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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.

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=\(\dfrac{\left(16+x\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{16+x}{\sqrt{x}+3}\)

15 tháng 10 2017

A=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

16 tháng 8 2023

\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))

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

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

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

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

c: loading...

b; 

 

Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

7 tháng 11 2018

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}.\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x\left(\sqrt{x}-1\right)+16.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{\left(\sqrt{x}-1\right).\left(x+16\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

7 tháng 11 2018

ĐKXĐ : \(x\ge0\) ; \(x\ne1\) ; \(x\ne9\)

18 tháng 11 2022

a: \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

b: \(P=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot5-6=10-6=4\)

Dấu = xảy ra khi x=4

a: \(P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

b: Khi \(x=7-4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{7-4\sqrt{3}+16}{2-\sqrt{3}+3}=\dfrac{23-4\sqrt{3}}{5-\sqrt{3}}\)

27 tháng 8 2023

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-3\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{\left(x\sqrt{x}-x\right)+\left(16\sqrt{x}-16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{x+16}{\sqrt{x}+3}\)

7 tháng 11 2018

ĐKXĐ :x\(\ge\)0;x\(\ne\)1;x\(\ne\)3

\(A=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{\left(\sqrt{x}-1\right)\left(x+16\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x+16}{\sqrt{x}+3}\)

7 tháng 11 2018

b, x =(\(\sqrt{2}-1)^2\)

Thay x =(\(\sqrt{2}-1)^2\)thỏa mãn đk vào a có:

A=\(\dfrac{\left(\sqrt{2}-1\right)^2+16}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

=\(\dfrac{2-2\sqrt{2}+1+16}{\sqrt{2}-1}\)

=\(\dfrac{19\sqrt{2}+19-4-2\sqrt{2}}{2-1}\)

=\(17\sqrt{2}+15\)