2x2+3x(x-1)x(x+1)-5x(x+1)=
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10 tháng 12 2021
\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)
Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)
Do đó \(x\in\left\{1;2\right\}\)
10 tháng 12 2021
\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)
Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)
Vậy PT có nghiệm \(x=4\)
2
10 tháng 1 2023
`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):}`
21 tháng 5 2022
a: \(f\left(x\right)+g\left(x\right)=2x^3-2x^2+4x\)
b: \(f\left(x\right)-g\left(x\right)=-2x^2+2x+2\)
19 tháng 10 2021
1, \(2x^2+4x=2x\left(x+2\right)\)
2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)
3) \(5x^2\left(x-2y\right)+15x\left(x-2y\right)=\left(5x^2+15x\right)\left(x-2y\right)=5x\left(x+3\right)\left(x-2y\right)\)
4) \(3\left(x-y\right)+5x\left(y-x\right)=\left(x-y\right)\left(3-5x\right)\)
5) \(5x^2-10x=5x\left(x-2\right)\)
6) \(3x-6y=3\left(x-2y\right)\)
7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)
8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)
9) \(x\left(y-1\right)-y\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)
10) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)
19 tháng 10 2021
\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)
19 tháng 2 2020
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
19 tháng 2 2020
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
30 tháng 4 2023
\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
30 tháng 4 2023
\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)
9 tháng 10 2021
a)=\(3x^3-15x^2+21x\)
b)\(=-2x^4y-10x^2y+2xy\)
c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)
d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)
e)\(=x^2-4y^2\)
f)\(=-2x^2y^3+y-3\)
g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)
h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)
i)\(=x^2-x-3\)
j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)
\(2x^2+3x\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)\)
\(=2x^2+3x^3-3x-5x^2-5x\)
\(=3x^3-3x^2-8x\)
`2x^2 + 3x (x-1) (x+1) -5x(x+1)`
`= 2x^2 +3x (x^2-1) - 5x^2 -5x`
`= 2x^2 +3x^3 - 3x - 5x^2 - 5x`
`= 3x^3 - 3x^2 - 8x`