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

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

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

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

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

\(=3-1=2\) là số hữu tỉ (đpcm)

Ta có :

\(K=\left(-x^2-9y^2-1+6xy+6y-2x\right)+\left(-y^2+4y-4\right)+2015\)

\(=-\left[x^2+\left(3y\right)^2+1^2+2.x.3y+2.x.\left(-1\right)+2.3y.1\right]-\left(y^2-4y+4\right)+2015\)

\(=-\left(x-3y+1\right)^2-\left(y-2\right)^2+2015\)

Ta thấy \(-\left(x-3y+1\right)^2\le0\forall x;y\text{ }\text{and}\text{ }-\left(y-2\right)^2\le0\forall y\)

\(\Rightarrow-\left(x-3y+1\right)^2-\left(y-2\right)^2\le0\forall x;y\)

\(\Rightarrow K=-\left(x-3y+1\right)^2-\left(y-2\right)^2+2015\le2015\forall x;y\)

K đạt GTLN là 2015 khi \(\hept{\begin{cases}x-3y+1=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\y=2\end{cases}}\)

38+2810=

đố bạn làm được câu này cho m thuộc N. cmr 5m^3+40m chia hết cho 15

\(x^2+2xy-7y-12=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)-\left(y^2+7y+12\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2=\left(y+3\right)\left(y+4\right)\) (1)

Ta thấy VT là số CP với mọi x;y nguyên ; VP là tích 2 số nguyên liên tiếp nên ko phải là số CP

=> (1) vô lý   Hay PT trên ko có nghiệm x;y nguyên

\(x^2+2xy-7y-12=0\)

=> \(x^2+y\left(2x-7\right)=12\)

=> \(y=\frac{12-x^2}{2x-7}=\frac{-\left(x^2-12\right)}{2x-7}\)

Vì y là số nguyên nên

\(x^2-12⋮2x-7\)

=> 2x - 7 \(\in\)Ư(1) 

=> x = -3 , 4

x=-3 cho y \(\notin\)Z

x= 4 cho y = -4 (t/m)

Vậy .........

\(C=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)

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

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

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

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

\(C=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(C=\frac{\sqrt{x}}{\sqrt{x}+1}\)

P/s tham khảo nha

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)

\(a+b+c+ab+ac+bc=6abc\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)

Hay \(x+y+z+xy+yz+xz=6\)

Cần chứng minh \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=x^2+y^2+z^2\ge3\)

Ta có : \(\left(x^2+1\right)+\left(y^2+1\right)+\left(z^2+1\right)\ge2\left(x+y+z\right)\) (BĐT Cosi)

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\) (BĐT Cosi)

\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)=12\)

\(\Rightarrow x^2+y^2+z^2\ge3\) (đpcm)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)