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`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
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`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
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`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)
\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)
\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)
\(\left(x+2\right)^2-\left(4x-2\right)^2=0\)
\(\left(x+2-4x+2\right)\left(x+2+4x-2\right)=0\)
\(6x\left(-3x+4\right)=0\)
\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)
*) \(6x=0\)
\(x=0\)
*) \(-3x+4=0\)
\(3x=4\)
\(x=\dfrac{4}{3}\)
Vậy \(x=0;x=\dfrac{4}{3}\)
b) \(4x\left(x-2019\right)-x+2019=0\)
\(4x\left(x-2019\right)-\left(x-2019\right)=0\)
\(\left(x-2019\right)\left(4x-1\right)=0\)
\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)
*) \(x-2019=0\)
\(x=2019\)
*) \(4x-1=0\)
\(4x=1\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4};x=2019\)
A= 2006 X 2008 - 20072
A = 2006 . 2008 - 2007 . 2007
A = 2006 . ( 2007 + 1 ) - 2007 . ( 2006 + 1 )
A = 2006 . 2007 + 2006 - 2007 . 2006 + 2007
A = -1
B= 2016 X 2018 - 20172
B= 2016 . 2018 - 2017 . 2017
B = 2016 . ( 2017 + 1 ) - 2017 . ( 2016 + 1 )
B = 2016 . 2017 + 2016 - 2017 . 2016 + 2017
B = -1
Bài 1:
a) x2 + y2 - 2x + 10y + 26 = 0
<=> (x2 - 2x + 1) + (y2 + 10y + 25) = 0
<=> (x - 1)2 + (y + 5)2 = 0 (*)
Vì (x - 1)2 \(\ge\)0; (y + 5)2 \(\ge\)0
(*) <=> x - 1 = 0 hay y + 5 = 0
<=> x = 1 I <=> y = -5
b) 64x3 + 48x2 + 12x + 1 = 27
<=> 64x3 - 32x2 + 80x2 - 40x + 52x + 1 - 27 = 0
<=> 64x3 - 32x2 + 80x2 - 40x + 52x - 26 = 0
<=> 64x2(x - \(\frac{1}{2}\)) + 80x(x - \(\frac{1}{2}\)) + 52(x - \(\frac{1}{2}\)) = 0
<=> (x - \(\frac{1}{2}\))(64x2 + 80x + 52) = 0
<=> (x - \(\frac{1}{2}\))[(8x)2 + 2.8x.5 + 52 + 27) = 0
<=> (x - \(\frac{1}{2}\))[(8x + 5)2 + 27) = 0
<=> x - \(\frac{1}{2}\)= 0 (vì (8x + 5)2 + 27 > 0
<=> x = \(\frac{1}{2}\)
Bài 2:
a) x2 + 2xy + y2
= (x + y)2
= 32 = 9
b) x2 - 2xy + y2
= x2 + 2xy + y2 - 4xy
= (x + y)2 - 4xy
= 32 - 4.(-10)
= 9 + 40 = 49
c) x2 + y2
= x2 + 2xy + y2 - 2xy
= (x + y)2 - 2xy
= 32 - 2.(-10)
= 9 + 20 = 29
5x2 - 4(x2 - 2x + 1) - 5 = 0
=> 5x2 - 4x2 + 8x - 4 - 5 = 0
=> x2 + 8x - 9 = 0
=> x2 + 9x - x - 9 = 0
=> x(x + 9) - (x + 9) = 0
=> (x + 9)(x - 1) = 0
=> x + 9 = 0 => x = -9
hoặc x - 1 = 0 = > x = 1
Vậy x = -9, x = 1
\(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\left(5x^2-5\right)-4\left(x^2-2.1.x+1^2\right)=0\)
\(5\left(x^2-1\right)-4\left(x-1\right)^2=0\)
\(5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)\left(x-1\right)=0\)
\(\left[5\left(x+1\right)-4\left(x-1\right)\right]\left(x-1\right)=0\)
\(\left(5x+5-4x+4\right)\left(x-1\right)=0\)
\(\left(x+9\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-9\\x=1\end{cases}}.\)
Kệ cái thằng ấy, nó có trả lời đc câu nào tử tế đâu. Câu **** ý mà, kệ nó đi
a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne\pm2\end{cases}}\)
b) \(D=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right)\div\left(\frac{x-3}{2-x}\right)\)
\(\Leftrightarrow D=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2-x}{x-3}\)
\(\Leftrightarrow D=\frac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2+x\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x^2+8x}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x}{x-3}\)
c) Để D = 0
\(\Leftrightarrow\frac{4x}{x-3}=0\)
\(\Leftrightarrow4x=0\)
\(\Leftrightarrow x=0\)
Vậy để D = 0 \(\Leftrightarrow\)x = 0
d) Khi \(\left|2x-1\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\1-2x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=-2\left(ktm\right)\end{cases}}\)
Vậy khi \(\left|2x-1\right|=5\Leftrightarrow D\in\varnothing\)
a) (x + 3)2 - (x - 2)2 = 2x
=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x
=> 5(2x + 1) = 2x
=> 10x + 5 = 2x
=> 10x - 2x = -5
=> 8x = -5
=> x = -5/8
b) 7x(x - 2) = x - 2
=> 7x(x - 2) - (x - 2) = 0
=> (7x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)
c) 8x3 - 12x2 + 6x - 1 = 0
=> (2x - 1)3 = 0
=> 2x - 1 = 0
=> 2x = 1
=> x = 1/2
a) (x - 4)^3 = (x + 4)(x^2 - x - 16)
<=> x^3 - 8x^2 + 16x - 4x^2 + 32x - 64 = x^3 - x^2 - 16x + 4x^2 - 4x - 64
<=> -12x^2 + 48x - 64 = 3x^2 - 20
<=> 12x^2 - 48x + 64 + 3x^2 - 20 = 0
<=> 15x^2 - 68x = 0
<=> x(15x - 68) = 0
<=> x = 0 hoặc 15x - 68 = 0
<=> x = 0 hoặc 15x = 68
<=> x = 0 hoặc x = 68/15
b) \(\frac{x+2}{x}=\frac{x^2+5x+4}{x^2+2x}+\frac{x}{x+2}\) (ĐKXĐ: x khác 0, x khác -2)
<=> \(\frac{x+2}{x}=\frac{\left(x+1\right)\left(x+4\right)}{x\left(x+2\right)}=\frac{x}{x+2}\)
<=> x(x + 2) + 2(x + 2) = (x + 1)(x + 4) + x^2
<=> x^2 + 2x + 2x + 4 = x^2 + 4x + x + 4 + x^2
<=> x^2 + 4x + 4 = 2x^2 + 5x + 4
<=> x^2 + 4x = 2x^2 + 5x
<=> x^2 + 4x - 2x^2 - 5x = 0
<=> -x^2 - x = 0
<=> x(x + 1) = 0
<=> x = 0 hoặc x + 1 = 0
<=> x = 0 (ktm) hoặc x = -1 (tm)
Vậy: nghiệm của phương trình là: -1
`(x^{2}+2x+3)(x^{2}+x-12)=0(***)`
Vì : `x^{2}+2x+3=(x+1)^{2}+2\ge2>0` với mọi `x`
Hay `x^{2}+2x+3\ne 0` với mọi `x`
Do đó `(***)` xảy ra `<=>x^{2}+x-12=0`
`<=>(x+4)(x-3)=0`
`=>x=-4` hoặc `x=3`
Vậy `S={-4;3}`