![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(-x^2+4x-7\)
\(=-\left(x^2-4x+7\right)\)
\(=-\left(x^2-2.x.2+2^2-4+7\right)\)
\(=-\left[\left(x-2\right)^2+3\right]\)
\(=-\left(x-2\right)^2-3\)
Do \(-\left(x-2\right)^2\le0\) với \(\forall x\)
\(\Rightarrow-\left(x-2\right)^2-3\le-3< 0\)
\(\Rightarrow-x^2+4x-7< 0\) (đpcm)
câu b,c đề sai bạn nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(4x^2+2x+1=\left(2x\right)^2+2\cdot2x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
b ) \(x^2+3x+4=\left(x^2+2\cdot\frac{3}{2}\cdot x+\frac{9}{4}\right)+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\forall x\)
c ) \(9x^2+3x+5=\left(3x\right)^2+2\cdot3x\cdot\frac{1}{2}+\frac{1}{4}+\frac{19}{4}=\left(3x+\frac{1}{2}\right)^2+\frac{19}{4}>0\forall x\)
Ta có : 4x2 + 2x + 1
= (2x)2 + 2.2x.\(\frac{1}{2}\) + \(\frac{1}{2}+\frac{3}{4}\)
= (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\)
Mà : (2x + \(\frac{1}{2}\))2 \(\ge0\forall x\)
=> (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) \(\ge\frac{3}{4}\forall x\)
Hay : (2x + \(\frac{1}{2}\))2 + \(\frac{3}{4}\) \(>0\forall x\)
Vậy 4x2 + 2x + 1 \(>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)
\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)
Bài 2
\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a , Ta có \(x^2+x+1=x^2+2x\frac{1}{2}+\left(\frac{1}{2}\right)^2+\)\(\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\) \(\ge\frac{3}{4}>0\left(đpcm\right)\)
b , Ta có : \(4x^2-2x+3\)= \(\left(2x\right)^2-2.2x.1+1^2+2\) = \(\left(2x-1\right)^2+2\ge2>0\left(đpcm\right)\)
c , Ta có \(3x^2+2x+1=x^2-\frac{2x}{3}+\frac{1}{9}+2x^2+\frac{8x}{3}+\frac{8}{9}\)
= \(\left(x-\frac{1}{3}\right)^2+2\left(x^2+\frac{4x}{3}+\frac{4}{9}\right)=\left(x-\frac{1}{3}\right)^2+2\left(x+\frac{2}{3}\right)^2\ge0\)
Vì Dấu "=" không thể xảy ra , do đó \(3x^2+2x+1>0\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
a) Ta có: \(-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\left(\forall x\right)\)
=> đpcm
b) \(x^4+3x^2+3=\left(x^4+3x^2+\frac{9}{4}\right)+\frac{3}{4}=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)
=> đpcm
a) -x2 + 4x - 5 = -x2 + 4x - 4 - 1
= -( x2 - 4x + 4 ) - 1
= -( x - 2 )2 - 1 ≤ -1 < 0 ∀ x ( đpcm )
b) x4 + 3x2 + 3 ( * )
Đặt t = x2
(*) <=> t2 + 3t + 3
<=> ( t2 + 3t + 9/4 ) + 3/4
<=> ( t + 3/2 )2 + 3/4
<=> ( x2 + 3/2 )2 + 3/4 ≥ 3/4 > 0 ∀ x ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a, 3x^2-6x+3=(x-1)(x+4)\)
\(⇔(3x^2-6x+3)-(x-1)(x+4)=0\)
\(⇔3x^2-6x+3-x^2-3x+4=0\)
\(⇔2x^2-9x+7=0\)
\(⇔2x^2-2x-7x+7=0\)
\(⇔2x(x-1)-7(x-1)=0\)
\(⇔(x-1)(2x-7)=0\)
\(⇔\left[\begin{array}{} x-1=0\\ 2x-7=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=1\\ x=\dfrac{7}{2} \end{array}\right.\)
Vậy pt có tập nghiệm là S={\(1; \dfrac{7}{2}\)}
\(b, 18x^2(x+4)-12(x^2+4x)=0\)
\(⇔18x^2(x+4)-12x(x+4)=0\)
\(⇔6x(x+4)(3x-2)=0\)
\(⇔\left[\begin{array}{} 6x=0\\\ x+4=0\\ 3x-2=0 \end{array}\right. \)
\(⇔\left[\begin{array}{} x=0\\\ x=-4\\ x=\dfrac{2}{3} \end{array}\right.\)
vậy pt có tập nghiệm là S={\(0;-4;\dfrac{2}{3}\)}
\(c,25x^2-10x+1=2(5x-1)(3x-4)\)
\(⇔(5x-1)^2=(5x-1)(6x-8)\)
\(⇔(5x-1)^2-(5x-1)(6x-8)=0\)
\(⇔(5x-1)[(5x-1)-(6x-8)]=0\)
\(⇔(5x-1)(7-x)=0\)
\(⇔\left[\begin{array}{} 5x-1=0\\ 7-x=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=\dfrac{1}{5}\\ x=7 \end{array}\right.\)
Vậy pt có tập nghiệm là S={\(\dfrac{1}{5};7\)}
\(3x^2-6x+3=\left(x-1\right)\left(x+4\right)\\ \Leftrightarrow3x^2-6x+3=x\left(x+4\right)-\left(x+4\right)\\ \Leftrightarrow3x^2-6x+3=x^2+4x-x-4\\ \Leftrightarrow3x^2-6x+3=x^2+3x-4\\ \Leftrightarrow3x^2-6x+3-x^2-3x+4=0\\ \Leftrightarrow2x^2-9x+7=0\\ \Leftrightarrow x\left(2x-9\right)=-7\)