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Bài 1:
=>x^4-x^3+5x^2+x^2-x+5+n-5 chia hết cho x^2-x+5
=>n-5=0
=>n=5
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
Bài 1:
a: \(=\dfrac{3x+5-5}{2x}=\dfrac{3x}{2x}=\dfrac{3}{2}\)
b: \(=\dfrac{2x}{x+3}\cdot\dfrac{\left(x+3\right)\left(x-3\right)}{x}=2\left(x-3\right)\)
Bài 2:
=>x^3+x+2x^2+2+a-2 chia hết cho x^2+1
=>a-2=0
=>a=2
\(x^4-x^3+6x^2-x+n\)\(:\)\(x^2-x+5\)\(=x^2+1\)dư \(n-5\)
Để \(x^4-x^3+6x^2-x+n\) \(⋮\)\(x^2-x+5\) thì \(n-5=0\)hay \(n=5\)
Bài 1:
Ta có: \(9(x-1)^2-4(2x+3)^2=(3x-3)^2-(4x+6)^2\)
\(=(3x-3-4x-6)(3x-3+4x+6)=-(x+9)(7x+3)\)
Bài 2:
Có: \(x^2-x+\frac{9}{20}=x^2-2x.\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\left(x-\frac{1}{2}\right)^2+\frac{1}{5}\)
Ta thấy \(\left(x-\frac{1}{2}\right)^2\geq 0\forall x\in\mathbb{R}\Rightarrow x^2-x+\frac{9}{20}\geq \frac{1}{5}>0\forall x\in\mathbb{R}\)
Ta có đpcm.
Bài 3:
Thực hiện phân tích:
\(f(x)=x^3-8x^2+ax-5=x(x^2-3x+1)-5(x^2-3x+1)+ax-16x\)
\(=(x-5)(x^2-3x+1)+ax-16x\)
Thấy rằng bậc của \(ax-16x\) nhỏ hơn bậc của $g(x)$ nên $ax-16x$ là dư của $f(x)$ cho $g(x)$
Để \(f(x)\vdots g(x)\Rightarrow ax-16x=0\forall x\Rightarrow a=16\)
Bài 4:
Để \(\overline{2017x}\vdots 12\Leftrightarrow \left\{\begin{matrix} \overline{2017x}\vdots 3(1)\\ \overline{2017x}\vdots 4(2)\end{matrix}\right.\)
\((1)\Leftrightarrow 2+0+1+7+x\vdots 3\Leftrightarrow 10+x\vdots 3\Leftrightarrow x+1\vdots 3\)
\((2)\Leftrightarrow \overline{7x}\vdots 4\Rightarrow x\in\left\{2;6\right\}\)
Từ hai điều trên suy ra \(x=2\)
Bài 5:
Ta có: \(x+\frac{1}{x}=\sqrt{2017}\Rightarrow \left(x+\frac{1}{x}\right)^2=2017\Leftrightarrow x^2+\frac{1}{x^2}+2=2017\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=2015\)
Như vậy: \(A=3x^2-5+\frac{3}{x^2}=3\left(x^2+\frac{1}{x^2}\right)-5=3.2015-5=6040\)
Bài 6:
Đặt \(\left\{\begin{matrix} x+y+z=a\\ xy+yz+xz=b\end{matrix}\right.\). ĐKĐB tương đương với:
\(\left\{\begin{matrix} a^2-2b=3\\ a+b=6\rightarrow b=6-a\end{matrix}\right.\)
\(\Rightarrow a^2-2(6-a)=3\Leftrightarrow a^2-2a+15=0\Leftrightarrow (a+5)(a-3)=0\Leftrightarrow a=3\)
(do \(a\in\mathbb{R}^+\))
Kéo theo \(b=6-a=3\Rightarrow x^2+y^2+z^2=xy+yz+xz\)
Theo BĐT AM-GM thì \(x^2+y^2+z^2\geq xy+yz+xz\)
Dấu bằng xảy ra khi \(x=y=z\Rightarrow x=y=z=1\) do \(x+y+z=3\)