chịu

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 \(B=-\dfrac{47}{5}\)

sos

A=(2x+1-2x+1)^2+xy

=xy+4

=2023+4

=2027

\(A=(2x+1-2x+1)^2+xy\)

\(=xy+4\)

\(=2023+4\)

\(=2027\)

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

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

\(=\sqrt{x}-1\)

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

                        Đk: \(x>0\) và \(x\ne1\)

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

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

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

        \(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)

b) Thay \(x=3+2\sqrt{2}\) vào A ta được:

  \(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)

      \(=\sqrt{2}+1-1=\sqrt{2}\)

(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))

Đặt \(A=\left(\dfrac{2}{5}x^3y\right)\cdot\left(-5xy\right)\)

\(=\left(\dfrac{2}{5}\cdot\left(-5\right)\right)\cdot x^3\cdot x\cdot y\cdot y\)

\(=-2x^4y^2\)

Thay x=-1 và y=1/2 vào A, ta được:

\(A=-2\cdot\left(-1\right)^4\cdot\left(\dfrac{1}{2}\right)^2=-2\cdot\dfrac{1}{4}=-\dfrac{1}{2}\)