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MN
17 tháng 2 2021
\(a.\)
\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)
\(b.\)
\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)
17 tháng 2 2021
a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
\(=\dfrac{4x+1}{4x-1}\)
b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)
\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{y-2x}{y+2x}\)
TH
0
TH
2
14 tháng 12 2021
a) x ≠ -5.
b) Ta có P = ( x + 5 ) 2 x + 5 = x + 5
c) Ta có P = 1 Û x = -4 (TMĐK)
d) Ta có P = 0 Û x = -5 (loại). Do vậy x ∈ ∅ .
NB
2
KV
13 tháng 1 2022
\(A=\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
a) ĐKXĐ:
\(\left(4x-1\right)^2\ne0\Leftrightarrow4x-1\ne0\Leftrightarrow x\ne\dfrac{1}{4}\)
b) \(A=\dfrac{\left(4x+1\right)\left(4x-1\right)}{\left(4x-1\right)^2}=\dfrac{4x+1}{4x-1}\)
DT
13 tháng 1 2022
a,đkxđ : \(16x^2\ne0\Leftrightarrow x\ne0\)
b, \(\dfrac{16x^2}{1}-\dfrac{1}{16x^2}-\dfrac{8x}{1}+1=\dfrac{256x^4}{16x^2}-\dfrac{1}{16x^2}-\dfrac{128x^3}{16x^2}+\dfrac{16x^2}{16x^2}\)
\(=\dfrac{256x^4-1-128x^3+16x^2}{16x^2}=\dfrac{256x^4-128x^3+16x^2-1}{16x^2}\)
\(=\dfrac{\left(256x^4-128x^3+16^2\right)-1}{16x^2}=\dfrac{16x^2\left(16x^2-8x+1\right)-1}{16x^2}\)
\(=\dfrac{\left(4x\right)^2.\left(\left(4x\right)^2-8x+1\right)-1}{16x^2}=\dfrac{\left(4x\right)^2.\left(4x-1\right)^2-1}{16x^2}\)
\(=\dfrac{\left(16x^2-4x\right)^2-1}{16x^2}=\dfrac{\left(16x^2-4x-1\right)\left(16x^2-4x+1\right)}{16x^2}\)
\(=\dfrac{\left(\left(4x\right)^2-4x-1\right)\left(\left(4x\right)^2-4x+1\right)}{\left(4x\right)^2}\)
19 tháng 7 2016
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
UN
0
2T
11 tháng 8 2019
\(A=16x^2+8x+3=\left(4x\right)^2+2.4x.1+1+2\)
\(=\left(4x+1\right)^2+2>0\forall x\)
NH
2
11 tháng 10 2021
\(=\left(x-3\right)\left(8x^3-16x^2\right)=8x^2\left(x-2\right)\left(x-3\right)\)
11 tháng 10 2021
\(8x^3\left(x-3\right)+16x^2\left(3-x\right)\)
\(=8x^3\left(x-3\right)-16x^2\left(x-3\right)\)
\(=8x^2\left(x-3\right)\left(x-2\right)\)