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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: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)
a) \(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1>0\forall x\)
b) \(4x-x^2-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\forall x\)
a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)
Với mọi x ta có :
\(\left(x-3\right)^2\ge0\)
\(\Leftrightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-6x+10>0\)
b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)
Với mọi x ta có :
\(\left(x-2\right)^2\ge0\)
\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)
\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)
c/ \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi x ta có :
\(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)
d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)
Với mọi x,y ta có :
\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)
\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)
\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)
2/ Ta có :
\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)
Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)
3/ \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)
Mà \(x+y=7;xy=-3\)
\(\Leftrightarrow x^2+y^2=7^2-2.\left(-3\right)=49+6=55\)
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)
b: \(4y^2+2y+1\)
\(=4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{4}\right)\)
\(=4\left(y^2+2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{3}{16}\right)\)
\(=4\left(y+\dfrac{1}{4}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall y\)
c: \(-2x^2+6x-10\)
\(=-2\left(x^2-3x+5\right)\)
\(=-2\left(x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)
\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{2}< =-\dfrac{11}{2}< 0\forall x\)
`#3107.101107`
a)
`x^2 + x + 1`
`= (x^2 + 2*x*1/2 + 1/4) + 3/4`
`= (x + 1/2)^2 + 3/4`
Vì `(x + 1/2)^2 \ge 0` `AA` `x`
`=> (x + 1/2)^2 + 3/4 \ge 3/4` `AA` `x`
Vậy, `x^2 + x + 1 > 0` `AA` `x`
b)
`4y^2 + 2y + 1`
`= [(2y)^2 + 2*2y*1/2 + 1/4] + 3/4`
`= (2y + 1/2)^2 + 3/4`
Vì `(2y + 1/2)^2 \ge 0` `AA` `y`
`=> (2y + 1/2)^2 + 3/4 \ge 3/4` `AA` `y`
Vậy, `4y^2 + 2y + 1 > 0` `AA` `y`
c)
`-2x^2 + 6x - 10`
`= -(2x^2 - 6x + 10)`
`= -2(x^2 - 3x + 5)`
`= -2[ (x^2 - 2*x*3/2 + 9/4) + 11/4]`
`= -2[ (x - 3/2)^2 + 11/4]`
`= -2(x - 3/2)^2 - 11/2`
Vì `-2(x - 3/2)^2 \le 0` `AA` `x`
`=> -2(x - 3/2)^2 - 11/2 \le 11/2` `AA` `x`
Vậy, `-2x^2 + 6x - 10 < 0` `AA `x.`
\(a,=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\\ b,=-\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\)
a,=(x2+3x+94)+194=(x+32)2+194≥194>0b,=−(x2−5x+254)−74=−(x−52)2−74≤−74<0
2.
Ta có hằng đẳng thức : \(\left(a-b\right)^2=a^2-2ab+b^2\left(1\right)\)
Lại có \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2-4ab=a^2+2ab-4ab+b^2\)
\(\Leftrightarrow\left(a+b\right)^2-4ab=a^2-2ab+b^2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(a-b\right)^2=\left(a+b\right)^2-4ab\)( đpcm )
3.
Ta có hằng đẳng thức \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\Rightarrow x^2+y^2=\left(x+y\right)^2-2xy\)
Thay \(x+y=7\)và \(xy=-3\)vào ta được :
\(x^2+y^2=7^2-2\left(-3\right)\)
\(\Leftrightarrow x^2+y^2=49+6=55\)
Vậy ...
1.
a) Đặt \(A=x^2-6x+10\)
\(A=\left(x^2-6x+9\right)+1\)
\(A=\left(x-3\right)^2+1\)
Mà \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow A\ge1>0\)
Vậy ...
b) Đặt \(B=x^2-4x+7\)
\(B=\left(x^2-4x+4\right)+3\)
\(B=\left(x-2\right)^2+3\)
Mà \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow B\ge3\)
Vậy ...