+1=-4\) 

            \(x=3\) hoặc \(x=-5\)

\(A=x^2+x+1=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

A= x² + x + 1

A = x² + 2. \(\dfrac{1}{2}\). x + (\(\dfrac{1}{2}\))² +\(\dfrac{3}{4}\)

A = ( x + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)

Vậy, x² + x + 1>0 với mọi x

Đúng thì like giúp mik nha. Thx bạn

         \(x^2+x+1\)

\(=\)   \(x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+1\)

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

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

\(=\)     \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\) luôn dương với mọi \(x\)     ( 1 )

mà cộng thêm 1 lượng \(\dfrac{3}{4}\)  luôn dương   ( 2 )

Từ ( 1 ) và ( 2 ):     ⇒    \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) luôn dương

⇒   \(x^2+x+1\) luôn dương với mọi giá trị của x

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

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

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

a: \(M=\dfrac{2\left(1-3x\right)\left(1+3x\right)}{3x\left(x+2\right)}\cdot\dfrac{3x}{2\left(1-3x\right)}=\dfrac{3x+1}{x+2}\)

Xem lại đề bạn nhé

\(=\dfrac{19}{23}\times\dfrac{23}{37}\times\dfrac{37}{19}=1\)