Mấy bạn bị lms í=)) dễ v cũng ko biết làm

Mình chỉ đăng lên để thử xem coi ai làm đc ko chứ mình cx ko biết làm. Ai jup mình vớiiiiii

ko biết

a) \(x^2-8x+20\)

\(=x^2-2.x.4+16+4\)

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

Có: \(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2+4>0\)

Hay:.............

b) \(x^2+11\)

Có: \(x^2\ge0\Rightarrow x^2+11>0\)

Hay:.............

c) \(4x^2-12x+11\)

\(=4\left(x^2-3x+\frac{11}{4}\right)\)

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

\(=4\left(x-\frac{3}{2}\right)^2+2>0\)

d) \(x^2+5y^2+2x+6y+34\)

\(=x^2+2.x.1+1+y^2+4y^2+2.y.3+9+24\)

\(=\left(x^2+2.x.1+1\right)+\left(y^2+2.y.3+9\right)+4y^2+24\)

\(=\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24\)

Ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\\\left(2y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24>0\)

f) \(x^2-2x+y^2+4y+6\)

\(=x^2-2.x.1+1+y^2+2.y.2+4+1\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\)

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

`B = x^2- 2xy + y^2 + 2x - 10y + 17

`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`

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

3) 5x² + y² -4xy - 2y + 8x + 2013

= ( 4x² + y² -4xy -2y + 8x ) + x² + 2013

= ( 2x - y +1)² + x² +2013

Vì ( 2x-y+1)² \(\ge\)0 \(\forall x,y\); x² \(\ge\)0\(\forall x\)

=> (2x - y+1)² + x² \(\ge\)0

=> ( 2x-y +1)² +x² + 2013\(\ge\)0

hay  A \(\ge0\)\(\forall x,y\)=> A ko âm

Giúp mk phần 1 và phần 2 vs!!!

HELP ME PLEASE!!!

1\(\left(15x-1\right)^2+3\left(7x+3\right)\left(x+1\right)-\left(x^2-73\right)\))

\(=\left(15x-1\right)^2+21x^2+30x+9-x^2+73\)

\(=\left(15x-1\right)^2+20x^2+30x+82\)

\(=\left(15-1\right)^2+20\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+\frac{283}{4}\)

\(=\left(15x-1\right)^2+20\left(x+\frac{3}{4}\right)^2+\frac{283}{4}\)

Vì \(\left(15x-1\right)^2;20\left(x+\frac{3}{4}\right)^2;\frac{283}{4}\ge0\forall x\)=> Biểu thức ko âm

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

chúc mng lm bài được

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