\(A=9x^2-6x+2=\left(3x\right)^2-2.3x+1+1=\left(3x-1\right)^2+1>0\forall x\)

Vậy ta có đpcm 

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

Vậy ta có đpcm 

Trả lời:

\(A=9x^2-6x+2=\left(3x\right)^2-2.3x.1+1+1=\left(3x-1\right)^2+1\ge1>0\forall x\)

Vậy A > 0 với mọi x 

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

Vậy B > 0 với mọi x;y

\(9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1>0\Rightarrowđpcm\)

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

\(25x^2-20x+7=25x^2-20x+4+3=\left(5x-2\right)^2+3>0\left(đpcm\right)\)

\(9x^2-6xy+2y^2+1=\left(9x^2+6xy+y^2\right)+y^2+1=\left(3x+y\right)^2+y^2+1>0\left(đpcm\right)\)

\(\Leftrightarrow x^2+y^2\ge xy;x^2+y^2\ge2\sqrt{x^2y^2}=2\left|xy\right|\ge\left|xy\right|\ge xy\Rightarrowđpcm\)

Cách khác câu e:

\(x^2-xy+y^2=x^2-2x.\frac{y}{2}+\frac{y^2}{4}+\frac{3y^2}{4}=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge0\forall xy\) (đpcm)

a. \(x^2+3x+5\)

\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

=> đpcm

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

a) \(x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy : \(S=\left\{2;-3\right\}\)

a) PT <=> \(\left(x^2-2x\right)+\left(3x-6\right)=0\)

<=> \(x\left(x-2\right)+3\left(x-2\right)=0\)

<=> \(\left(x-2\right)\left(x+3\right)=0\)

<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

KL: ...

b) \(PT< =>\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\)

<=> \(\left(x+\frac{1}{2}\right)^2=\frac{-15}{4}\)

<=> x = \(\varnothing\)

c) PT <=> \(\left(t^2-6t\right)+\left(12t-72\right)=0\)

<=> \(t\left(t-6\right)+12\left(t-6\right)=0\)

<=> \(\left(t+12\right)\left(t-6\right)=0\)

<=> \(\left[{}\begin{matrix}t=-12\\t=6\end{matrix}\right.\)

d) PT <=> \(\left(x^2-x\right)-\left(8x-8\right)=0\)

<=> \(x\left(x-1\right)-8\left(x-1\right)=0\)

<=> \(\left(x-1\right)\left(x-8\right)=0\)

<=> \(\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

e) PT <=> \(\left(x^2-9x+\frac{81}{4}\right)+\frac{23}{4}\)

<=> \(\left(x-\frac{9}{2}\right)^2=\frac{-23}{4}\)

<=> x = \(\varnothing\)

a, x² - 2x + 3 > 0

Xét : VT = x² - 2x + 1 + 2 = ( x - 1 )² + 2 .

Có : ( x - 1 )² \(\ge\) 0 với mọi x \(\Rightarrow\) ( x - 1 )² + 2 > 0 với mọi x hay

VT > 0 .

Vậy BĐT x² - 2x + 3 > 0 đúng .

Các câu còn lại tương tự .

Chúc bn học tốt !!!!!!!!

a) 

Đặt \(A=9x^2-6x+2\)

\(=\left(3x\right)^2-2.3x+1+1\)

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

Ta có: \(\left(3x+1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(3x+1\right)^2+1\ge0+1;\forall x\)

Hay \(A\ge1>0;\forall x\)

Các phần khác tương tự cứ việc biến đổi thành hằng đẳng thức

\(a,9x^2-6x+2\)

\(=\left(3x\right)^2-2.3x.1+1^2+1\)

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

Vì\(\left(3x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(3x-1\right)^2+1\ge1>0\forall x\)

\(\Rightarrow9x^2-6x+2>0\forall x\)

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

Vì\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

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

\(\Rightarrow x^2+x+1>0\forall x\)

mình cũng muốn lắm nhưng mình mới lớp 7

a,\(a^2-6a+10=a^2-6a+9+1=\left(a-3\right)^2+1\)

Mà \(\left(a-3\right)^2\ge0=>\left(a-3\right)^2+1>0\)

\(=>a^2-6a+10>0\)

b, \(a^2+a+1=a^2+2a\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(a+\frac{1}{2}\right)^2\ge0=>\left(a+\frac{1}{2}\right)+\frac{3}{4}>0\)

\(=>a^2+a+1>0\)

\(\left(x-3\right)\left(x-5\right)+4=x^2-8x+15+4\)

\(=x^2+8x+16+3=\left(x+4\right)^2+3\)

Vì \(\left(x+4\right)^2\ge0=>\left(x+4\right)^2+3>0\)

\(=>\left(x-3\right)\left(x-5\right)+4>0\)

chứng minh:4x²-5x+13>0