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Ta có
\(x^2-2x+2=\left(x-1\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
=>\(\left(x-1\right)^2+1>0\) với mọi x
tick nha
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- x2.(x3-x2+x-1)
- x.( x3-3x2-1)+3
- x.(x2-xy-y2)
Tìm x:
x3-16x = 0
=> x.(x2-16) = 0
=> x = 0 hay x2-16 = 0
=> x = 0 hay x2 = 0+16
=> x = 0 hay x2 = 16
=> x = 0 hay x = 4 hay x = -4
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a) \(x^4-x^2+3=\left[\left(x^2\right)^2-2\cdot x^2\cdot\frac{1}{2}+\frac{1}{4}\right]+\frac{11}{4}=\left(x^2-\frac{1}{2}\right)^2+\frac{11}{4}>0\)
=>đpcm
b) \(x^2-x+1=\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=>đpcm
c) \(x^2+x+2=\left(x^2+2\cdot x+\frac{1}{2}+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=>đpcm
d) \(\left(x+3\right)\left(x-11\right)+20\)
\(=x^2-11x+3x-33+20\)
\(=x^2-8x-13\)
\(=\left(x^2-8x+16\right)-29=\left(x+4\right)^2-29\)
Xem lại đề
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(x-3)(x-5)+4
=x2-8x+19
=x2-8x+16+3
=(x-4)2+3
ta thấy
(x-4)2 lon hon hoac bang 0 voi moi x
=>(x-4)2+3 lon hon hoac bang 3 voi moi x
=>(x-4)2+3 >0
xét x<3\(\Rightarrow\left(X-3\right)< 0,\left(x-5\right)< 0\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)>O\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)+4>4\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)>0\)
xét x=3\(\Rightarrow\left(x-3\right)\left(x-5\right)+4=0+4=4>0\)
xét x=4\(\Rightarrow\left(x-3\right)\left(x-5\right)+4=1+4=5>0\)
Xét x=5\(\Rightarrow\left(x-3\right)\left(x-5\right)+4=0+4>0\)
XÉt x>5\(\Rightarrow\left(X-3\right)\left(x-5\right)>0\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)+4>0+4=4\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)+4>0\)VỚI \(\forall x\in Z\)