\(P=\dfrac{13}{3}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{13}{3}=0\left(\text{Đ}KC\text{Đ}:x>0\right)\)

\(\Leftrightarrow\dfrac{3x+3\sqrt{x}+3-13\sqrt{x}}{3\sqrt{x}}=0\)

\(\Rightarrow3x-10\sqrt{x}+3=0\)

Đặt \(\sqrt{x}=t\left(t>0\right)\)

\(\Rightarrow3t^2-10t+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{1}{3}\end{matrix}\right.\)( TM)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\)(TM)

\(ĐK:x>0\\ P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{13}{3}< =>3x+3\sqrt{x}+3=13\sqrt{x}\\ < =>3x-10\sqrt{x}+3=0\\ < =>\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=9\\x=\dfrac{1}{9}\end{matrix}\right.\left(TMDK\right)}}\)

\(T=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}+1\right)}+\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+1\right)}}{\sqrt{5}+1}-\sqrt{2}+1\)

\(=\dfrac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-\sqrt{2}+1\)

\(\sqrt{2}T=\dfrac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{5}+1}-2+\sqrt{2}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}+1}-2+\sqrt{2}\)

\(=\dfrac{2+2\sqrt{5}}{\sqrt{5}+1}-2+\sqrt{2}=\sqrt{2}\Rightarrow T=\dfrac{\sqrt{2}}{\sqrt{2}}=1\)

\(T=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}}{\sqrt{\sqrt{5}+1}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\dfrac{\sqrt{\sqrt{5}+2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)+\sqrt{5}-2}}}{\sqrt{\sqrt{5}+1}}-\sqrt{2}+1\)

\(=\dfrac{\sqrt{2\sqrt{5}+2\sqrt{5-4}}}{\sqrt{\sqrt{5}+1}}-\sqrt{2}+1\)

\(=\dfrac{\sqrt{2\sqrt{5}+2}}{\sqrt{\sqrt{5}+1}}-\sqrt{2}+1\)

\(=\dfrac{\sqrt{2}.\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{2}+1\)

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

có:

\(\dfrac{1}{2}=0,500000000......\)

theo mk ngĩ số bạn cần tìm là 0 nhé

a) Ta có \(\sqrt{x-4\sqrt{x-4}}=\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}\) \(=\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}-2\right|\) 

b) Ta có \(\sqrt{x-2+2\sqrt{x-3}}=\sqrt{\left(x-3\right)+2\sqrt{x-3}+1}\) \(=\sqrt{\left(\sqrt{x-3}+1\right)^2}=\sqrt{x-3}+1\) (vì \(\sqrt{x-3}+1>0\) với \(x\ge3\))

c) Ta có \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\) \(=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\) \(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)

d) Ta có \(\sqrt{x-2\sqrt{x}+1}+\sqrt{x+2\sqrt{x}+1}\) \(=\sqrt{\left(\sqrt{x}-1\right)^2}+\sqrt{\left(\sqrt{x}+1\right)^2}\) \(=\left|\sqrt{x}-1\right|+\sqrt{x}+1\)

Đặt \(A=2\sqrt{2}+\sqrt{3}\)

Giả sử A là một số hữu tỉ\(\Rightarrow\) A = \(\dfrac{x}{y}\) (tối giản, \(y\ne0\))

\(\Rightarrow2\sqrt{2}+\sqrt{3}=\dfrac{x}{y}\)\(\Rightarrow\left(2\sqrt{2}+\sqrt{3}\right)^2=\dfrac{x^2}{y^2}\Leftrightarrow11+4\sqrt{6}=\dfrac{x^2}{y^2}\)

\(\Leftrightarrow\dfrac{x^2}{y^2}-11=4\sqrt{6}\)

Ta thấy: \(\dfrac{x^2}{y^2};11\) là các số hữu tỉ nên \(\dfrac{x^2}{y^2}-11\) là số hữu tỉ

Mặt khác: \(4\sqrt{6}\) là số vô tỉ

=> \(\dfrac{x^2}{y^2}-11\ne4\sqrt{6}\)

=> Giả sử là sai

=> A là một số vô tỉ

=> \(2\sqrt{2}+\sqrt{3}\) là số vô tỉ

mình chỉ cm đc diều sau:

\(a^2=b^2+c^2-2bc.cosA\) bạn có viết nhầm ko

cách CM:

lần lượt hạ các đường cao AD,BE,CF

ta dễ cm:\(AE.EC+AB.FB=BC^2\) và  \(AE.AC=AB.AF\)

\(\Rightarrow AC.EC+AB.BF-AC.AE-AB.AF=BC^2\)

\(\Leftrightarrow b^2+c^2-2AC.AE=a^2\)

\(\Leftrightarrow b^2+c^2-2AC.AB.\dfrac{AE}{AB}=a^2\)

\(\Leftrightarrow a^2=b^2+c^2-2bc.cosA\)(đfcm)

a) \(\sqrt{x-4\sqrt{x-4}}=\sqrt{\left(x-4\right)-4\sqrt{x-4}+4}=\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}-2\right|\)

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

c) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

d) \(\sqrt{x-2\sqrt{x}+1}+\sqrt{x+2\sqrt{x}+1}\)

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

a, đk x >= 4 

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

b, đk x >= 3 

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

c, đk x >= 1 

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

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

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

d, tương tự 

đk x > = 2 

\(x^2-3x+2=x-1\Leftrightarrow\left(x-1\right)\left(x-2\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=1\left(loai\right);x=3\)