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\(C=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Ta co:
\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)
\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)
B=\(\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)\(-\frac{\sqrt{x}}{x+\sqrt{x}+1}\))\(\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)=\(\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)\(\left(x-2\sqrt{x}+1\right)\)=\(\sqrt{x}-1\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
1)
a) Ta có : \(\frac{x^2+5}{\sqrt{x^2+4}}=\frac{\left(x^2+4\right)+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\frac{1}{\sqrt{x^2+4}}\). Đến đây áp dụng bđt \(a+\frac{1}{a}>2\)là ra nhé :)
b) Ta sẽ chứng minh bằng biến đổi tương đương :
\(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\)
\(\Leftrightarrow\left(a+c\right)\left(b+d\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)
\(\Leftrightarrow ab+ad+bc+cd\ge ab+cd+2\sqrt{abcd}\)
\(\Leftrightarrow ad-2\sqrt{abcd}+bc\ge0\)
\(\Leftrightarrow\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)(luôn đúng)
Vì bđt cuối luôn đúng nên bđt ban đầu được chứng minh.
2) Mình làm tóm tắt thôi nhé , do đề dài...
a) \(\sqrt{2x+\sqrt{4x-1}}-\sqrt{2x-\sqrt{4x-1}}\)
\(=\frac{\sqrt{\left(4x-1\right)+2\sqrt{4x-1}+1}+\sqrt{\left(4x-1\right)-2\sqrt{4x-1}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{4x-1}-1\right|+\left|\sqrt{4x-1}+1\right|}{\sqrt{2}}\)
b) \(\frac{x-y+3\sqrt{x}+3\sqrt{y}}{\sqrt{x}-\sqrt{y}+3}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}+3}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+3\right)}{\sqrt{x}-\sqrt{y}+3}=\sqrt{x}+\sqrt{y}\)
c) Biến đổi : \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}-1\right|\)
d) Biến đổi tương tự c)
e) \(\sqrt{x+\sqrt{x^2-4}}.\sqrt{x-\sqrt{x^2-4}}=\sqrt{x^2-\left(x^2-4\right)}=\sqrt{4}=2\)
\(P=\frac{2}{\sqrt{x}-1}+\frac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(=\frac{2\left(x+\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2x+2\sqrt{x}+2+2x-2+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
Với \(x\ge0;x\ne1\), ta có:
\(P=\frac{2}{\sqrt{x}-1}+\frac{2.\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}\)
\(P=\frac{2.\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{2.\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}+\frac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{2x+2\sqrt{x}+2+2.\left(x-1\right)+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{3x-8\sqrt{x}+5+2x-2}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-\sqrt{8x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{5x-5\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right).\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
Vậy với \(x\ge0;x\ne1\) ta có: \(P=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)
TÍNH : \(\left(\sqrt{2}-1\right)^2-\frac{3}{2}\sqrt{\left(-2\right)^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.\sqrt{2}\)
\(=\left(\sqrt{2}-1\right)^2-\frac{3}{2}.2+\frac{4\sqrt{2}}{5}+\sqrt{\frac{36}{25}}.\sqrt{2}\)
\(=3-2\sqrt{2}-3+\frac{4\sqrt{2}}{5}+\frac{6\sqrt{2}}{5}=\frac{10\sqrt{2}}{5}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0\)
CHỨNG MINH :
Ta có : \(\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right]+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)với mọi \(x\ge0\)
Vậy ta có điều phải chứng minh.
\(\left(\frac{2x+1}{\sqrt{x}^3-1}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right).\frac{1+\sqrt{x}^3}{1+\sqrt{x}}-\sqrt{x}\)
\(=\left(\frac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\)
\(=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{1}{\sqrt{x}-1}.\left(1-\sqrt{x}+x\right)-\sqrt{x}\)
\(=\frac{1-\sqrt{x}+x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{1-\sqrt{x}+x-x+\sqrt{x}}{\sqrt{x}-1}=\frac{1}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{x-1}\)
CAm on ban