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Lời giải:
a. ĐKXĐ: $a\neq \pm 2$
\(M=\frac{(2+a)^2}{(2-a)(2+a)}+\frac{4a^2}{(2-a)(2+a)}-\frac{(2-a)^2}{(2+a)(2-a)}\)
\(=\frac{(2+a)^2+4a^2-(2-a)^2}{(2-a)(2+a)}=\frac{4a(a+2)}{(2-a)(2+a)}=\frac{4a}{2-a}\)
b.
$|a+1|=3\Rightarrow a+1=\pm 3\Rightarrow a=-2$ hoặc $a=-4$
Vì $a\neq \pm 2$ nên $a=-4$
Khi đó: $M=\frac{4a}{2-a}=\frac{4(-4)}{2-(-4)}=\frac{-8}{3}$
c.
Trước tiên cần tìm $a$ để $M$ nguyên đã.
$M=\frac{4a}{2-a}=\frac{8-4(2-a)}{2-a}=\frac{8}{2-a}-4$ nguyên khi $\frac{8}{2-a}$ nguyên
$\Rightarrow 2-a\in\left\{\pm 1; \pm 2; \pm 4; \pm 8\right\}$
$\Rightarrow a\in\left\{1; 3; 0; 4; -2; 6; 10; -6\right\}$.
Thử lại thấy $a\in\left\{1; 3; 0; 4\right\}$ thỏa mãn $M$ là số nguyên chia hết cho $4$
a: ĐKXĐ: a<>3; a<>-3; a<>-1
b: \(P=\dfrac{2a^2-3a+3a+9-2a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)
\(=\dfrac{6}{\left(a+3\right)\left(a+1\right)}\)
c: |a|=2
=>a=2 hoặc a=-2
Khi a=-2 thì \(P=\dfrac{6}{\left(-2+3\right)\left(-2+1\right)}=-6\)
Khi a=2 thì \(P=\dfrac{6}{\left(2+3\right)\left(2+1\right)}=\dfrac{6}{5\cdot3}=\dfrac{2}{5}\)
a) ĐKXĐ: \(x\ne3;x\ne\pm2\)
\(C=\frac{2a-a^2}{a+3}\cdot\left(\frac{a-2}{a+2}-\frac{a+2}{a-2}+\frac{4a^2}{4-a^2}\right)\)
\(C=\frac{-a^2+2a}{a+3}\cdot\left(-\frac{4a}{a-2}\right)\)
\(C=-\frac{2a-a^2}{a+3}\cdot\frac{4a}{a-2}\)
\(C=-\frac{\left(2a-a^2\right)\cdot4a}{\left(a+3\right)\left(a-2\right)}\)
\(C=\frac{4a^2}{a+3}\)
b) \(C=\frac{4.4^2}{4+3}=\frac{46}{7}\)
c) \(\frac{4a^2}{a+3}=1\)
<=> 4a2 = a + 3
<=> 4a2 - a - 3 = 0
<=> 4a2 - 3a - 4a - 3 = 0
<=> a(4a + 3) - (4a + 3) = 0
<=> (4a + 3)(a - 1) = 0
<=> 4a + 3 = 0 hoặc a - 1 = 0
<=> a = -3/4 hoặc a = 1
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
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 <=> \(x\in\left\{2;0\right\}\)
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)
Bạn xem thử tại đây:
https://hoc24.vn/cau-hoi/cho-bieu-thucm-dfrac2a2-a-dfrac4a2a2-4-dfrac2-a2aa-rut-gon-mb-tinh-gia-tri-cua-m-khi-a13c-tim-a-z-de-m-la-so-nguyen-chia-het-cho-4.7975358921144