Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

đkxđ:\(x\ne5,x\ne-5\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)

\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5x+25}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)

thay x=1 vào bt A, ta được:

\(-\dfrac{4}{1-5}=1\)

Ta có: \(Q=\left(\dfrac{1}{x+5}+\dfrac{1}{x-5}\right):\dfrac{2x}{x^2-25}\)

\(=\left(\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}+\dfrac{x+5}{\left(x-5\right)\left(x+5\right)}\right):\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x-5+x+5}{\left(x+5\right)\left(x-5\right)}:\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x}{\left(x+5\right)\left(x-5\right)}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{2x}\)

\(=1\)

Có: \(x^2-25=\left(x-5\right)\left(x+5\right)\)

ĐKXĐ của Q là x ≠ 5; x ≠ -5

Mà theo đề: x = 5; x = -5

=> Ko có giá trị của Q tìm đc

đk : x >= 0 ; x khác 25 ; -3 

\(\left(\dfrac{\sqrt{x}+5-\sqrt{x}+5}{x-25}\right):\dfrac{x+3}{x-25}=\dfrac{25}{x+3}\)

a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)

= x³ - 125 - x² + 4 + x³ + x² + 4x

= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)

= 2x³ + 4x - 121

b) Tại x = -2 ta có:

A = 2.(-2)³ + 4.(-2) - 121

= 2.(-8) - 8 - 121

= -16 - 129

= -145

c) x² - 1 = 0

x² = 1

x = -1; x = 1

*) Tại x = -1 ta có:

A = 2.(-1)³ + 4.(-1) - 121

= 2.(-1) - 4 - 121

= -2 - 125

= -127

*) Tại x = 1 ta có:

A = 2.1³ + 4.1 - 121

= 2.1 + 4 - 121

= 2 - 117

= -115

Ta có: \(P=\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(=\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`

`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`

`=5x^2-5x^2-4x+6x+1+9+245`

`=2x+255`

`b)(x-2)(x^2+2x+4)-(25+x^3)`

`=x^3-8-x^3-25=-33`

Lời giải:

a. 

$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$

$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$

$=5x^2+2x+10-(5x^2-245)=2x+255$

b.

$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$

$=-8-25=-33$

a: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)\)

\(=4x^2-4x+1+x^2+6x+9-5x^2+245\)

\(=2x+255\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

a: \(M=\dfrac{2x^2-10x-x^2+x+30-x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)

b: Để M là số nguyên thì \(x+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(x\in\left\{-4;-6;-3;-7;0;-10;-15\right\}\)

a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)

\(=\dfrac{x^2+5x+100}{5x}\)

b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:

\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)

Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)