Question

Bài 1: Rút gọn biểu thức : \(\left(2-\sqrt{2}\right).\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)

Bài 2: Cho biểu thức: P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

a) Rút gọn biểu thức (tìm đk)

b) Tìm x để P = 2

BÀi 3: Q = \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

a) Rút gọn biểu thức (tìm đk)

b) Tìm giá trị của a để Q > 0

Answer 1

Bài 1 : Rút gọn biểu thức :

\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)

\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)

\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)

\(=-10\sqrt{2}+10-7+30\sqrt{2}\)

\(=20\sqrt{2}+3\)

Answer 2

Bài 2:

a) ĐKXĐ : x # 4 ; x # - 4

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

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

P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\)

Vậy, để P = 2 thì x = 16.

Answer 3

Bài 3 :

a) ĐKXĐ : a # 1 ; a # 0, a # 4

\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(Q=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{1}\)

\(Q=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)

b) Để \(Q>0\) thì \(\dfrac{\sqrt{a}-2}{\sqrt{a}}>0\)

\(\Leftrightarrow\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>4\)

Vậy, để Q > 0 thì a > 4

Answer 4

\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)

= \(-10\sqrt{2}+10-\left(18-30\sqrt{2}+25\right)\)

= \(-10\sqrt{2}+10-18+30\sqrt{2}-25\)

= \(20\sqrt{2}-33\)

Answer 5

Câu 2:

a) P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

ĐK : \(x\ge0;x\ne4\)

Ta có: P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

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

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

= \(\dfrac{x+2\sqrt{x}+\sqrt{x}+3+2-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

= \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Để P = 2 <=> \(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)

=> \(3\sqrt{x}=2\sqrt{x}+4\)

<=> \(\sqrt{x}=4\) <=> \(x=16\left(TM\right)\)

Answer 6

bài 2 : a) điều kiện : \(x\ne4;x\ge0\)

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2+5\sqrt{x}}{x-4}\)

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

P = \(\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

b) P = 2 \(\Leftrightarrow\) \(\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\) \(\Leftrightarrow\) \(2\sqrt{x}+4=3\sqrt{x}\) \(\Leftrightarrow\) \(\sqrt{x}=4\) \(\Leftrightarrow\) \(x=16\)

Answer 7

T_T sorry bn nha, mik lm sai nhiều quá