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Lời giải:
Đặt \(x=2t+1\). Khi đó, \(q(x)=10^{6x+2}+10^{6t+4}+1\)
Ta thấy: \(10^6\equiv 1\pmod {91}\). Do đó:
\(\left\{\begin{matrix} 10^{6k}\equiv 1\pmod {91}\\ 10^{6t}\equiv 1\pmod {91}\end{matrix}\right.\)
\(\Rightarrow q(x)\equiv 10^2+10^4+1\equiv 10101\equiv 0\pmod {91}\)
Do đó, \(q(x)\vdots 91\) với \(x\in\mathbb{N}\) lẻ.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x2 + x + 1 = (x2 + x + 1/4) + 3/4 = (x + 1/2)2 + 3/4 > 0 => đa thức vô nghiệm
b) x2 - x + 1 = (x2 - x + 1/4) + 3/4 = (x - 1/2)2 + 3/4 > 0 => đa thức vô nghiệm
c) x2 - 6x + 10 = (x2 - 6x + 9) + 1 = (x - 3)2 + 1 > 0 => đa thức vô nghiệm
d) 9x2 + 6x + 2 = (9x2 + 6x + 1) + 1 = (3x + 1)2 + 1 > 0 => đa thức vô nghiệm
e) -2x2 + 8x - 11 = -2(x2 - 4x + 4) -3 = -2(x - 2)2 - 3 < 0 => đa thức vô nghiệm
g) -3x2 + 2x - 4 = -3(x2 - 2/3x + 1/9) - 11/3 < 0 => đa thức vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1>0\) mọi x
p/s: chúc bạn hk tốt
\(x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)+1>0\)
Code : Breacker
![](https://rs.olm.vn/images/avt/0.png?1311)
6. \(-2x^2+2x-4=-2\left(x^2-x+2\right)\)
\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+2\right)\)
\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{2}\le\dfrac{-7}{2}< 0\)
-> ĐPCM.
7. 8. 9 Tương tự
10. \(6x^2+15x-21\)
\(=6\left(x^2+\dfrac{15}{6}x-\dfrac{7}{2}\right)\)
\(=6\left(x^2+2.x.\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}-\dfrac{7}{2}\right)\)
\(=6\left[\left(x+\dfrac{5}{4}\right)^2-\dfrac{81}{16}\right]\)
\(=6\left(x+\dfrac{5}{4}\right)^2-\dfrac{243}{8}\)
\(=\dfrac{243}{8}-6\left(x+\dfrac{5}{4}\right)^2\)
....
1)
-2x2+2x-4
= -2(x2 -x +2)
= -2(x2 - 2.\(\dfrac{1}{2}\).x + \(\dfrac{1}{4}\)+\(\dfrac{7}{4}\))
= -2.(x-\(\dfrac{1}{2}\))2 - \(\dfrac{7}{2}\) \(\le\) - \(\dfrac{7}{2}\) với \(\forall\) x
=> -2x2+2x-4 luôn âm
=>đpcm
2)
-2x2 +6x -8
= -2 (x2 -3x + 4)
= -2(x2 - 2.\(\dfrac{3}{2}\).x +\(\dfrac{9}{4}\)+\(\dfrac{7}{4}\))
= -2.(x-\(\dfrac{3}{2}\))2 - \(\dfrac{7}{2}\) \(\le\) - \(\dfrac{7}{2}\) với \(\forall\) x
=> -2x2 +6x -8 luôn âm
=>đpcm
3)
-x2 + 4x -1
= - (x2 - 4x +1)
= -(x2 - 2.2.x + 4 -3)
= -(x - 2)2 +3 \(\le\) 3 với \(\forall\) x
=> -x2 + 4x -1 có thể không âm
=> sai đề
4)
-2x2 +6x -12
= -2(x2- 3x + 6)
= -2(x2 - 2.\(\dfrac{3}{2}\).x + \(\dfrac{9}{4}+\dfrac{15}{4}\))
= -2(x-\(\dfrac{3}{2}\))2 - \(\dfrac{15}{2}\) \(\le\) - \(\dfrac{15}{2}\) với \(\forall\) x
=> -2x2 +6x -12 luôn âm
=>đpcm
5)
6x2 +15x - 21
= 6(x2 + 2.\(\dfrac{15}{2}\)x + \(\dfrac{225}{4}\)- \(\dfrac{309}{4}\))
= 6.(x-\(\dfrac{15}{2}\))2 - \(\dfrac{927}{2}\) \(\ge\) - \(\dfrac{927}{2}\) với \(\forall\) x
=> 6x2 +15x - 21 có thể không âm
=> đề sai
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=x.5x-x.3-x^2.x+x^2.1+x.x^2-x.6x-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2+10+3x\)
\(=-10\)
Biểu thức trên kết quả là -10 => ĐPCM
\(x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
=\(5x^2-3x-x^3+x^2+x^3-6x^2-10+3x\)
=\(\left(x^3-x^3\right)+\left(5x^2+x^2-6x^2\right)+\left(-3x+3x\right)-10\)
=-10
=> ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)
Vậy A > 0 với mọi x.
\(B=x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\)
\(\left(x-y\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+1\ge1>0\)
Vậy B > 0 với mọi x, y.
\(M=x^2-6x+12\)
\(=x^2-6x+9+3\)
\(=\left(x-3\right)^2+3\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+3\ge3\)
\(MinB=3\Leftrightarrow x=3\)
\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7\)
\(2x^2+6x+5-2x^2+4x-2=7\)
\(10x=7+3\)
\(10x=10\)
\(x=1\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
\(x^3-\frac{1}{4}x=0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
\(\left(x+10\right)^2-\left(x^2+2x\right)\)
\(=x^2+20x+100-x^2-2x\)
\(=18x+100\)
\(\left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+x\right)\)
\(=x^2-4+x^3-1-x^3-x^2\)
\(=-5\)
sai đề ròi bạn ơi
mik nghĩ vậy...
\(x^2+6x+1\ge10\)
\(\Rightarrow x^2+6x\ge9\)
\(\Rightarrow x\left(x+6\right)\ge9\)
\(x^2+6x+9\ge18\)
\(\Rightarrow\frac{x^2+6x+9}{18}\ge1\)
\(\Leftrightarrow\frac{1}{18}\left(x+3\right)^2\ge1\)
Theo bạn dưới nói đề sai thì có vẻ đúng đấy bạn