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+) \(E=x^2-6x+9+x^2-22x+121=2x^2-28x+130\)
\(\Rightarrow2E=4x^2-56x+242=\left(4x^2-56x+196\right)+46=\left(2x-14\right)^2+46\)
Vì \(\left(2x-14\right)^2\ge0\Rightarrow2E=\left(2x-14\right)^2+46\ge46\Rightarrow E\ge23\)
Dấu "=" xảy ra khi x=7
Vậy Emin=23 khi x=7
+) \(F=\frac{-2}{x^2-2x+5}=\frac{-2}{x^2-2x+1+4}=\frac{-2}{\left(x-1\right)^2+4}\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow F=\frac{-2}{\left(x-1\right)^2+4}\le-\frac{2}{4}=-\frac{1}{2}\)
Dấu "=" xảy ra khi x=1
Vậy Fmin=-1/2 khi x=1
+) \(G=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-6x+x-6\right)\left(x^2-3x-2x+6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt x2-5x=t, ta được:
\(G=\left(t-6\right)\left(t+6\right)=t^2-36=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\Rightarrow G=\left(x^2-5x\right)^2-36\ge36\)
Dấu "=" xảy ra khi x=0 hoặc x=5
Vậy Gmin=36 khi x=0 hoặc x=5
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
a: \(E=x^2-6x+9+x^2-22x+121\)
\(=2x^2-28x+130\)
\(=2\left(x^2-14x+65\right)=2\left(x-7\right)^2+32>=32\)
Dấu '=' xảy ra khi x=7
b: \(x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4>=4\)
=>2/x2-2x+5<=2/4=1/2
=>A>=-1/2
Dấu '=' xảy ra khi x=1
\(a,x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
\(b,\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
\(c,\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
\(d,\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3-x^2\)
\(e,\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
\(f,\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x-12\)
\(g,\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^5-4x^4+8x^3-12x^2-5x^4+x^3-2x^2+3x+10x^3-2x^2+4x-6\)
\(=20x^5-9x^4+19x^3-16x^2+7x-6\)
a. x2(x−2x3)= x3-2x5
b. (x−2)(x−x2+4)= x2-x3+4x-2x+2x2-8= -x3+3x2+2x-8
c. (x2−1)(x2+2x)= x4+2x3-x2-2x
d. (2x−1)(3x+2)(3−x) = (6x2+x-2)(3-x)=18x2-6x3+3x-x2-6+2x =-6x3+17x2+5x-6
e. (x+3)(x2+3x−5)= x3+3x2-5x+3x2+9x-15= x3+6x2+4x-15
f. (xy−2)(x3−2x−6)= x4y-2x2y-6xy-2x3+4x+12
g. (5x3−x2+2x−3)(4x2−x+2)= 20x5-9x4+19x3-12x2+7x-6
b: =>(2x-1)(2x-1+4-2x)=0
=>3(2x-1)=0
=>2x-1=0
=>x=1/2
c: =>(x+1)(x^2-x+1)-x(x+1)=0
=>(x+1)(x-1)^2=0
=>x=1 hoặc x=-1
e: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
h: =>x[(x^2-5)^2-4]=0
=>x(x^2-7)(x^2-3)=0
=>\(x\in\left\{0;\pm\sqrt{7};\pm\sqrt{3}\right\}\)
k: =>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>x=1 hoặc x=-11/2
l: =>x^2(x+1)+(x+1)=0
=>(x+1)(x^2+1)=0
=>x+1=0
=>x=-1
e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)