a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

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

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

b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)

\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)

\(=\frac{27\sqrt{2}}{4}\cdot8\)

\(=54\sqrt{2}\)

Ta có: 

\(2x=3y=6z\)

\(\Rightarrow\)\(\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}=\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=9\\\frac{y}{3}=9\\\frac{2z}{2}=9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=27\\y=18\\z=9\end{cases}}\)

xin cho tui sửa lại tí @@

Ta có: \(2x=3y=6z\)

\(=>\frac{2x}{6}=\frac{3y}{6}=\frac{6z}{6}\)

\(=>\frac{x}{3}=\frac{y}{2}=\frac{2z}{2}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{x+y-2z}{3+2-2}=\frac{27}{3}=9\)

\(=>\hept{\begin{cases}\frac{x}{3}=9=>x=9\cdot3=27\\\frac{y}{2}=9=>y=9\cdot2=18\\\frac{2z}{2}=9=>z=9\cdot2:2=9\end{cases}}\)

Vậy: x = 27

        y = 18

        z = 9

Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\)

Mặt khác : \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)\(\Rightarrow\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\ge\frac{3}{2}\)

Dự đoán \(MinL=\frac{3}{2}\)khi a = b = c

Ta cần chứng minh \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\Leftrightarrow\left(\frac{a}{a+b}-\frac{1}{2}\right)+\left(\frac{b}{b+c}-\frac{1}{2}\right)+\left(\frac{c}{c+a}-\frac{1}{2}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{\left(a-b\right)+\left(c-a\right)}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{a-b}{2\left(b+c\right)}-\frac{c-a}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2}\left(\frac{1}{a+b}-\frac{1}{b+c}\right)-\frac{c-a}{2}\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2}.\frac{c-a}{\left(a+b\right)\left(b+c\right)}-\frac{c-a}{2}.\frac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{\left(a-b\right)\left(c-a\right)\left(a+b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đúng do \(a\ge b\ge c>0\))

Đẳng thức xảy ra khi a = b = c

\(x\left(x+1\right)=y^2+1\Leftrightarrow4x^2+4x+1=4y^2+5\)\(\Leftrightarrow\left(2x+1\right)^2-\left(2y\right)^2=5\Leftrightarrow\left(2x+2y+1\right)\left(2x-2y+1\right)=5\)

Vì \(x,y\in Z\)\(\Rightarrow2x+2y+1;2x-2y+1\)là ước của 5 nên ta có:

\(TH1:\hept{\begin{cases}2x+2y+1=5\\2x-2y+1=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)

\(TH2:\hept{\begin{cases}2x+2y+1=-1\\2x-2y+1=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}}\)

\(TH3:\hept{\begin{cases}2x+2y+1=5\\2x-2y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)

\(TH4:\hept{\begin{cases}2x+2y+1=-5\\2x-2y+1=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)

Vậy các cặp số (x;y) phải tìm là: (1;1);(1;-1);(-2;1);(-2;-1)

Tự vẽ hình:

a) Ta có: Áp dụng định lý Pytago:

\(AB^2+AC^2=BC^2\)

\(\Rightarrow AC^2=BC^2-AB^2=5^2-3^2=16=4^2\)

\(\Rightarrow AC=4\left(cm\right)\)

Từ đó ta dễ dàng tính được: \(AH.BC=AB.AC\)

\(\Rightarrow AH=\frac{AB.AC}{BC}=\frac{3.4}{5}=\frac{12}{5}\left(cm\right)\)

\(ĐKXĐ:x\ge1,y\ge2\)

Ta có : \(C=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}\)

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

Áp dụng BĐT Cô - si ta có :

\(\sqrt{1.\left(x-1\right)}\le\frac{1+x-1}{2}=\frac{x}{2}\Rightarrow\frac{\sqrt{1.\left(x-1\right)}}{x}\le\frac{1}{2}\)

\(\sqrt{2.\left(y-2\right)}\le\frac{2+y-2}{2}=\frac{y}{2}\Rightarrow\frac{\sqrt{2\left(y-2\right)}}{y}\le\frac{1}{2\sqrt{2}}\)

\(\)Do đó \(C\le\frac{2+\sqrt{2}}{4}\)

Dấu "=" xảy ra khi x = 2, y = 4