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ĐKXĐ: \(x\notin\left\{2;-2;-1\right\}\)
a) Ta có: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\dfrac{3\left(x+1\right)}{x\left(x+2\right)}\)
\(=\left(\dfrac{x+4x+8}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{5x+8+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6x+6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)}{x-2}\cdot\dfrac{x}{3\left(x+1\right)}\)
\(=\dfrac{2x}{x-2}\)
b) Để A nguyên thì \(2x⋮x-2\)
\(\Leftrightarrow2x-4+4⋮x-2\)
mà \(2x-4⋮x-2\)
nên \(4⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(4\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{3;1;4;0;6;-2\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{0;1;3;4;6\right\}\)
Vậy: Khi \(x\in\left\{0;1;3;4;6\right\}\) thì A nguyên
a) đk: x khác 0;1
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
= \(\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b) Để \(\left|2x-5\right|=3\)
<=> \(\left[{}\begin{matrix}2x-5=3< =>2x=8< =>x=4\left(c\right)\\2x-5=-3< =>2x=2< =>x=1\left(l\right)\end{matrix}\right.\)
Thay x = 4 vào A, ta có:
\(A=\dfrac{4^2}{4-1}=\dfrac{16}{3}\)
c) Để A = 4
<=> \(\dfrac{x^2}{x-1}=4\)
<=> \(\dfrac{x^2}{x-1}-4=0< =>\dfrac{x^2-4x+4}{x-1}=0\)
<=> \(\left(x-2\right)^2=0\)
<=> x = 2 (T/m)
d) Để A < 2
<=> \(\dfrac{x^2}{x-1}< 2< =>\dfrac{x^2}{x-1}-2< 0< =>\dfrac{x^2-2x+2}{x-1}< 0\)
<=> \(\dfrac{\left(x-1\right)^2+1}{x-1}< 0\)
Mà \(\left(x-1\right)^2+1>0\)
<=> x - 1 < 0 <=> x < 1
KHĐK: x < 1 ( x khác 0)
e) Để A thuộc Z
<=> \(\dfrac{x^2}{x-1}\in Z\)
<=> \(x^2⋮x-1\)
<=> \(x^2-x\left(x-1\right)-\left(x-1\right)⋮x-1\)
<=> \(1⋮x-1\)
Ta có bảng:
| x-1 | 1 | -1 |
| x | 2 | 0 |
| T/m | T/m |
KL: Để A thuộc Z <=> \(x\in\left\{2;0\right\}\)
f) Để A thuộc N <=> \(x\in\left\{2;0\right\}\)
`B17:`
`a)` Với `x \ne +-3` có:
`A=[x+15]/[x^2-9]+2/[x+3]`
`A=[x+15+2(x-3)]/[(x-3)(x+3)]`
`A=[x+15+2x-6]/[(x-3)(x+3)]`
`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`
`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)
`=>` Ko có gtr nào của `x` t/m
`c)A in ZZ<=>3/[x-3] in ZZ`
`=>x-3 in Ư_3`
Mà `Ư_3={+-1;+-3}`
`@x-3=1=>x=4`
`@x-3=-1=>x=2`
`@x-3=3=>x=6`
`@x-3=-3=>x=0`
________________________________
`B18:`
`a)M=1/3` `ĐK: x \ne +-4`
`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`
`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`
`<=>32/[x-4].[x+4]/32=1/3`
`<=>3x+12=x-4`
`<=>x=-8` (t/m)
Cậu xem lại đề hộ tớ xem là nhân \(\dfrac{2x}{5x-5}\)hay là chia cho \(\dfrac{2x}{5x-5}\)nhé :>
Lời giải:
ĐK: $x\neq \pm 1$
a)
\(A=\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}.\frac{2x}{5(x-1)}=\frac{4x}{(x-1)(x+1)}.\frac{2x}{5(x-1)}=\frac{8x^2}{5(x-1)^2(x+1)}\)
b) Tại $x=4$ thì: $A=\frac{8.4^2}{5(4-1)^2(4+1)}=\frac{128}{225}$
c)
$A=5$
$\Leftrightarrow \frac{8x^2}{5(x-1)^2(x+1)}=5$
$\Leftrightarrow 8x^2=25(x-1)^2(x+1)$
PT này không phải không giải được nhưng giải phức tạp và nghiệm cực xấu. Nên mình nghĩ bạn đã viết sai đề?
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x-1}\right)^2}\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
b) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}+1-6\sqrt{x}=0\)
\(\Leftrightarrow-5\sqrt{x}+1=0\)
\(\Leftrightarrow-5\sqrt{x}=-1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{25}\)(nhận)
Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=\dfrac{1}{25}\)
a, ĐK : \(x\ne0;1\)
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x^2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x^2}{x-1}\)
b, Thay x = 3 vào A ta được : \(\dfrac{9}{2}\)
c, \(A=4\Rightarrow\dfrac{x^2}{x-1}=4\Rightarrow x^2=4x-4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
d, \(A< 2\Rightarrow\dfrac{x^2}{x-1}-2< 0\Leftrightarrow\dfrac{x^2-2x+1}{x-1}< 0\Rightarrow x-1< 0\Leftrightarrow x>1\)
a,\(\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x+1}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)