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Đặt x+1995=a , ta có\(A=\frac{a-1995}{a^2}=\frac{1}{a}-\frac{1995}{a^2}=-1995\left(\frac{1}{a^2}-\frac{1}{1995a}\right)=-1995\left(\frac{1}{a^2}-2\cdot\frac{1}{a}\cdot\frac{1}{3990}+\frac{1}{3990^2}\right)+\frac{1995}{3990^2}\)

\(=-1995\cdot\left(\frac{1}{a}-\frac{1}{3990}\right)^2+\frac{1}{7980}\)

Vì \(-1995\cdot\left(\frac{1}{a}-\frac{1}{3990}\right)^2\le0\forall a\)

\(\Rightarrow A\le\frac{1}{7980}\)

 => GTNN của A=\(\frac{1}{7980}\)

\(\Leftrightarrow\frac{1}{a}-\frac{1}{3990}=0\Leftrightarrow a=3990\Rightarrow x+1995=3990\Leftrightarrow x=1995\)

nhầm đề ak

Xin phép được sủa đề một chút nhé :)

\(\left\{{}\begin{matrix}x+y=z=a\\x^2+y^2+z^2=b\\a^2=b+4034\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\\x^2+y^2+z^2=b\\a^2-b=4034\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-b=2\left(xy+yz+zx\right)\\a^2-b=4034\end{matrix}\right.\Leftrightarrow xy+yz+zx=2017\)

\(M=x\sqrt{\frac{\left(2017+y^2\right)\left(2017+z^2\right)}{2017+x^2}}+y\sqrt{\frac{\left(2017+x^2\right)\left(2017+z^2\right)}{2017+y^2}}+z\sqrt{\frac{\left(2017+y^2\right)\left(2017+x^2\right)}{2017+z^2}}\)

\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\frac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\frac{\left(x+y\right)\left(z+x\right)\left(x+y\right)\left(y+z\right)}{\left(y+z\right)\left(z+x\right)}}\)

\(=2\left(xy+yz+zx\right)=4034\)

\(ĐKXĐ:x\ne1;x\ge0\)

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

\(P=\frac{x\sqrt{x}+2x+\sqrt{x}-2x-4\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}.\frac{x-2\sqrt{x}+1}{2}\)

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

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

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

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

b,dễ thấy \(\sqrt{x}+1>0\left(\forall x\right)\)

\(< =>\sqrt{x}-1>0\)

\(x>1\)

\(c,P=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)

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

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

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

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

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

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

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

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

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

\(\sqrt{x}+1+\frac{3}{\sqrt{x}+1}\ge\sqrt{\sqrt{x}+1.\frac{3}{\sqrt{x}+1}}=\sqrt{3}\)

\(P\le3-\sqrt{3}\)dấu "=" xảy ra khi và chỉ khi \(\sqrt{x}+1=\frac{3}{\sqrt{x}+1}\)

\(\sqrt{x}+1=\sqrt{3}\)

\(\sqrt{x}=\sqrt{3}-1\)

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

\(< =>MAX:P=\sqrt{3}\)

ĐK : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

a) \(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\cdot\frac{\left(x-1\right)^2}{2}\)

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

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

b) Để P > 0 thì \(-2\left(\sqrt{x}-1\right)>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)

Kết hợp với ĐK => Với 0 ≤ x < 1 thì P > 0

c) Ta có : \(P=-2\left(\sqrt{x}-1\right)=-2\sqrt{x}+2\le2\forall x\ge0\)

Dấu "=" xảy ra <=> x = 0 (tm)

Vậy MaxP = 2

Ta có:\(|x+2017|+|x-2|\)

         \(=|x+2017|+|2-x|\ge|x+2017+2-x|\)

\(\Rightarrow\frac{1}{|x+2017|+|2-x|}\le\frac{1}{2015}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2017\right).\left(2-x\right)\ge0\) 

Tự làm típ nha gợi í có 2 Th là 2 cái lớn hơn hoặc bằng 0 và TH2 là 2 cái nhỏ hơn 0

\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2017\ge0\\2-x\ge0\end{cases}}\\\hept{\begin{cases}x+2017< 0\\2-x< 0\end{cases}}\end{cases}}\)

Để A có GTLN thì mẫu số phải có GTNN

Áp dụng bất đẳng thức: \(|x|+|y|\ge|x+y|\)

Ta có: \(|x+2017|+|x-2|=|x+2017|+|2-x|\ge|x+2017+2-x|=2019\)

Dấu "=" xảy ra \(\Leftrightarrow xy\ge0\)

\(\Leftrightarrow-2017\le x\le2\)

Vậy GTLN của \(A=\frac{1}{2019}\Leftrightarrow-2017\le x\le2\)

Cảm ơn mọi người!

\(a)\)\(x+xy+y=-6\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(y+1\right)=-5\)

Lập bảng xét TH ra là xong 

\(b)\) CM : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\)\(\frac{x+y}{xy}\ge\frac{4}{x+y}\)

\(\Leftrightarrow\)\(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\)\(x^2+2xy+y^2-4xy\ge0\)

\(\Leftrightarrow\)\(\left(x-y\right)^2\ge0\) ( luôn đúng ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)

Xin thêm 1 slot đi hok về làm cho -,- 

\(b)\) CM : \(x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\)

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{1+1}=\frac{1}{2}\left(x+y\right)^2\) ( bđt Cauchy-Schawarz dạng Engel ) 

Ta có : 

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+2017\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}+2017\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}+2017=\frac{\left(2+\frac{4}{2}\right)^2}{2}+2017=2025\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1\)

Bài này còn có cách khác là sử dụng tính chất tổng 2 phân số nghịch đảo nhau nhá :)) 

Chúc bạn học tốt ~ 

\(c)\)\(x^2+y^2+z^2< xy+3y+2z-3\)

\(\Leftrightarrow\)\(x^2+y^2+z^2-xy-3y-2z+3< 0\)

Mà x, y, z nguyên nên \(x^2+y^2+z^2-xy-3y-2z+3\le-1\)

\(\Leftrightarrow\)\(\left(x^2-xy+\frac{y^2}{4}\right)+3\left(\frac{y^2}{4}-y+1\right)+\left(z^2-2z+1\right)\le-1+1\)

\(\Leftrightarrow\)\(\left(x-\frac{y}{2}\right)^2+3\left(\frac{y}{2}-1\right)^2+\left(z-1\right)^2\le0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-\frac{y}{2}\right)^2=0\\3\left(\frac{y}{2}-1\right)^2=0\\\left(z-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{y}{2}=\frac{2}{2}=1\\y=2\\z=2\end{cases}}}\)

Vậy \(x=;y=2;z=2\)

Chúc bạn học tốt ~ 

Answer:

a. \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)   ĐK: \(x\ge0;x\ne1\)

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

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

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

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

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

b. Vì \(0< x< 1\Rightarrow\hept{\begin{cases}\sqrt{x}\ge0\\1-\sqrt{x}>0\end{cases}}\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)

Do vậy \(\sqrt{x}\left(1-\sqrt{x}\right)>0\)

c. \(P=\sqrt{x}\left(1-\sqrt{x}\right)\)

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

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

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

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\Rightarrow x=\frac{1}{4}\)