\(C=\frac{6\sqrt{x}}{3\sqrt{x}+5}=\frac{2\left(3\sqrt{x}+5\right)-10}{3\sqrt{x}+5}=2-\frac{10}{3\sqrt{x}+5}\)

\(\Rightarrow3\sqrt{x}+5\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

tôi cũng vậy

Đặt \(\hept{\begin{cases}\sqrt[4]{100-x}=a\\\sqrt[4]{x-18}=b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a+b=4\\a^4+b^4=82\end{cases}}\)

Làm nốt

\(a,\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.2\sqrt{2}3\sqrt{5}+8}\)

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

\(2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)

\(-4\sqrt{5}\)

\(b,\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}\)

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}\)

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)

\(\sqrt{6-2\sqrt{4+\sqrt{12}}}\)

\(\sqrt{6-2\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)

\(\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(\sqrt{6-2\left(\sqrt{3}+1\right)}\)

\(\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)

\(\)

Ta có:

\(VT^2=\left(13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}\right)^2=\left(\sqrt{13}.\sqrt{13}.\sqrt{x^2-x^4}+3.\sqrt{3}.\sqrt{3}.\sqrt{x^2+x^4}\right)^2\)

\(\le\left(13+27\right)\left(13\left(x^2-x^4\right)+3\left(x^2+x^4\right)\right)\)

\(=40\left(16x^2-10x^4\right)=16^2-16\left(25x^4-40x^2+16\right)\)

\(=16^2-16\left(5x^2-4\right)^2\le16^2=VP^2\)

Làm nốt

đk: \(\hept{\begin{cases}x+x^2\ge0\\x-x^2\ge0\\x+1\ge0\end{cases}}\)

Sử dụng bđt Cô si cho 2 số không âm có:

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

\(\Leftrightarrow\sqrt{x+x^2}+\sqrt{x-x^2}\le x+1\)

Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+x^2=1\\x-x^2=1\end{cases}\Leftrightarrow\hept{\begin{cases}x^2+x-1=0\\x^2-x+1=0\end{cases}}\Leftrightarrow x\in\varnothing}\)

Dấu '=' không xảy ra, vì: \(\sqrt{x+x^2}+\sqrt{x-x^2}< x+1\)

Điều kiện: \(x\ge\frac{1}{5}\)

Ta có: \(x^3-2x^2+6x+3-4\sqrt{5x-1}\)

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

\(=x\left(x-1\right)^2+\left(\sqrt{5x-1}-2\right)^2\ge0\)

Làm nốt

đè có sai ko nhỉ

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

đk: \(x\ge0;x^2-4x+1\ge0\)

Bình phương 2 vế pt ta có:

\(x^2+2x+1+2\left(x+1\right)\sqrt{x^2-4x+1}+x^2-4x+1=9x\)

\(\Leftrightarrow2x^2-11x+2+2\sqrt{\left(x^2+2x+1\right)\left(x^2-4x+1\right)}=0\)

giả sử \(2x^2-11x+2=m\left(x^2+2x+1\right)+n\left(x^2-4x+1\right)\Rightarrow\hept{\begin{cases}m+n=2\\2m-4n=-11\\m+n=2\end{cases}\Leftrightarrow\hept{\begin{cases}m=-\frac{1}{2}\\n=\frac{5}{2}\end{cases}}}\)

pt trở thành: \(-\frac{1}{2}\left(x^2+2x+1\right)+\frac{5}{2}\left(x^2-4x+1\right)+2\sqrt{\left(x^2+2x+1\right)\left(x^2-4x+1\right)}=0\)

Chia pt cho \(x^2+2x+1>0\) ta được:

\(-1+5\left(\frac{x^2-4x+1}{x^2+2x+1}\right)+4\sqrt{\left(\frac{x^2-4x+1}{x^2+2x+1}\right)}=0\)

Đặt \(t=\sqrt{\left(\frac{x^2-4x+1}{x^2+2x+1}\right)}\ge0\)

Ta có: \(5t^2+4t-1=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=\frac{1}{5}\end{cases}}\Leftrightarrow t=\frac{1}{5}\Leftrightarrow\left(\frac{x^2-4x+1}{x^2+2x+1}\right)=\frac{1}{25}\)

\(\Leftrightarrow24x^2-102x+24=0\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=4\end{cases}}\)

ĐK: \(x\ge1;x\le0\)

\(pt\Leftrightarrow x^2-x-2=\sqrt{x^2-x}\)

ĐẶT \(t=\sqrt{x^2-x}\Rightarrow t^2-2=t\)      hay \(\left(t+1\right)\left(t-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}t=-1\left(loại\right)\\t=2\end{cases}}\)     

Ta có: \(x^2-x=4\Leftrightarrow\orbr{\begin{cases}x=\frac{1-\sqrt{17}}{2}\\\frac{1+\sqrt{17}}{2}\end{cases}}\)

bạn ơi đây là \(4\sqrt{x^2-1}\) chứ kh phải \(4\sqrt{x^2-x}\)đâu ạ