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\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
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 \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\ge-1\)
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.
Do \(0\le x\le1\Rightarrow\left(1+x\right)^2\ge\left(x+x\right)^2=4x^2\) (đpcm)
Dấu "=" xảy ra khi \(x=1\)
b.
Do \(x;y\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}x^2\le x\\y^2\le y\end{matrix}\right.\) \(\Rightarrow x+y\ge x^2+y^2\)
\(\Rightarrow\left(1+x+y\right)^2\ge4\left(x+y\right)\ge4\left(x^2+y^2\right)\) (đpcm)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)
(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)=(x2-5x+4)(x2-5x+6)
đặt a=x2-5x+4
=>(x-1)(x-2)(x-3)(x-4)=a(a+2)=a2+2a
=>(x-1)(x-2)(x-3)(x-4)+1=a2+2a+1=(a+1)2>=0
=>(x-1)(x-2)(x-3)(x-4)>=-1 (dpcm)
b)a2+b2+c2≥ab+bc+aca2+b2+c2≥ab+bc+ac
⇔2(a2+b2+c2)≥2(ab+bc+ac)⇔2(a2+b2+c2)≥2(ab+bc+ac)
⇔2a2+2b2+2c2−2ab−2bc−2ac≥0⇔2a2+2b2+2c2−2ab−2bc−2ac≥0
⇔(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)≥0⇔(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ac+a2)≥0
⇔(a−b)2+(b−c)2+(c−a)2≥0⇔(a−b)2+(b−c)2+(c−a)2≥0 (luôn đúng)
Dấu ''='' xảy ra khi a=b=c