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a2 - 6a + 9 = (a-3)2
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1-4x+4x2 = (1-2x)2
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(a+b)2 - 1 = (a+b-1)(a+b+1)
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4x2- 9 = (2x-3)(2x+3)
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25x2 - 20xy + 4y2 = (5x - 2y)2
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\times\left(2x+3\right)\)
\(\frac{\left(x-3\right)}{x^2+4x+9}+2+\frac{x^2+4x+9}{x-3}=0\)
\(x^2+4x+9=\left(x+2\right)^2+5\ge5\)
x>3 hiển nhiên vô nghiệm
xét x<3
\(\frac{!\left(x-3\right)!}{x^2+4x+9}+\frac{x^2+4x+9}{!x-3!}\ge2\)
vậy pt chỉ nghiệm
khi \(\frac{!\left(x-3\right)!}{x^2+4x+9}=\frac{x^2+4x+9}{!x-3!}\Leftrightarrow x^2+4x+9=!x-3!\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\)
25-24=1
=>
x=-3 loại
x=-2 nhận
Đk:....
Đặt \(\hept{\begin{cases}a=x-3\\b=x^2+4x+9\end{cases}}\) pt trở thành
\(\frac{a}{b}+2+\frac{b}{a}=0\)\(\Leftrightarrow\frac{a^2}{ab}+\frac{2ab}{ab}+\frac{b^2}{ab}=0\)
\(\Leftrightarrow\frac{a^2+2ab+b^2}{ab}=0\)\(\Leftrightarrow\left(a+b\right)^2=0\)
\(\Leftrightarrow a=-b\)\(\Leftrightarrow x-3=-\left(x^2+4x+9\right)\)
\(\Leftrightarrow x-3=-x^2-4x-9\)\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)
a) \(x^2-4x=0\)
\(x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
b) \(4x^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\left(2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)
c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
d) \(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-2\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)
\(\left(x-3\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)
\(x^2-4x=0\)
\(x.\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)
\(4x^2-9=0\)
\(2^2x^2-9=0\)
\(\left(2x\right)^2-9=0\)
\(\left(2x\right)^2-3^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)
\(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\left(x-3\right)\cdot\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)
\(x\left(2x+9\right)-4x-18=0\)
\(x\left(2x+9\right)-\left(4x+18\right)=0\)
\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)
\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)
\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)
\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)
\(\)
\(\frac{1}{4x^2-12x+9}-\frac{3}{9-4x^2}=\frac{4}{4x^2+12x+9}\)
\(\Leftrightarrow\frac{-1}{\left(3-2x\right)^2}-\frac{3}{\left(3-2x\right)\left(3+2x\right)}=\frac{4}{\left(2x+3\right)^2}\)
\(\Leftrightarrow-4x^2-12x-9-27+12x^2-16x^2+48x-36=0\)
\(\Leftrightarrow-8x^2+36x-72=0\)
Rút -4 ra ngoài \(\Leftrightarrow2x^2-9x+18=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x-6=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=3\\x=6\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=6\end{cases}\left(tmđk\right)}\)
a) (x2-6xy+9y2):(3y-x)
= (x-3y)2:(3y-x)
=(3y-x)2:(3y-x)
= 3y-x
b) (8x3-1):(4x2+2x+1)
=[(2x)3-1]:(4x2+2x+1)
= (2x-1)(4x2+2x+1):(4x2+2x+1)
= 2x-1
c) (4x4-9):(2x2-3)
=(2x2-3)(2x2+3):(2x2-3)
=2x2+3
d) (8x3-27):(4x2+6x+9)
=(2x-3)(4x2+6x+9):(4x2+6x+9)
=2x-3
x2 +6xx+9=4x2-4x+1
<=>3x2-10x -8=0
<=>3x2-12x+2x-8 =0
<=>3x(x-4)+2(x-4)=0
<=>(3x+2)(x-4)=0
<=>x =-2/3 hoặc x=4
x2 +6xx+9=4x2-4x+1
<=>3x2-10x -8=0
<=>3x2-12x+2x-8 =0
<=>3x(x-4)+2(x-4)=0
<=>(3x+2)(x-4)=0
<=>x =-2/3 hoặc x=4
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
4x2 - 9 =0
<=> ( 2x +3 ) ( 2x- 3 ) =0
<=> \(\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}2x=-3\\2x=3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}\)