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DT
1
7 tháng 12 2016
\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x^2-4\right)}{x\left(x+2\right)}=\frac{A}{x}\)
\(\Leftrightarrow\frac{4\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x-2\right)}{x}=\frac{A}{x}\)
\(\Leftrightarrow4\left(x-2\right)=A\Leftrightarrow A=4x-8\)
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
TG
0
DT
3
6 tháng 12 2016
\(\frac{4x^2}{x^2+2x}=\frac{A}{x}\)\(\Rightarrow\frac{x\cdot4x}{x\left(x+2\right)}=\frac{A}{x}\)
\(\Rightarrow\frac{4x}{x+2}=\frac{A}{x}\Rightarrow4x^2=A\left(x+2\right)\)\(\Rightarrow A=\frac{4x^2}{x+2}\)
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:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
2 tháng 12 2019
a) MTC: 2xy
Quy đồng: \(\frac{2x-3y}{2xy}\) giữ nguyên
\(\frac{x+2y}{x}=\frac{2y\left(x+2y\right)}{2xy}=\frac{2xy+y^2}{2xy}\)
b) \(\frac{2}{x^2-4x}=\frac{2}{x\left(x-4\right)};\frac{x}{x^2-16}=\frac{x}{\left(x-4\right)\left(x+4\right)}\)
MTC: x (x-4)(x+4)
Quy đồng : \(\frac{2}{x\left(x-4\right)}=\frac{2\left(x+4\right)}{x\left(x-4\right)\left(x+4\right)}=\frac{2x+8}{x\left(x-4\right)\left(x+4\right)}\)
\(\frac{x}{\left(x+4\right)\left(x-4\right)}=\frac{x^2}{x\left(x-4\right)\left(x+4\right)}\)
Học tốt nhé ^3^
18 tháng 11 2019
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
18 tháng 11 2019
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
24 tháng 6 2017
a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:
\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)
\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)
b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)
=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)
c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)
d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6
\(\frac{x}{x^2-2x}=\frac{B}{4x^2-16}\Leftrightarrow\frac{x}{x\left(x-2\right)}=\frac{B}{\left(2x+4\right)\left(2x-4\right)}\)
\(\Leftrightarrow x\left(2x+4\right)\left(2x-4\right)=x\left(x-2\right).B\)
\(\Rightarrow B=\frac{x.\left[2\left(x+2\right)\right].\left[2\left(x-2\right)\right]}{x\left(x-2\right)}=\frac{x.2\left(x+2\right).2\left(x-2\right)}{x\left(x-2\right)}\)
\(B=\frac{x.4\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}=4\left(x+2\right)\)
\(\frac{x}{x^2-2x}=\frac{B}{4x^2-16}\)
\(\frac{x}{x\left(x-2\right)}=\frac{B}{4.\left(x^2-4\right)}\)
\(\frac{1}{x-2}=\frac{B}{4.\left(x^2-4\right)}\)
\(\Rightarrow B.\left(x-2\right)=4.\left(x-2\right)\left(x+2\right)\)
\(B=4.\left(x+2\right)\)
\(B=4x+8\)