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![](https://rs.olm.vn/images/avt/0.png?1311)
Có : \(\left|x+1\right|+\left|x+2\right|+.....+\left|x+9\right|\ge0\)
<=> \(10x\ge0\)
<=> \(x\ge0\)
Vậy , ta có thể phá trị tuyệt đối vì trị của nó không âm
=> \(x+1+x+2+x+3+.....+x+9=10x\)
=> \(9x+45=10x\)
<=> x = 45
Dễ thấy: \(VT\ge0\Rightarrow VP\ge0\Rightarrow10x\ge0\Rightarrow x\ge0\)
\(pt\Leftrightarrow\left(x+1\right)+\left(x+2\right)+...+\left(x+9\right)=10x\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+9\right)=10x\)
\(\Leftrightarrow9x+45=10x\)
\(\Leftrightarrow9x-10x=-45\Leftrightarrow x=45\) (thỏa)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : x^3-x^2+2x-8
= (x^3-2x^2)+(x^2-2x)+(4x-8)
= (x-2).(x^2+x+4)
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
( x - 2 )( x - 4 ) + 3
<=> x2 - 6x + 8 + 3
<=> ( x2 - 6x + 9 ) + 2
<=> ( x - 3 )2 + 2 ≥ 2 > 0 ∀ x ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)
\(=25x^2+10x+1-\left(2xy-3\right)^2\)
\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)
\(=25x^2+10x+1-4x^2y^2+12xy-9\)
\(=25x^2-4x^2y^2+10x+12xy-8\)
Bài 2:
\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)
\(=x^3-1=x^3-9x^2+2x+6\)
\(=x^3-9x^2+2x+6=x^3-1\)
\(=x^3-9x^2+2x+6+1=x^3-1+1\)
\(=x^3-9x^2+2x+7=x^3\)
\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)
\(=-9x^2+2x+7=0\)
\(\Rightarrow x=-\frac{7}{9};x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{\left(\frac{1}{4}x^2-\frac{1}{2}x+\frac{1}{4}\right)+\left(\frac{3}{4}x^2+\frac{3}{2}x+\frac{3}{4}\right)}{x^2-2x+1}=\frac{\frac{1}{4}\left(x-1\right)^2+\frac{3}{4}\left(x+1\right)^2}{\left(x-1\right)^2}=\frac{1}{4}+\frac{\frac{3}{4}\left(x+1\right)^2}{\left(x-1\right)^2}\)
Ta thấy : \(\frac{\frac{3}{4}\left(x+1\right)^2}{\left(x-1\right)^2}\ge0\forall x\) nên \(\frac{1}{4}+\frac{\frac{3}{4}\left(x+1\right)^2}{\left(x-1\right)^2}\ge\frac{1}{4}\forall x\) có GTNN là \(\frac{1}{4}\) tại x = - 1
Vậy \(P_{min}=\frac{1}{4}\) tại \(x=-1\)
\(P=\frac{\left(x^2-2x+1\right)+\left(3x-3\right)+3}{\left(x-1\right)^2}=\frac{\left(x-1\right)^2+3\left(x-1\right)+3}{\left(x-1\right)^2}=1+\frac{3}{x-1}+\frac{3}{\left(x-1\right)^2}\)
đặt \(y=\frac{1}{x-1}\Rightarrow P=1+3y+3y^2=3\left(y+\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
vậy \(MinP=\frac{1}{4}\Leftrightarrow y=-\frac{1}{2}\Leftrightarrow\frac{1}{x-1}=-\frac{1}{2}\Leftrightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X/2=y/6
X=y/6 . 2 = y/3=1/3 y
Thay vào ta có
1/3y+ y=6
4/3y=6
Y=18/4
X=18/4 . 1/3=18/12=3/2
Đến đây bạn tự tính x-y nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(1-x\right)\left(6a+2a^2\right)=2a\left(3+2a\right)\left(1-x\right)\\ 2,=\left(x-5\right)\left(x-3-2\right)=\left(x-5\right)^2\)
Chứng minh rằng :
4 x mũ 2 trừ 4 x + 3 lớn hơn 0 với mọi x
(giúp mình với, mình đang cần gấp lắm!!!!!)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2-4x+3\)
\(=\left(4x^2-4x+1\right)+2\)
\(=\left(2x+1\right)^2+2>0\)với mọi x
vậy \(4x^2-4x+3>0\)với mọi x
\(4x^2-4x+3=4x^2-4x+1+2=\left(2x-1\right)^2+2\)
Vì \(\left(2x-1\right)^2\ge0\forall x\)\(\Rightarrow4x^2-4x+3\ge2\forall x\)
hay \(4x^2-4x+3>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x+1+x-1\right)\left(x+1-x+1\right)-3\left(x^2-1\right)=4\)
\(\Leftrightarrow2x.2-3x^2+3=4\)
\(\Leftrightarrow-3x^2-4x-1=0\)
\(\Leftrightarrow-3x^2-3x-x-1=0\)
\(\Leftrightarrow-3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(-3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{3}\end{cases}}\)
(x + 2)(x - 1) - x(x + 3)
= x^2 - x + 2x - 2 - x^2 - 3x
= -2x - 2
giúp mình giải câu này vs
\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\)