RG: A = \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}\) - \(\dfrac{2\sqrt{x}}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\) ; ĐKXĐ: x ≥ 0
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H9
HT.Phong (9A5)
CTVHS
16 tháng 8 2023
\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))
\(=\left(\dfrac{1}{\sqrt{x}}-\dfrac{x}{\sqrt{x}}\right)\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{-\sqrt{x}}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{-\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\left(\sqrt{x}+1\right)^2\)
17 tháng 8 2023
c: 
b;
Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
AH
Akai Haruma
Giáo viên
31 tháng 3 2023
Yêu cầu đề bài là gì bạn nên ghi đầy đủ để được hỗ trợ tốt hơn.
15 tháng 6 2023
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
Khi x=căn 2 thì \(A=\dfrac{\sqrt{2}+16}{\sqrt{\sqrt{2}}+3}\)
14 tháng 12 2022
a: \(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
b: Để A=4 thì \(x+16-4\sqrt{x}-12=0\)
=>x=4
27 tháng 5 2022
a: \(A=\dfrac{x+4\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}-2-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{4\sqrt{x}-1+x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x+4\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b: \(B=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
UV
11 tháng 12 2018
\(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}
\)P=\(\dfrac{-x+16\sqrt{x}+x\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
P=\(\dfrac{\left(16+x\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{16+x}{\sqrt{x}+3}\)
17 tháng 4 2022
a.\(P=\dfrac{3\left(x+\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(ĐK:x\ge0;x\ne1;x\ne-2\)
\(P=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3x+3\sqrt{x}-9+x-\sqrt{x}+3\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b.\(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}+2}\)
\(=1+1+1+\dfrac{2}{\sqrt{x}+2}\)
Để P lớn nhất thì \(\sqrt{x}+2\) nhỏ nhất
Mà \(\sqrt{x}+2\ge2\) \(\Rightarrow Min=2\)
\(\Rightarrow P\le1+1+1+\dfrac{2}{2}=1+1+1+1=4\)
Vậy \(P_{max}=4\) khi \(x=0\)
LD
15 tháng 10 2017
A=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
=\(\dfrac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(x\sqrt{x}-x\right)+\left(16\sqrt{x}-16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{x+16}{\sqrt{x}+3}\)