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1,\(x^4-x=0\\ ->x\left(x-1\right)\left(x^2+x+1\right)=0\\ ->\left(......\right)\)
2\(x^4-x^2=0\\ ->x^2\left(x^2-1\right)\\ ->x^2\left(x-1\right)\left(x+1\right)\\ ->......\)
3,\(x^5+x^2\\ ->x^2\left(x^3+1\right)\\ ->x^2\left(x+1\right)\left(x^2-x+1\right)\\ ->.......\)
4\(3x\left(x-20\right)-x+20=0->\left(3x-1\right)\left(x-20\right)=0->.....\)
1: \(\Leftrightarrow3x+4x=4\)
=>7x=4
hay x=4/7
2: \(\Leftrightarrow3x-5x-5^3:5^2=0\)
=>-2x=5
=>x=-5/2
1) \(-6x^2-x+7=0\)
\(\Leftrightarrow-6x^2+6x-7x+7=0\)
\(\Leftrightarrow\left(-6x^2+6x\right)-\left(7x-7\right)=0\)
\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(-6x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-6x-7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)
2) \(-4x^2-5x+9=0\)
\(\Leftrightarrow-4x^2+4x-9x+9=0\)
\(\Leftrightarrow\left(-4x^2+4x\right)-\left(9x-9\right)=0\)
\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(-4x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=1\end{matrix}\right.\)
3) \(x^2+3x-4=0\)
\(\Leftrightarrow x^2-x+4x-4=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
4) \(x^2-6x-7=0\)
\(\Leftrightarrow x^2+x-7x-7=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(7x+7\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
5) \(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow\left(x^2+x\right)+\left(4x+4\right)=0\)
\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
a) \(-6x^2-x+7=0\)
\(\Leftrightarrow-6x^2+6x-7x+7=0\)
\(\Leftrightarrow-6x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-7}{6}\end{matrix}\right.\)
b) \(-4x^2-5x+9=0\)
\(\Leftrightarrow-4x^2+4x-9x+9=0\)
\(\Leftrightarrow-4x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2,25\end{matrix}\right.\)
c) \(x^2+3x-4=0\)
\(\Leftrightarrow x^2-x+4x-4=0\)
\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
d) \(x^2-6x-7=0\)
\(\Leftrightarrow x^2+x-7x-7=0\)
\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=7\end{matrix}\right.\)
e) \(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
GIẢI PHƯƠNG TRÌNH :
\(\left(3x+5\right)^2-\left(2x+1\right)^2=0\)
giải hộ e vs ạ !!!
e cảm ơn nhìu :3
(3x + 5)2 - (2x + 1)2 = 0
<=> (3x + 5 + 2x + 1)(3x + 5 - 2x - 1) = 0
<=> (5x + 6)(x + 4) = 0
<=> \(\orbr{\begin{cases}x=-\frac{6}{5}\\x=-4\end{cases}}\)
Vậy \(x\in\left\{-\frac{6}{5};-4\right\}\)là nghiệm phương trình
\(\left(3x+5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(3x+5+2x+1\right)\left(3x+5-2x-1\right)=0\)
\(\Leftrightarrow\left(5x+6\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=-\frac{6}{5}\)
Vậy tập nghiệm của phương trình là S = { -4 ; -6/5 }
`x^2+2x+3>2`
`<=>x^2+2x+1>0`
`<=>(x+1)^2>0`
`<=>x+1 ne 0`
`<=>x ne -1`
`(x+5)(3x^2+2)>0`
Vì `3x^2+2>=2>0`
`=>x+5>0<=>x>-5`
c) Ta có: \(21x-10x^2+9< 0\)
\(\Leftrightarrow10x^2-21x-9>0\)
\(\Leftrightarrow x^2-\dfrac{21}{10}x-\dfrac{9}{10}>0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{21}{20}+\dfrac{441}{400}>\dfrac{801}{400}\)
\(\Leftrightarrow\left(x-\dfrac{21}{20}\right)^2>\dfrac{801}{400}\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3\sqrt{89}+21}{20}\\x< \dfrac{-3\sqrt{89}+21}{20}\end{matrix}\right.\)
A=26x2+y(2x+y)-10x(x+y)
A=26x2+2xy+y2-10x2-10xy
A=16x2-8xy+y2 =(4x)2-2.4x.y+y2 =(4x-y)2
Thay x=0,25y,ta có: A=(4.0,25y - y)2=(y-y)2=0
B=x3+6x2y+12xy2+8y3
B=x3+3x22y+3x(2y)2+(2y)3 =(x+2y)3
Có x+2y=-5 ⇒ x=-5-2y
Thay x=-5-2y vào, ta có B=(-5-2y+2y)3=(-5)3=-125
xét :
|2x - 1| = 2x - 1 nếu 2x - 1 >0 hay x > \(\frac{1}{2}\)
=> 4(2x - 1) - x = 2 <=> 8x - 4 -x = 2 <=> 7x = 6 => x = \(\frac{6}{7}\)
( thỏa mãn ĐK )
|2x -1| = 1 - 2x nếu 2x - 1 < 0 hay x < \(\frac{1}{2}\)
=> 4.( 1 - 2x) - x = 2 <=> 4 - 8x -x = 2<=> 2 =9x => x = \(\frac{2}{9}\) (thỏa mãn ĐK)
vậy phương trình có nghiệm s = { \(\frac{1}{2},\frac{2}{9}\) }
g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
\(x^4-10x^3+35x^2+24>0\)
\(\Leftrightarrow x^4-2.5.x^3+\left(5x\right)^2+10x^2+24>0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+10x^2+24>0\)
\(\Leftrightarrow x^2\left(x-5\right)^2+10x^2+24>0\)(luôn đúng)
Vậy nghiệm của bất phương trình \(x\in R\)
Ta có 27^5=3^3^5=3^15
243^3=3^5^3=3^15
Vậy A=B
2^300=2^(3.100)=2^3^100=8^100
3^200=3^(2.100)=3^2^100=9^100
Vậy A<B