a) \(x^4+2x^3+x^2=\left(x^2+x\right)^2\)

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

c) \(5x^2-10xy+5y^2-20z^2=\left(\sqrt{5x}-\sqrt{5y}\right)^2-20z^2\)

d) \(x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)

Mình nghĩ đề là Cho \(x+\dfrac{1}{x}=3\) phù hợp hơn nhé.

Ta có: \(x+\dfrac{1}{x}=3\)\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=3^2\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=9\Leftrightarrow x^2+2+\dfrac{1}{x^2}=9\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}=7\)

Vậy \(x^2+\dfrac{1}{x^2}=7\)

:) 

x = 2,5

thay vào, dùng máy tính bấm ra nhé 

2,5 ² + 1: 2,5 ² 

\(x^2+10x+25\)

\(=x^2+2.5x+5^2\)

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

`x^2 +10x + 25`

`=x^2 + 2.5.x + 25`

`=(x+5)^2`

giúp mình với

\(4)25-1^2=\left(5-1\right)\left(5+1\right)=4.6=24\\ 5)a^2-9=a^2-3^2=\left(a-3\right)\left(a+3\right)\)

Ta có \(P=2x^2-4x+7=2\left(x^2-2x+\dfrac{7}{2}\right)=2\left(x^2-2x+1+\dfrac{5}{2}\right)\) \(=2\left(x-1\right)^2+5\)

Mà \(2\left(x-1\right)^2\ge0\Leftrightarrow2\left(x-1\right)^2+5\ge5>0\) hay \(P>0\) (đpcm)

Mặt khác \(H=-2x^2-16x+38=-2\left(x^2+8x-19\right)\) \(=-2\left(x^2+8x+16-35\right)=-2\left(x+4\right)^2-70\)

Mà \(-2\left(x+4\right)^2\le0\Leftrightarrow-2\left(x+4\right)^2-70\le-70< 0\) nên ta có \(H< 0\) (đpcm)

`2x^2 - 4x + 7`

`<=> 2x^2 - 4x  + 2 + 5`

`<=> 2.(x-1)^2 + 5`

Mà : \(2\left(x-1\right)^2\ge0\forall x\)

`=>` \(2\left(x-1\right)^2+5\ge0\forall x\)

Vậy `2x^2 -4x+7` luôn dương với mọi `x`

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`-2x^2 - 16x + 38`

`<=> -2.(x^2+8x-19)`

Mà :\(x^2+8x-19\ge0\forall x\)

`=>` \(-2x.\left(x^2+8x-19\right)\le0\forall x\)

Vậy `-2x^2-16x+38` luôn âm với mọi `x`

\(\left(x+2\right)^2=x^2+4x+4\)  

Áp dụng hằng đẳng thức 

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

`(x+2)^2 = x^2 + 2.2.x + 2^2 = x^2 + 4x + 4`

Áp dụng HĐT số `1` 

`(x+y)^2 = x^2 + 2.x.y + y^2`

a) Điều kiện \(x\ne0\). Ta có 

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

Đặt \(\dfrac{1}{x}=p\) , khi đó \(A=2p^2-p+1\) 

Lại có \(A=2p^2-p+1=2\left(p^2-\dfrac{p}{2}+\dfrac{1}{2}\right)\) \(=2\left(p^2-2p.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{7}{16}\right)=2\left[\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{16}\right]\) \(=2\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\)

Mà \(2\left(p-\dfrac{1}{4}\right)^2\ge0\Leftrightarrow2\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\) hay \(min_A=\dfrac{7}{8}\)

Dấu "=" xảy ra khi \(p-\dfrac{1}{4}=0\Leftrightarrow p=\dfrac{1}{4}\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{4}\Leftrightarrow x=4\)

Vậy GTNN của A là \(\dfrac{7}{8}\) khi \(x=4\)

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

Đến đây đặt \(\dfrac{1}{x+1}=t\) thì \(B=2t^2-4t+1\) và làm tương tự như câu a.

\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x\right)^2-2.\left(\dfrac{1}{2}x\right).\left(\dfrac{1}{3}y\right)+\left(\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)

\(4-6x+\dfrac{9}{4}x^2=2^2-2.2.\dfrac{2}{3}x+\left(\dfrac{2}{3}x\right)^2=\left(2-\dfrac{2}{3}x\right)^2\)

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

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

a, \(8-12x+6x^2-x^3=\left(2-x\right)^3\)

b, \(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)

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

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