Đặt \(\hept{\begin{cases}x-1=a\\y-1=b\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ab\left(a+b\right)=6\\a^2+b^2=5\end{cases}}\)

Làm nốt

đề tìm max hả bạn ? 

\(C=-x^2+6x+13=-\left(x^2-6x-13\right)=-\left(x^2-6x+9-22\right)\)

\(=-\left(x-3\right)^2+22\le22\)

Dấu ''='' xảy ra khi x = 3

Vậy GTLN C là 22 khi x = 3 

x =3 nha

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

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

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

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

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

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