25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10

C = (-15 + |x|) + (25 - |-x|)

= -15 + |x| + 25 - |-x|

= (25 - 15) + |x| - |-x|

= 10 + |x| - |-x|

Ta có: |x| = |-x| => |x| - |-x| = 0

=> C = 10 + |x|

C = (-15 + IxI ) + ( 25 - I-xI)

C = (-15 + x ) + ( 25 - x )

C = -15 + x + 25 - x

C = ( -15 + 25 ) + ( x - x )

C = 10 + 0

C = 10

Vậy C = 10

P = ( - 15 + | x | ) + ( 25 - |  - x | )

P = ( - 15 + x ) + ( 25 - x )

P = - 15 + x + 25 - x

P = - 15  + 25 + x - x

P = 25 - 15 + 0

P = 10

P=(-15)+x+25-x

P=[(-15)+25]+(x-x)

P=10+0

P=10

\(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\left(đk:x\ne25,x\ge0\right)\)

\(=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+3\right)}=\dfrac{1}{\sqrt{x}+3}\)

Ta có: \(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\cdot\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)

\(=\dfrac{1}{\sqrt{x}+3}\)

a) 2x + (-61) - (21 - 61) = 2x - 21 + (61 - 61) = 2x - 21.

b) (- 3 - x + 5) + 3 = (- 3 + 3) + 5 - x = 5 - x.

c) 11- (13 - x) + (13 - 11) = (11- 11) + (13- 13) + x = x

d) 25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10

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.

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

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

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)