tim x de x-2/x+2 <0
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![](https://rs.olm.vn/images/avt/0.png?1311)
1.x=1;5
2.x=11
3.x=1;y=4
4.a)a=2;12 b)a=1;2
nho h cho minh nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\text{Δ}=\left(4m-4\right)^2-4\left(-4m+10\right)\)
\(=16m^2-32m+16+16m-40\)
\(=16m^2-16m-24\)
\(=8\left(2m^2-2m-3\right)\)
Để pT có nghiệm kép thì \(2m^2-2m-3=0\)
hay \(m\in\left\{\dfrac{1+\sqrt{7}}{2};\dfrac{1-\sqrt{7}}{2}\right\}\)
b: Thay x=2 vào PT, ta được:
\(4+8\left(m-1\right)-4m+10=0\)
=>8m-8-4m+14=0
=>4m+6=0
hay m=-3/2
Theo VI-et, ta được: \(x_1+x_2=-4\left(m-1\right)=-4\cdot\dfrac{-5}{2}=10\)
=>x2=8
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có \(\frac{x-3}{2-x}\)=\(\frac{2}{3}\)nên (x-3)3=(2-x)2
--->3x-9=4-2x
---->5x = 36 x =\(\frac{36}{5}\)
\(\frac{x-3}{2-x}=\frac{2}{3}\)
\(\Rightarrow\)\(3\left(x-3\right)\)\(=2\left(2-x\right)\)
\(\Rightarrow3x-9=4-2x\)
\(\Rightarrow3x+2x=4+9\)
\(\Rightarrow5x=13\)
\(\Rightarrow x=\frac{13}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
dkxd \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)
b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)
\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)
tới khúc này bí rồi ^^
a,ĐKXĐ của A là:\(x\ne+2;-2\)
b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)
c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)
Ta có bảng
\(x^2-4\) | x |
4 | \(\sqrt{8}\) |
-4 | 0 |
2 | \(\sqrt{6}\) |
-2 | \(\sqrt{2}\) |
1 | \(\sqrt{5}\) |
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
\(\frac{x-2}{x+2}\)< 0 <=> \(\hept{\begin{cases}x-2< 0\\x+2>0\end{cases}}\)<=> \(\hept{\begin{cases}x< 2\\x>-2\end{cases}}\)<=> x\(\in\){-1;0;1}
\(\frac{x-2}{x+2}\)<0
x-2 <0
x < 2