Cho x+\(\dfrac{1}{2}\) =3. Tính x\(^2\)+\(\dfrac{1}{x^2}\)?
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\(4)25-1^2=\left(5-1\right)\left(5+1\right)=4.6=24\\ 5)a^2-9=a^2-3^2=\left(a-3\right)\left(a+3\right)\)
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Ta có \(P=2x^2-4x+7=2\left(x^2-2x+\dfrac{7}{2}\right)=2\left(x^2-2x+1+\dfrac{5}{2}\right)\) \(=2\left(x-1\right)^2+5\)
Mà \(2\left(x-1\right)^2\ge0\Leftrightarrow2\left(x-1\right)^2+5\ge5>0\) hay \(P>0\) (đpcm)
Mặt khác \(H=-2x^2-16x+38=-2\left(x^2+8x-19\right)\) \(=-2\left(x^2+8x+16-35\right)=-2\left(x+4\right)^2-70\)
Mà \(-2\left(x+4\right)^2\le0\Leftrightarrow-2\left(x+4\right)^2-70\le-70< 0\) nên ta có \(H< 0\) (đpcm)
`2x^2 - 4x + 7`
`<=> 2x^2 - 4x + 2 + 5`
`<=> 2.(x-1)^2 + 5`
Mà : \(2\left(x-1\right)^2\ge0\forall x\)
`=>` \(2\left(x-1\right)^2+5\ge0\forall x\)
Vậy `2x^2 -4x+7` luôn dương với mọi `x`
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`-2x^2 - 16x + 38`
`<=> -2.(x^2+8x-19)`
Mà :\(x^2+8x-19\ge0\forall x\)
`=>` \(-2x.\left(x^2+8x-19\right)\le0\forall x\)
Vậy `-2x^2-16x+38` luôn âm với mọi `x`
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\(\left(x+2\right)^2=x^2+4x+4\)
Áp dụng hằng đẳng thức
\(\left(x+y\right)^2=x^2+2xy+y^2\)
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a) Điều kiện \(x\ne0\). Ta có
\(A=\dfrac{x^2-x+2}{x^2}=1-\dfrac{1}{x}+\dfrac{2}{x^2}\)
Đặt \(\dfrac{1}{x}=p\) , khi đó \(A=2p^2-p+1\)
Lại có \(A=2p^2-p+1=2\left(p^2-\dfrac{p}{2}+\dfrac{1}{2}\right)\) \(=2\left(p^2-2p.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{7}{16}\right)=2\left[\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{16}\right]\) \(=2\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\)
Mà \(2\left(p-\dfrac{1}{4}\right)^2\ge0\Leftrightarrow2\left(p-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\) hay \(min_A=\dfrac{7}{8}\)
Dấu "=" xảy ra khi \(p-\dfrac{1}{4}=0\Leftrightarrow p=\dfrac{1}{4}\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{4}\Leftrightarrow x=4\)
Vậy GTNN của A là \(\dfrac{7}{8}\) khi \(x=4\)
b) \(B=\dfrac{x^2-2x+3}{\left(x+1\right)^2}=\dfrac{\left(x+1\right)^2-4x+2}{\left(x+1\right)^2}=1-\dfrac{4x+2}{\left(x+1\right)^2}\) \(=1-\dfrac{4\left(x+1\right)-2}{\left(x+1\right)^2}=1-\dfrac{4}{x+1}+\dfrac{2}{\left(x+1\right)^2}\)
Đến đây đặt \(\dfrac{1}{x+1}=t\) thì \(B=2t^2-4t+1\) và làm tương tự như câu a.
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\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x\right)^2-2.\left(\dfrac{1}{2}x\right).\left(\dfrac{1}{3}y\right)+\left(\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)
\(4-6x+\dfrac{9}{4}x^2=2^2-2.2.\dfrac{2}{3}x+\left(\dfrac{2}{3}x\right)^2=\left(2-\dfrac{2}{3}x\right)^2\)
\(x^3+3x^2+3x+1=\left(x+1\right)^3\)
\(x^6+3x^5+3x^4+x^3=x^3\left(x^3+3x^2+3x+1\right)=x^3\left(x+1\right)^3\)
a, \(8-12x+6x^2-x^3=\left(2-x\right)^3\)
b, \(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)
d, \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
e, \(x^3\left(x^3+3x^2+3x+1\right)=x^3\left(x+1\right)^3\)
Mình nghĩ đề là Cho \(x+\dfrac{1}{x}=3\) phù hợp hơn nhé.
Ta có: \(x+\dfrac{1}{x}=3\)\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=3^2\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=9\Leftrightarrow x^2+2+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=7\)
Vậy \(x^2+\dfrac{1}{x^2}=7\)
:)
x = 2,5
thay vào, dùng máy tính bấm ra nhé
2,5 2 + 1: 2,5 2