Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét hiệu : \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(-1\right)=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)
Đặt \(x^2-5x+5=y\). Biểu thức trên bằng :\(\left(y-1\right)\left(y+1\right)+1=y^2\ge0\)
Vậy \(\text{ ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ≥ − 1}\)
\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)
a) \(A=\left(x+4\right)\left(x-4\right)-2x\left(3+x\right)+\left(x+3\right)^2\)
\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b) \(B=\left(x^2+4\right)\left(x+2\right)\left(x-2\right)-\left(x^2+3\right)\left(x^2-3\right)\)
\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
\(x\left(x+1\right)^4+x\left(x+1\right)^3+x\left(x+1\right)^2+\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)^2+x\left(x+1\right)+x+1\right]\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x\left(x+1\right)+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x^2+x+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left[\left(x+1\right)\left(x^2+2x+1\right)\right]\)
\(=\left(x+1\right)^2\cdot\left(x+1\right)^3\)
\(=\left(x+1\right)^5\left(đpcm\right)\)
a) \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2\ge\left(x+y\right)^2\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\left(đúng\right)\)
b) \(x^3+y^3\ge\dfrac{\left(x+y\right)^3}{4}\)
\(\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow3x^3+3y^3\ge3x^2y+3xy^2\)
\(\Leftrightarrow3x^2\left(x-y\right)-3y^2\left(x-y\right)\ge0\)
\(\Leftrightarrow3\left(x-y\right)\left(x^2-y^2\right)\ge0\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\left(đúng\right)\)
a: Ta có: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Leftrightarrow2x^2+2y^2-x^2-2xy-y^2\ge0\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\)(luôn đúng)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
Xét hiệu :
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-\left(-1\right)=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1.\)
Đặt \(x^2-5x+=y.\) Biểu thức trên bằng \(\left(y-1\right)\left(y+1\right)+1=y^2\ge0\)
Vậy \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)