a)  \(x^2-8x+20=\left(x-4\right)^2+4>0\)

b)  \(4x^2-12x+11=\left(2x-3\right)^2+2>0\)

c)  \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

d)  \(x^2-2x+y^2+4y+6=\left(x-1\right)^2+\left(y+2\right)^2+1>0\)

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

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

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

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

\(=2\left(x^2-4x+10\right)\)

\(=2\left(x^2-4x+4+6\right)\)

\(=2\left[\left(x-2\right)^2+6\right]\)

\(=2\left(x-2\right)^2+12>0\forall x\)

b) \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

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

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

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

a) \(A=x^2+2x+2\)

\(=x^2+2x+1+1\)

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

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

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

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

\(=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4>0\forall x,y\)

\(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

\(C=x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(D=x^2-2x+y^2+4y+6=x^2-2x+1+y^2+4y+4+1\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1>0\)

\(E=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

Cm: Ta có: 

a) A = x² - 8x + 20 = (x² - 8x + 16) + 4 = (x - 4)² +  4 > 0 \(\forall\) x(vì (x - 4)² \(\ge\)0 \(\forall\)x ; 4 > 0)

=> A luôn dương với mọi x

b) B = 4x² - 12x + 11 = [(2x)² - 12x + 9] + 2 = (2x - 3)² + 2 > 0 \(\forall\)x (vì (2x - 3)² \(\ge\)0 \(\forall\)x; 2 > 0)

=> B luôn dương với mọi x

c) C = x² - x + 1 = (x² - x + 1/4) + 3/4 = (x -  1/2)² + 3/4 > 0 \(\forall\)x (vì (x - 1/2)² \(\ge\)0 \(\forall\)x; 3/4 > 0)

=> C luôn dương với mọi x

* Tìm x

3(x + 2)² + (2x - 1)² - 7(x + 3)(x - 3) = 36

=> 3(x² + 4x + 4) + 4x² - 4x + 1 - 7(x² - 9) = 36

=> 3x² + 12x + 12 + 4x² - 4x + 1 - 7x² + 63 = 36

=> 8x + 76 = 36

=> 8x = 36 - 76

=> 8x = -40

=> x = -40 : 8 = -5

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

Ta có : A = x² - 8x + 20

=> A = x² - 8x + 16 + 4

=> A = (x - 4)² + 4

Mà ; (x - 4)² \(\ge0\forall x\)

Nên : A  = (x - 4)² + 4 \(\ge4\forall x\)

Vậy A_min = 4 , dấu "=" xảy ra khi và chỉ khi x = 4

Ta có : 4x² - 12x + 11

= (2x)² - 12x + 9 + 2

= (2x - 3)² + 2

Mà : (2x - 3)² \(\ge0\forall x\) 

Nên : (2x - 3)² + 2 \(\ge2\forall x\)

Vậy (2x - 3)² + 2 \(>0\forall x\)