\(x^4+3x^2y^2+4y^4\)

\(x^4+4y^4-2xy^3+2xy^3+2x^2y^2+2x^2y^2-x^2y^2\)

\(+x^3y-x^3y\)

\(=\left(4y^4-2xy^3+2x^2y^2\right)+\left(2xy^3-x^2y^2+x^3y\right)\)

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

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

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

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

1-D

2-D

3-A

4-C

5-D

6-C

Công thức đúng nhé bạn

✞๖ۣۜ ☾ ɪ’ʟʟ ɓє ƴσυʀ ɓєѕтƒʀɪєη∂,⁀ᶜᵘᵗᵉ(♥)ღ༻ bạn phân tích đúng rồi nha !

Ta có:

A = \(\frac{1}{2x^2+2x-5}\)

A = \(\frac{1}{2\left(x^2+x+\frac{1}{4}\right)-\frac{11}{2}}\)

A = \(\frac{1}{2\left(x+\frac{1}{2}\right)^2-\frac{11}{2}}\)

Do \(2\left(x+\frac{1}{2}\right)^2\ge0\forall x\)=> \(2\left(x+\frac{1}{2}\right)^2-\frac{11}{2}\ge-\frac{11}{2}\forall x\)

=> \(\frac{1}{2\left(x+\frac{1}{2}\right)^2-\frac{11}{2}}\le\frac{1}{-\frac{11}{2}}=-\frac{2}{11}\forall x\)

Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\) <=> \(x=-\frac{1}{2}\)

Vậy MaxA = -2/11 <=> x = -1/2

Ta có: \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)(1)

Đặt \(x^2+10x+16=a\)

\(\Rightarrow\left(1\right)=a\left(a+8\right)+16\)

\(=a^2+8a+16=\left(a+4\right)^2\)(2)

Mà \(x^2+10x+16=a\)(theo cách đặt) nên :

\(\left(2\right)=\left(x^2+10x+20\right)^2\)(là bình phương của 1 số)

Vậy (x+2)(x+4)(x+6)(x+8)+16 là scp

đề có vấn đề

                                                        Bài giải

Nếu 

\(\left(x+5\right)\left(x-2\right)-x\left(x-1\right)=2=P\)

\(x^2+5x-2x-10-x^2+1=2\)

\(3x-9=2\)

\(3x=2+9\)

\(3x=11\)

\(x=\frac{3}{11}\)

Vậy \(P=2\) khi \(x=\frac{3}{11}\)

ME=Math Error