B3 mk tìm đc cách giải r nhưng bạn nào muốn thì trả lời cg đc

Các bạn giải giúp mình B2 và B5 nhé. Mấy bài kia mình giải được rồi.

a, Với \(a\ge0;a\ne1\)

\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1}{\sqrt{a}+1}\right)^2\)

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

b, Với \(a\ge0;a\ne4;9\)

 \(B=\frac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{2\sqrt{a}+1}{3-\sqrt{a}}\)

\(=\frac{2\sqrt{a}-9-\left(a-9\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{a-5\sqrt{a}+6}\)

\(=\frac{2\sqrt{a}-9-a+9+2a-3\sqrt{a}-2}{a-5\sqrt{a}+6}=\frac{a-\sqrt{a}-2}{a-5\sqrt{a}+6}\)

\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)

Bài 18 :

a, \(\sqrt{-2x^2+6}=x-1\)ĐK : \(-\sqrt{3}\le x\le\sqrt{3}\)

\(\Leftrightarrow6-2x^2=x^2-2x+1\Leftrightarrow3x^2-2x-5=0\)

\(\Leftrightarrow\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1>0\right)=0\Leftrightarrow x=\frac{25}{9}\)( ktm )

b, \(\sqrt{t-5}+\sqrt{4t-20}-\frac{1}{5}\sqrt{9t-45}=3\)ĐK : t >= 5

\(\Leftrightarrow\sqrt{t-5}+2\sqrt{t-5}-\frac{3}{5}\sqrt{t-5}=3\)

\(\Leftrightarrow\frac{12}{5}\sqrt{t-5}=3\Leftrightarrow\sqrt{t-5}=\frac{5}{4}\Leftrightarrow t-5=\frac{25}{16}\Leftrightarrow t=\frac{105}{16}\)( tm ) 

\(\sqrt{3x-12}=x-2\)ĐK : x >= 4 

\(\Leftrightarrow3x-12=x^2-4x+4\Leftrightarrow x^2-7x+16=0\)

\(\Delta=49-4.16=49-64< 0\)

Vậy pt vô nghiệm 

\(\sqrt{3x-12}=x-2\left(x\ge4\right)\)

\(\Leftrightarrow3x-12=x^2-4x+4\Leftrightarrow x^2-7x+18=0\)

Dễ thấy pt trên vô nghiệm

Tìm đk , rút gọn

ĐK : x > 2 

\(\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\frac{1}{x-1}\right)\)

\(=\frac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}\left(\frac{x-1-1}{x-1}\right)\)

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

Với x > 2 

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

ĐK : x > 1

\(D=\left[\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}\right]+\frac{x\sqrt{x}-x}{\sqrt{x}-1}\)

\(=\left[\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}\right]+\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

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

\(=\frac{-2\sqrt{x-1}}{x-x+1}+x=-2\sqrt{x-1}+x\)

Để D > 0 thì \(-2\sqrt{x-1}+x>0\Leftrightarrow-2\sqrt{x-1}>0\Leftrightarrow\sqrt{x-1}< 0\left(voli\right)\)

Vậy không có gtri x thỏa mãn D > 0

a, Với \(x\ne\pm\sqrt{2}\)

\(M=\frac{x^2-2x\sqrt{2}+2}{x^2-2}=\frac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\frac{x-\sqrt{2}}{x+\sqrt{2}}\)

b, Với \(x\ne-\sqrt{5}\)

\(N=\frac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\frac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\frac{1}{x+\sqrt{5}}\)

chăm chỉ hem học lúc 2h sáng -.-