K mình nha

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

\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

phương trình <=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)

<=> \(\frac{1}{2}x+1=\sqrt{5}-1\) hoặc \(\frac{1}{2}x+1=-\sqrt{5}+1\)

+)  \(\frac{1}{2}x+1=\sqrt{5}-1\)<=> \(x=2\sqrt{5}+4\)

+) \(\frac{1}{2}x+1=-\sqrt{5}+1\) <=> \(x=-2\sqrt{5}\)

Vậy pt có 2 nghiệm \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)

a/ \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=4\)

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

\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=4\)

Làm nốt

b/ \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

Làm nốt

xét x âm dương rồi nhân 2 vế thêm x

\(\Leftrightarrow\left(x-\frac{1}{x}\right)+\sqrt{x-\frac{1}{x}}=\left(2x-\frac{5}{x}\right)+\sqrt{2x-\frac{5}{x}}\)

\(a=\sqrt{x-\frac{1}{x}};\text{ }b=\sqrt{2x-\frac{5}{2}};\text{ }a,\text{ }b>0\)

\(a^2+a=b^2+b\Leftrightarrow\left(a-b\right)\left(a+b\right)+\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow a=b\text{ }\left(do\text{ }a+b+1\ge1>0\right)\)

\(\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\Leftrightarrow x-\frac{4}{x}=0\Leftrightarrow x^2-4=0\Leftrightarrow x=\pm2\)

2/ x² - 6x + 4 + \(2\sqrt{2x-1}\)= 0

<=> (x² - 4x + 4) - (2x - 1 - \(2\sqrt{2x-1}\)+1) = 0

<=> (x - 2)² - (1 - \(\sqrt{2x-1}\))² = 0

\(\Leftrightarrow\left(x-1-\sqrt{2x-1}\right)\left(x-3+\sqrt{2x-1}\right)=0\)

Làm tiếp nhé

câu mik muốn hỏi là câu 1 bn giúp mik

Câu 1/ 

\(\frac{x\sqrt{y}+y\sqrt{x}}{x+y}-\frac{x+y}{2}\le\frac{x\sqrt{y}+y\sqrt{x}}{2\sqrt{xy}}-\frac{x+y}{2}\)

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

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

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

\(\le\frac{\sqrt{x}+\sqrt{y}}{2}-\frac{\sqrt{x}+\sqrt{y}}{2}+\frac{1}{4}=\frac{1}{4}\)

ĐKXĐ:\(\hept{\begin{cases}x-2>0\\y-1>0\\z-5>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>2\\y>1\\z>5\end{cases}}\)

pt\(\Leftrightarrow\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}+\sqrt{x-2}+\sqrt{y-1}+\sqrt{z-5}=16\)

Áp dụng BĐT Cauchy:

\(\frac{4}{\sqrt{x-2}}+\sqrt{x-2}+\frac{1}{\sqrt{y-1}}+\sqrt{y-1}+\frac{25}{\sqrt{z-5}}+\sqrt{z-5}\)

\(\ge2\sqrt{\frac{4}{\sqrt{x-2}}.\sqrt{x-2}}+2\sqrt{\frac{1}{\sqrt{y-1}}.\sqrt{y-1}}+2\sqrt{\frac{25}{\sqrt{z-5}}.\sqrt{z-5}}\)

\(=2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2.2+2.1+2.5\)

\(=4+2+10=16\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=4\\y-1=1\\z-5=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\\z=30\end{cases}}\)

Cần gì phải thế.

Đặt \(\sqrt{x-\frac{1}{x}}=a\ge0;\sqrt{2x-\frac{5}{x}}=b\ge0\Rightarrow x-\frac{4}{x}=b^2-a^2\)

\(\Rightarrow a=b^2-a^2+b\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

Đến đây tự làm tiếp

đây là toàn lp 3 hả bn

đây ko phải toán lớp 3

quên đây là toán lớp 1