Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=x^3-27-x^3+3x^2-3x+1-4\left(x^2-4\right)-x\)
\(=3x^2-4x-26-4x^2+16\)
\(=-x^2-4x-10\)
a: Ta có: \(-x^2+4x-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-4x+4+1\right)\)
\(=-\left(x-2\right)^2-1< 0\forall x\)
b: Ta có: \(x^4\ge0\forall x\)
\(3x^2\ge0\forall x\)
Do đó: \(x^4+3x^2\ge0\forall x\)
\(\Leftrightarrow x^4+3x^2+3>0\forall x\)
c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)
\(x^2+2x+4=\left(x+1\right)^2+3>0\forall x\)
Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)
1, 2x2-6x+1=0
\(\Leftrightarrow\) 2(x2-3x+\(\dfrac{1}{2}\))=0
\(\Leftrightarrow\)x2-3x+\(\dfrac{1}{2}\)=0(vì 2 \(\ne\) 0)
\(\Leftrightarrow\)x2-2.\(\dfrac{3}{2}.x+\dfrac{9}{4}+\dfrac{1}{2}-\dfrac{9}{4}\)=0
\(\Leftrightarrow\)(x-\(\dfrac{3}{2}\))2-\(\dfrac{7}{4}\)=0
\(\Leftrightarrow\)(x-\(\dfrac{3+\sqrt{7}}{2}\))(x-\(\dfrac{3-\sqrt{7}}{2}\))=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{7}}{2}\\x=\dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\)
Vậy tập nghiệm bạn tự giải nhé
2a, -x2+4x-9\(\le\)5
\(\Leftrightarrow\)-x2+4x-4\(\le\)0
\(\Leftrightarrow\)-(x-2)2\(\le\)0
\(\Leftrightarrow\)(x-2)2\(\ge\)0 đúng \(\forall\) x
Vậy dfcm
tại x = 18; y = 4
a) Ta có: -\(x^2\)+4x - 9
<=> - ( \(x^2\)- 4x + 4 ) - 5
<=> - ( x - 2 )\(^2\) - 5
Vì - ( x - 2 )\(^2\)\(\le\)0 <=> - ( x - 2 )\(^2\) - 5 \(\le\)-5 với mọi x
b) Ta có x\(^2\)- 2x + 9
<=> ( x\(^2\) - 2x +1 ) + 8
<=> ( x - 1 ) \(^2\)+ 8
Vì ( x - 1 ) \(^2\)\(\ge\) 0 <=> ( x - 1 ) \(^2\)+ 8 \(\ge\) 8 với mọi thực x
a,Ta có:\(-x^2+4x-9\)
\(\Leftrightarrow-\left(x^2-4x+4\right)-5\)
\(\Leftrightarrow-\left(x-2\right)^2-5\)
Vì \(-\left(x-2\right)^2\le0\Leftrightarrow-\left(x-2\right)^2-5\le-5\forall x\)
b.Ta có:\(x^2-2x+9\)
\(\Leftrightarrow\left(x^2-2x+1\right)+8\)
\(\Leftrightarrow\left(x-1\right)^2+8\)
Vì \(\left(x-1\right)^2\ge0\Leftrightarrow\left(x-1\right)^2+8\ge8\forall x\)
a, Ta có: \(-x^2+4x-9+5=-x^2+4x-4\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\le0\)
=> \(-x^2+4x-9\le-5\)
b, Ta có: \(x^2-2x+9-8=x^2-2x+1=\left(x-1\right)^2\ge0\)
=> \(x^2-2x+9\ge8\)
a, Ta có: −x2+4x−9+5=−x2+4x−4−x2+4x−9+5=−x2+4x−4
=−(x2−4x+4)=−(x2−4x+4)
=−(x−2)2≤0=−(x−2)2≤0
=> −x2+4x−9≤−5−x2+4x−9≤−5
b, Ta có: x2−2x+9−8=x2−2x+1=(x−1)2≥0x2−2x+9−8=x2−2x+1=(x−1)2≥0
=> x2−2x+9≥8