cậu chia từng câu ra cho mình nhé

a )\(A=\frac{x^2+4x+4}{x^2-4}=\frac{\left(x+2\right)^2}{x^2-2^2}=\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{x+2}{x-2}=\frac{5}{3}\)

<=> (x + 2).3 = (x - 2).5

<=> 3x + 6 = 5x - 10

<=> 3x - 5x = - 10 - 6

<=> - 2x = - 16

=> x = 8

b ) \(\frac{x+2}{x-2}=\frac{\left(x-2\right)+4}{x-2}=1+\frac{4}{x-2}\)

đến đây tự tìm đc 

Bài 2 lớp 8 ko làm đc thì đi chết đi

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

=(x+1+x-1)(x+1-x+1)-3x^2-3

=2x^2-3x^2-3

=-x^2-3

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2\)

\(=x^2-25-\left(x^2+6x+9\right)+3\left(x^2-4x+4\right)\)

\(=x^2-25-x^2-6x-9+3x^2-12x+12\)

\(=3x^2-18x-22\)

Ta có: \(\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

\(=x^2+2x+1-\left(x^2-16\right)+3x^2\)

\(=4x^2+2x+1-x^2+16=3x^2+2x+17\)

Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^2+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x-4\right)\left(x+4\right)+3x^2\)

=>\(3x^2-18x-22=3x^2+2x+17\)

=>-20x=39

=>\(x=-\frac{39}{20}\)

Lời giải:

$(x-5)(x+5)-(x+3)^2+3(x-2)^2=(x+1)^2-(x-4)(x+4)+3x^2$

$\Leftrightarrow x^2-25-(x^2+6x+9)+3(x^2-4x+4)=(x^2+2x+1)-(x^2-16)+3x^2$
$\Leftrightarrow 3x^2-18x-22=3x^2+2x+17$

$\Leftrightarrow -18x-22=2x+17$

$\Leftrightarrow 20x=-39$

$\Leftrightarrow x=\frac{-39}{20}$

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

\(\Leftrightarrow x=-1\left(tm\right)\)

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)

c) \(A=\dfrac{1}{2}\) khi:

\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}

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nhìn hơi đau mắt nhá bạn hoa mắt quá

Phương pháp giải:

 - Muốn tìm số bị trừ ta lấy hiệu cộng với số trừ.

 - Muốn tìm số bị chia ta lấy thương nhân với số chia. 

Lời giải chi tiết:

a)

● x − 4 = 2

    x = 2 + 4

    x = 6

● x : 4 = 2

    x = 2 × 4

    x = 8

b)

● x − 5 = 4

    x = 4 + 5

    x = 9

● x : 5 = 4

    x = 4 × 5

    x = 20

c)

● x − 3 = 3

    x = 3 + 3

    x = 6

● x : 3 = 3

    x = 3 × 3

    x = 9