\(T=x^4+y^4+z^4\)

áp dụng bđt bunhia cốp -xki với bộ số \(\left(x^2,y^2,z^2\right);\left(1,1,1\right)\)

\(\left(\left[x^2\right]^2+\left[y^2\right]^2+\left[z^2\right]^2\right)\left(1^2+1^2+1^2\right)\ge\left(x^2+y^2+z^2\right)^2\)

\(\left(x^4+y^4+z^4\right)\ge\frac{\left(x^2+y^2+z^2\right)^2}{3}\)

\(\left(x^4+y^4+z^4\right)\ge\frac{\left(2xy+2yz+2xz\right)^2}{3}\)(bđt tương đương)

\(\left(x^4+y^4+z^4\right)\ge\frac{4}{3}\)

dấu "=" xảy rakhi và chỉ khi

\(\hept{\begin{cases}\frac{x^2}{1}=\frac{y^2}{1}=\frac{z^2}{1}\\x=y=z=1\end{cases}< =>\frac{1^2}{1}=\frac{1^2}{1}=\frac{1^2}{1}}\)(luôn đúng)

vậy dấu "=" có xảy ra

\(< =>MIN:T=\frac{4}{3}\)

sửa dòng 3 dưới lên 

\(T\ge\frac{\left(xy+yz+xz\right)^2}{3}=\frac{1}{3}\)

Dấu ''='' xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

Vậy GTNN T là 1/3 khi \(x=y=z=\frac{\sqrt{3}}{3}\)

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

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

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

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

\(=2\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)

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

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

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