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a.
= (2x)2 2.2x+1 +2
=(2x+1)2+2(luôn dương)
b. =x2 +2x.1/2 +1/4+3/4
= (x+1/2)2+3/4 (luôn dương)
c. 2C=(2x)2-4x1/2 +1/4+7/4
= (2x-1/2)2+7/4
r bạn suy ra C luôn dương :>
A= 4x2-4x+3 = 4x2-4x+1+2 = (4x2-4x+1)+2 = (2x-1)2 +2
Vì (2x-1)2 >=0 với mọi x nên (2x-1)2 +2 >0 với mọi x
B= x2+x+1 = x2+x+1/4 +3/4 = (x2+x+1/4) +3/4 = (x+1/2)2 +3/4
Vì (x+1/2)2 >=0 với mọi x nên (x+1/2)2 +3/4 > 0 với mọi x
C=2x2-x+2 = 2(x2-1/2x+1) = 2(x2-1/2x + 1/16 +15/16) = 2[(x-1/4)2 + 15/16] = 2(x-1/4)2 + 15/8
Vì 2(x-1/4)2 >=0 với mọi x nên 2(x-1/4)2 + 15/8 > 0 với mọi x
a.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
1)
a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^3-x+3=3\)
=>đpcm
b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2=-24\)
=>đpcm
2,
a) \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
\(\Leftrightarrow x=-2\)
b) \(0,6x\left(x-0,5\right)-0,3x\left(2x+1,3\right)=0,138\)
\(\Leftrightarrow0,6x^2-0,3x-0,6x^2-0,39x=0,138\)
\(\Leftrightarrow-0,69x=0,138\)
\(\Leftrightarrow x=-0,2\)
Câu 1:
a)\(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^2-x+3\right)\)
\(=2x^2+x-x^3-2x^2+x^2-x+3\)
\(=x^3+3\)(ko thể CM)
b)\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)(đpcm)
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
\(2005^3-1=\left(2005-1\right)\left(2005^2+2005+1\right)=2004\times\left(2005^2+2005+1\right)⋮2004\left(\text{đ}pcm\right)\)
\(2005^3+125=\left(2005+5\right)\left(2005^2-2005\times5+5^2\right)=2010\times\left(2005^2-2005\times5+5^2\right)⋮2010\)
\(x^6+1=\left(x^2+1\right)\left(x^4-x^2+1\right)⋮x^2+1\left(\text{đ}pcm\right)\)
\(x^6-y^6=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)⋮x-y;x+y\left(\text{đ}pcm\right)\)
\(A=\frac{2}{x}+\frac{3}{x+2}+\frac{x-4}{x^2+2x}\)
\(=\frac{2}{x}+\frac{3}{x+2}+\frac{x-4}{x\left(x+2\right)}\)
\(=\frac{2\left(x+2\right)}{x\left(x+2\right)}+\frac{3x}{x\left(x+2\right)}+\frac{x-4}{x\left(x+2\right)}\)
\(=\frac{2x+4+3x+x-4}{x\left(x+2\right)}\)
\(=\frac{6x}{x\left(x+2\right)}=\frac{6}{x+2}=B\)
=> đpcm
Bài làm :
Ta có :
\(A=\frac{2}{x}+\frac{3}{x+2}+\frac{x-4}{x^2+2x}\)
\(=\frac{2\left(x+2\right)}{x\left(x+2\right)}+\frac{3x}{x\left(x+2\right)}+\frac{x-4}{x\left(x+2\right)}\)
\(=\frac{2x+4+3x+x-4}{x\left(x+2\right)}\)
\(=\frac{6x}{x\left(x+2\right)}\)
\(=\frac{6}{x+2}=B\)
=> Điều phải chứng minh