\(\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne4\right)\\ =\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-2-2\sqrt{x}-4-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =-\dfrac{2}{\sqrt{x}-2}\)

dk là x khác 4 mới đúng nhee 

\(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\left(ĐKXĐ:x\ge0\right)\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)}+\dfrac{10-x}{\sqrt{x}+2}\)

\(=\dfrac{x-4+10-x}{\sqrt{x}+2}\)

\(=\dfrac{6}{\sqrt{x}+2}\)

\(=\dfrac{6\left(\sqrt{x}-2\right)}{x-4}\)

2a + 1 = -2

2a = -2 -1

2a = -3

a = \(-\dfrac{3}{2}\)

\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}+1\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=\dfrac{\sqrt{x}+1-\sqrt{x}+1+x+\sqrt{x}-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\\=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\=1\)

Gọi số thứ 1 là a ( a > 0 )

số thứ 2 : a : \(\dfrac{1}{2}\) = 2 x a

số thứ 3 : 2a : \(\dfrac{2}{5}\) = 5 x a

Ta có tổng 3 số là 120 nên st1 + st2 + st3 = 120

→ a + 2 x a + 5 x a = 120

→ 8 x a = 120

→ a = 15

Số thứ 1 là 15

Số thứ 2 là 2 x a = 2 x 15 = 30

Số thứ 3 là 5 x a = 5 x 15 = 75

\(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{\sqrt{3}-1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(\Leftrightarrow\dfrac{2}{3-1}\)

\(\Leftrightarrow1\)

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

\(x^4-4x^3-2x^2-3x+2\)

\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)

\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)

\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)

Xin tick ạ !!!