x(x-1)+(x+1)(3-x)=2
tìm x
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11 tháng 4 2023
P>1/3
=>P-1/3>0
=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{3}>0\)
=>\(\dfrac{3\sqrt{x}-3-\sqrt{x}-2}{3\left(\sqrt{x}+2\right)}>0\)
=>2 căn x-5>0
=>x>25/4
TT
1
TD
2
L
15 tháng 12 2021
\(a,2x\left(x^3-3\right)-2x^4=18\\ 2x^4-6x-2x^4=18\\ -6x=18\\ x=-3\)
\(b,9x\left(4-x\right)+\left(3x+1\right)^2=2\\ 36x-9x^2+9x^2+6x+1=2\\ 42x=2-1\\ 42x=1\\ x=\dfrac{1}{42}\)
15 tháng 12 2021
\(a,\Leftrightarrow2x^4-3x-2x^4=18\Leftrightarrow-3x=18\Leftrightarrow x=-6\\ b,\Leftrightarrow36x-9x^2+9x^2+6x+1=2\\ \Leftrightarrow42x=1\Leftrightarrow x=\dfrac{1}{42}\)
6 tháng 2 2022
\(\Leftrightarrow a\cdot\dfrac{13}{15}=\dfrac{28}{13}:2=\dfrac{14}{13}\)
=>\(a=\dfrac{14}{13}:\dfrac{13}{15}=\dfrac{210}{169}\)
NH
11 tháng 1 2023
\(\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{2x+4}{4-x^2}\\ =\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x+4-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x}{x+2}\)
\(\left|x+1\right|=3\\ \left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=2\left(loai\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
với x=-4 thì
\(\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)
TC
11 tháng 1 2023
\(=>P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{-2x-4}{x^2-4}\)`(x ne +-2)`
\(P=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-2x-4}{\left(x+2\right)\left(x-2\right)}\)
\(P=\dfrac{x^2-2x+2x+4-2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(P=\dfrac{x}{x+2}\)
`|x+1| =3`
`=>[(x+1=3),(x+1=-3):}`
`=> [(x=3-1=2(ktm) ),(x=-3-1=-4(t/m)):}`
Thay `x=-4` vào `P` ta đc
`P= (-4)/(-4+2) = 2`
12 tháng 7 2023
2:
a: 5/x-y/3=1/6
=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)
=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)
=>30-2xy=x
=>x(2y+1)=30
=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}
=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}
b: x/6-2/y=1/30
=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)
=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)
=>5xy-60=y
=>y(5x-1)=60
=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)
=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}
TC
18 tháng 5 2022
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-5.\left(\dfrac{1}{2}\right)^3+3\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5-5\left(\dfrac{1}{2}\right)^3+6\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5.1}{8}+\dfrac{3.1}{4}+6-\dfrac{5.1}{8}+\dfrac{6.1}{4}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5}{8}+\dfrac{3}{4}+6-\dfrac{5}{8}+\dfrac{3}{2}+6\)
\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=13\)
\(\Leftrightarrow x^2-x-x^2+2x+3=2\)
=>x=-1