a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)

\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)

\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

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

nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)

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

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

Vì \(\left(x-y\right)^2\ge0\forall x;y\)

\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)

nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)

\(x-x^2-1\\ =-\left(x^2-x+1\right)\\ =-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\\ \left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\in R\\ \Rightarrow-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]< 0\forall x\in R\\ \Leftrightarrow x-x^2-1< 0\forall x\in R\)

Vậy \(x-x^2-1< 0\forall x\in R\)

Ta có: \(x-x^2-1\)

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

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

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

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

Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\in R\)

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le\dfrac{-3}{4}< 0\forall x\in R\)

-> ĐPCM.

Tự nhiên quên mất cách làm rồi, mọi người giúp với!

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

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

\(=\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{1}{2}>0\)(đúng \(\forall x;y\in R\))

ta có : \(x-x^2-1=-\left(x^2-x+1\right)=-\left(x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right)\)

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

ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi \(x\)

\(\Leftrightarrow-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le\dfrac{-3}{4}< 0\) với mọi \(x\)

vậy \(x-x^2-1< 0\) với mọi số thực \(x\) (đpcm)

Dễ mak

Ta có:

x luôn bé hơn hoặc bằng x²

=>x-x²\(\le0\)

\(\Rightarrow x-x^{2^{ }}-1\le0\forall x\in Q\)

Ta có: \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\)

Thấy \(x^8\ge0;x^5< x^8\Rightarrow x^8-x^5\ge0\)

\(\Rightarrow x^8-x^5+x^2-x+1>0\forall x\in R.\)(đpcm) 

sai rồi bạn ơi

x = 0 

co dung KO thi nho ket ban voi minh nha ^_^

\(=-\left(9x^2+2\cdot3\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{1}{4}-1=-\left(3x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\)

\(-9x^2+3x-1\)

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

\(=-9\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{1}{12}\right)\)

\(=-9\left(x-\dfrac{1}{6}\right)^2-\dfrac{3}{4}< 0\forall x\)

b) Câu hỏi của Nguyễn Hoàng Thanh Trúc - Toán lớp 8 - Học toán với OnlineMath tham khảo

a ) Đề sai

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

c ) \(x-x^2-2=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\forall x\left(đpcm\right)\)

\(a,x^2-6xy+9y^2+1=\left(x-3y\right)^2+1\ge1>0\\ b,-25x^2+5x-1=-\left(25x^2+2\cdot5\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(5x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}< 0\)