Đa thức A trong đẳng thức
\(\frac{x^2-2}{x+1}\)=\(\frac{A}{2x+2}\)
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LC
1
24 tháng 3 2020
\(\frac{x^2-2}{x+1}=\frac{A}{2x+2}\Rightarrow A=\frac{\left(x^2-2\right)\left(2x+2\right)}{x+1}\Leftrightarrow A=\frac{\left(x^2-2\right)\left(x+1\right)2}{x+1}\Rightarrow A=2\left(x^2-2\right)\)
24 tháng 11 2016
a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:
\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)
b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)
\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
4 tháng 11 2016
a)\(\frac{\left(x+2\right)P}{x-2}=\frac{\left(x+2\right)^2P}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)^2P}{x^2-4}=\frac{\left(x-1\right)Q}{x^2-4}\Rightarrow\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Rightarrow\frac{P}{Q}=\frac{x-1}{\left(x+2\right)^2}\)
b) Từ gt,ta có :\(\left(x+2\right)\left(x^2-2x+1\right)P=\left(x^2-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2P=\left(x-1\right)\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Rightarrow\frac{P}{Q}=\frac{\left(x+1\right)\left(x-2\right)}{\left(x+2\right)\left(x-1\right)}=\frac{x^2-x-2}{x^2+x-2}\)
Ở đây có nhiều cặp đa thức (P ; Q) thỏa mãn lắm ! Mình xét P/Q để chỉ rằng chúng tỉ lệ với 2 đa thức ở vế phải
Ví dụ : Câu a : P = 2 - 2x thì Q = -2x2 - 8x - 8
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
\(\begin{array}{l}\frac{{2{{\rm{x}}^2} + 1}}{{4{\rm{x}} - 1}} = \frac{{8{{\rm{x}}^3} + 4{\rm{x}}}}{Q}\\ \Rightarrow Q = \frac{{\left( {8{{\rm{x}}^3} + 4{\rm{x}}} \right)\left( {4{\rm{x}} - 1} \right)}}{{2{{\rm{x}}^2} + 1}}\\Q = \frac{{4{\rm{x}}\left( {2{{\rm{x}}^2} + 1} \right)\left( {4{\rm{x}} - 1} \right)}}{{2{{\rm{x}}^2} + 1}}\\Q = 4{\rm{x}}\left( {4{\rm{x}} - 1} \right) = 16{{\rm{x}}^2} - 4{\rm{x}}\end{array}\)
Đáp án D
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
a)
\(\begin{array}{l}P + \frac{1}{{x + 2}} = \frac{x}{{{x^2} - 2{\rm{x}} + 4}}\\P = \frac{x}{{{x^2} - 2{\rm{x}} + 4}} - \frac{1}{{x + 2}}\\P = \frac{{x\left( {x + 2} \right) - {x^2} + 2{\rm{x}} - 4}}{{\left( {{x^2} - 2{\rm{x}} + 4} \right)\left( {x + 2} \right)}}\\P = \frac{{{x^2} + 2{\rm{x}} - {x^2} + 2{\rm{x}} + 4}}{{{x^3} + 8}}\\P = \frac{{4{\rm{x}} - 4}}{{{x^3} + 8}}\end{array}\)
b)
\(\begin{array}{l}P - \frac{{4\left( {x - 2} \right)}}{{x + 2}} = \frac{{16}}{{x - 2}}\\P = \frac{{16}}{{x - 2}} + \frac{{4\left( {x - 2} \right)}}{{x + 2}}\\P = \frac{{16\left( {x + 2} \right) + 4\left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{16{\rm{x}} + 32 + 4{{\rm{x}}^2} - 16{\rm{x}} + 16}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\P = \frac{{4{{\rm{x}}^2} + 48}}{{{x^2} - 4}}\end{array}\)
c)
\(\begin{array}{l}P.\frac{{x - 2}}{{x + 3}} = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}\\ \Rightarrow P = \frac{{{x^2} - 4{\rm{x}} + 4}}{{{x^2} - 9}}.\frac{{x + 3}}{{x - 2}}\\P = \frac{{{{(x - 2)}^2}(x + 3)}}{{(x - 3)(x + 3)(x - 2)}} = \frac{{x - 2}}{{x - 3}}\end{array}\)\(\)
d)
\(\begin{array}{l}P:\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}} = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}\\ \Rightarrow P = \frac{{{x^2} - 4}}{{{x^2} + 3{\rm{x}}}}.\frac{{{x^2} - 9}}{{2{\rm{x}} + 4}}\\P = \frac{{(x - 2)(x + 2)(x - 3)(x + 3)}}{{2{\rm{x}}(x + 3)(x + 2)}}\\P = \frac{{(x - 2)(x - 3)}}{{2{\rm{x}}}}\end{array}\)
9 tháng 9 2023
a) P=\(\dfrac{4x-4}{x^3-8}\)( lấy VP-VT)
b)P=\(\dfrac{4x^2+48}{x^2-4}\) ( chuyển VT và thành VP+VT)
c) P=\(\dfrac{x-2}{x-3}\) ( chuyển VT thành VP.VT là ra)
d) \(\dfrac{\left(x-2\right)\left(x-3\right)}{2x}\)( lấy VP.VT)
16 tháng 12 2020
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