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Mình làm xong rồi nhưng ko gửi đc lên olm

Thông cảm nhé!!!!!

a + b, Với \(x>0;x\ne1\)

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

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

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

Thay \(x=4\Rightarrow\sqrt{x}=2\)vào biểu thức A ta được 

\(\frac{-4+1}{2}=-\frac{3}{2}\)

c, Ta có : \(P< 0\Rightarrow\frac{-x+1}{\sqrt{x}}< 0\Rightarrow-x+1< 0\Leftrightarrow x< 1\)

\(\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)2

<=>\(\sqrt{\frac{x^2+3}{x}}-2=\frac{x^2-4x+3}{2\left(x+1\right)}\)

<=>\(\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x+3}{2\left(x+1\right)}\)

<=>\(\frac{x^2-4x+3}{x\left(\sqrt{x+\frac{3}{x}}+2\right)}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)

<=>

con lai ban tu giai nha

\(\left(x^2+1\right)\left(x-1\right)\left(x-3\right)=15\left(2x-1\right)^2\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-4x+3\right)=15\left(2x-1\right)^2\)(1)

Đặt \(\hept{\begin{cases}x^2+1=a\\2x-1=b\end{cases}}\)

\(\Rightarrow\left(1\right)\Leftrightarrow a\left(a-2b\right)=15b^2\)

\(\Leftrightarrow a^2-2ab=15b^2\)

\(\Leftrightarrow\left(a-b\right)^2=16b^2\)

\(\Leftrightarrow\orbr{\begin{cases}a-b=4b\\a-b=-4b\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=5b\\a=-3b\end{cases}}\)

TH1: a=5b

\(\Rightarrow x^2+1=10x-5\)

\(\Leftrightarrow x^2-10x+6=0\)

\(\Delta=100-24=76\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{10+\sqrt{76}}{2}=5+\sqrt{19}\\x=5-\sqrt{19}\end{cases}}\)

TH2: a= -3b tương tự 

là sao vậy bạn ???

KO CÓ J 

Biểu thức thứ 2 trong HPT chưa chính xác \(\left(y-1\right)\)xem lại đề bài nha bạn !

mik bt nè

a, Điều kiện : \(a\ge0,a\ne1\)

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

\(=\left(\frac{a+1+\sqrt{a}}{a+1}\right):\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)

\(=\left(\frac{a+\sqrt{a}+1}{a+1}\right):\left[\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)

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

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

a, \(M=\frac{\sqrt{x}}{\sqrt{x}+6}+\frac{1}{\sqrt{x}-6}+\frac{17\sqrt{x}+30}{\left(\sqrt{x}+6\right)\left(\sqrt{x}-6\right)}\)

\(=\frac{x-6\sqrt{x}+\sqrt{x}+6+17\sqrt{x}+30}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{12\sqrt{x}+x+36}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-6}\)

b, Ta có : \(L=N.M\Rightarrow L=\frac{\sqrt{x}+6}{\sqrt{x}-6}.\frac{24}{\sqrt{x}+6}=\frac{24}{\sqrt{x}+6}\)

Vì \(\sqrt{x}+6\ge6\)

\(\Rightarrow\frac{24}{\sqrt{x}+6}\le\frac{24}{6}=4\)

Dấu ''='' xảy ra khi \(\sqrt{x}+6=6\Leftrightarrow x=0\)

Vậy GTLN L là 4 khi x = 0