|2×X-1|=(-4)2
Tìm x
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\(a,2x\left(x^3-3\right)-2x^4=18\\ 2x^4-6x-2x^4=18\\ -6x=18\\ x=-3\)
\(b,9x\left(4-x\right)+\left(3x+1\right)^2=2\\ 36x-9x^2+9x^2+6x+1=2\\ 42x=2-1\\ 42x=1\\ x=\dfrac{1}{42}\)
\(a,\Leftrightarrow2x^4-3x-2x^4=18\Leftrightarrow-3x=18\Leftrightarrow x=-6\\ b,\Leftrightarrow36x-9x^2+9x^2+6x+1=2\\ \Leftrightarrow42x=1\Leftrightarrow x=\dfrac{1}{42}\)
\(\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{2x+4}{4-x^2}\\ =\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x+4-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x}{x+2}\)
\(\left|x+1\right|=3\\ \left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=2\left(loai\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
với x=-4 thì
\(\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)
\(=>P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{-2x-4}{x^2-4}\)`(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`
\(\left(3x+1\right)^2=9\left(x-2\right)^2\)
\(\Leftrightarrow9x^2+6x+1=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2+6x+1=9x^2-36x+36\)
\(\Leftrightarrow9x^2+6x+1-9x^2+36x-36=0\)
\(\Leftrightarrow42x-35=0\)
\(\Leftrightarrow42x=35\)
\(\Leftrightarrow x=\dfrac{35}{42}=\dfrac{5}{6}\)
Vậy: \(S=\left\{\dfrac{5}{6}\right\}\)
ĐK: x≥2
\(\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow x^2-4=x^2-4x+4\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(tm\right)\)
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)}
a, \(Chof\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
- Lập bảng xét dấu :
Vậy \(\left\{{}\begin{matrix}f\left(x\right)>0\Leftrightarrow x\in\left(3;4\right)\\f\left(x\right)< 0\Leftrightarrow x\in\left(-\infty;3\right)\cup\left(4;+\infty\right)\\f\left(x\right)=0\Leftrightarrow x\in\left\{3;4\right\}\end{matrix}\right.\)
b, \(f\left(x\right)=\left(x-1\right)\left(x+6\right)\)
( Làm tương tự câu a )
\(\Leftrightarrow a\cdot\dfrac{13}{15}=\dfrac{28}{13}:2=\dfrac{14}{13}\)
=>\(a=\dfrac{14}{13}:\dfrac{13}{15}=\dfrac{210}{169}\)
|2xX-1|=(-4)2
=>|2xX-1|=16
ta có: 2xX-1= 16 hoặc 2xX-1= -16
*nếu 2xX-1=16=>X=8.5
*nếu 2xX-1= -16=>X= -7.5
vậy X= 8.5 hoặc -7.5
X=8,5 Hoặc -7.5
Nha bạn Nguyễn Anh Tú