rút gọn biểu thức này chi tiết giùm em với ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\dfrac{-8xy\left(1-3x\right)^3:4x\left(1-3x\right)}{12x^3\left(1-3x\right):4x\left(1-3x\right)}=\dfrac{-2y\left(1-3x\right)^2}{3x^2}\)
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(C=\left(\dfrac{3}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
2P = 24.(5^2 + 1 )( 5^4 + 1 ) (5^8 + 1 )(5^16 + 1 )
2P = ( 5^2 - 1 )((5^2 + 1 )( 5^4 + 1 ) (5^ 8 + 1 )( 5^ 16 + 1)
2P = ( 5 ^ 4 - 1 )( 5 ^ 4 + 1 ) (5^8 + 1 )(5^16 + 1 )
2P = ( 5 ^8 - 1 )( 5^8 + 1 )( 5^16 + 1)
2P = ( 5 ^16 - 1 )( 5^16 + 1 )
2P = 5^32 - 1
=> P = \(\frac{5^{32}-1}{2}\)
a) Ta có: \(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b) Để B=16 thì \(4\sqrt{x+1}=16\)
\(\Leftrightarrow x+1=16\)
hay x=15
ĐKXĐ: \(x>2\)
\(A=\dfrac{\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2+4\sqrt{x-2}+4}}{\sqrt{\left(\dfrac{2}{x}-1\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}}{\left|\dfrac{2}{x}-1\right|}=\dfrac{\left|\sqrt{x-2}-2\right|+\left|\sqrt{x+2}+2\right|}{1-\dfrac{2}{x}}\)
- Với \(x\ge6\Rightarrow A=\dfrac{\sqrt{x-2}-2+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{2x\sqrt{x-2}}{x-2}=\dfrac{2x}{\sqrt{x-2}}\)
- Với \(2< x< 6\Rightarrow A=\dfrac{2-\sqrt{x-2}+\sqrt{x-2}+2}{\dfrac{x-2}{x}}=\dfrac{4x}{x-2}\)
em cảm ơn ạ