BT A - B hay A.B vậy bn ?

\(\)với \(x\ge0,x\ne4\) 

\(=>B=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{3}{\sqrt{x}+1}\)

\(B=\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}+1}{3}\)

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

\(=>P=\dfrac{18}{A.B}=\dfrac{18}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}=\dfrac{18}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}\)

\(=\dfrac{18\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)\(=\dfrac{18\sqrt{x}-18}{\sqrt{x}+2}=\dfrac{18\sqrt{x}}{\sqrt{x}+2}-\dfrac{18}{\sqrt{x}+2}\ge-9\)

dấu"=" xảy ra<=>x=0(tm)

\(\)

Mình không thấy đề bạn nha

đề đâu v bn

đề đou aa:)

Câu 1: C

Câu 2: B

Câu 3: A

Câu 4: C

Câu 5: B

Câu 6: A

Câu 7: B

Câu 8: A

Câu 9: D

Câu 10: B

Câu 11: B

Câu 12: D

Câu 13: A

Câu 14: B

Câu 15: B

Câu 16: C

Câu 17: C 
Câu 18: \(\left\{{}\begin{matrix}a:D\\b:B\end{matrix}\right.\)

c) Ta có: \(\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{\sqrt{6}}{\sqrt{3}}-3\cdot\sqrt{\dfrac{1}{3}}\)

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

=0

A nhé

through and through = completely: hoàn toàn

Chọn A

Câu 3: 

a: Ta có: \(\left(1-4x\right)\left(x-1\right)+\left(2x+1\right)\left(2x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4x^2+6x+2x+3=38\)

\(\Leftrightarrow13x=36\)

hay \(x=\dfrac{36}{13}\)

b: Ta có: \(\left(2x+3\right)\left(x+2\right)-\left(x-4\right)\left(2x-1\right)=75\)

\(\Leftrightarrow2x^2+4x+3x+6-2x^2+x+8x-4=75\)

\(\Leftrightarrow15x=73\)

hay \(x=\dfrac{73}{15}\)

Câu 2: 

a: \(\sqrt{9x-9}+1=7\)

\(\Leftrightarrow3\sqrt{x-1}=6\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: \(\sqrt{9x+27}-\dfrac{1}{4}\sqrt{16x+48}+\sqrt{x+3}=9\)

\(\Leftrightarrow\sqrt{x+3}=3\)

hay x=6