Question

Chứng minh rằng:

a, \(25x^2-10x+3>0\)

b, \(y^2-y+2>0\)

c,\(y^2-3y+5>0\)

d, \(16y^2-6y+9>0\)

Answer 1

a) \(25x^2-10x+3=25x^2-10x+1+2\)

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

Vì \(\left(5x-1\right)^2\ge0\forall x\)

Nên \(\left(5x-1\right)^2+2>0\forall x\)

Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.

b) \(y^2-y+2=y^2-y+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\)

Nên \(\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)

Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.

c) \(y^2-3y+5=y^2-3y+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(y-\dfrac{3}{2}\right)^2\ge0\forall x\)

Nên \(\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\forall x\)

Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.

d) \(16y^2-6y+9=16y^2-6y+\dfrac{9}{16}+\dfrac{135}{16}\)

\(=\left(4x-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\)

Vì \(\left(4x-\dfrac{3}{4}\right)^2\ge0\forall x\)

Nên \(\left(4x-\dfrac{3}{4}\right)^2+\dfrac{135}{16}>0\forall x\)

Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.

Answer 2

a,

\(25x^2-10x+3\\ =\left(5x\right)^2-10x+1+2\\ =\left(5x-1\right)^2+2\\ \left(5x-1\right)^2\ge0\forall x\\ \Rightarrow\left(5x-1\right)^2+2\ge2\forall x\\ \Rightarrow\left(5x-1\right)^2+2>0\forall x\)

b,

\(y^2-y+2\\ =y^2-y+\dfrac{1}{4}+\dfrac{7}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\\ \left(y-\dfrac{1}{2}\right)^2\ge0\forall y\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall y\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall y\)

c,

\(y^2-3y+5\\ =y^2-3y+\dfrac{9}{4}+\dfrac{11}{4}\\ =\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\\ \left(y-\dfrac{3}{2}\right)^2\ge0\forall y\\ \Rightarrow\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall y\\ \Rightarrow\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\forall y\)

d,

\(16y^2-6y+9\\ =\left(4y\right)^2-6y+\dfrac{9}{16}+\dfrac{135}{16}\\ =\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\\ \left(4y-\dfrac{3}{4}\right)^2\ge0\forall y\\ \Rightarrow\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\ge\dfrac{135}{16}\forall y\\ \Rightarrow\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}>0\forall y\)