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1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).
Ta có (x+y)xy=x2+y2-xy
=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)
<=>\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2+\frac{3}{4}\left(\frac{1}{x}-\frac{1}{y}\right)^2\ge\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
<=> \(0\le\frac{1}{x}+\frac{1}{y}\le4\)
mà \(A=\frac{1}{x^3+y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)
Vậy Max A =16 khi \(x=y=\frac{1}{2}\)
\(P=\frac{1}{xy-xyz-z}+\frac{1}{yz-xyz-x}+\frac{1}{xz-xzy-y}\) .Do xyz=-z =>-xyz=1 và x+y+z=0 . Thế vào P ta được \(P=\frac{1}{xy+1+x+y}+\frac{1}{yz+1+y+z}+\frac{1}{xz+1+x+z}\)\(P=\frac{1}{\left(x+1\right)\left(y+1\right)}+\frac{1}{\left(y+1\right)\left(z+1\right)}+\frac{1}{\left(x+1\right)\left(z+1\right)}\) =\(\frac{z+1+x+1+y+1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(P=\frac{3}{xyz+z+xz+yz+xy+1+x+y}\) =\(\frac{3}{xy+yz+xz}\) (Do x+y+z=0; xyz=-1)
x+y+z=0 => (x+y+z)2=0 => x2+y2+z2 +2(xy+yz+xz)=0 => 2(xy+yz+xz)=-6 => xy+yz+xz=-3 Thế vào P ta được :
\(P=\frac{3}{-3}=-1\) . Chúc bạn học tốt
Ta có: \(x^2+\frac{1}{x^2}=14\)(1)
=> \(x^2+\frac{1}{x^2}+2=16\)
<=> \(\left(x+\frac{1}{x}\right)^2=16\)
<=> \(x+\frac{1}{x}=4\) (Vì x > 0)
<=> \(\left(x+\frac{1}{x}\right)^3=4^3\)
<=> \(x^3+3x+\frac{3}{x}+\frac{1}{x^3}=64\)
<=> \(x^3+\frac{1}{x^3}=64-3\left(x+\frac{1}{x}\right)\)
<=> \(x^3+\frac{1}{x^3}=64-3.4=52\) (2)
Từ (1) và (2) nhân vế theo vế:
\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=14.52=728\)
=> \(x^5+\frac{1}{x}+x+\frac{1}{x^5}=728\)
=> \(x^5+\frac{1}{x^5}=728-4=724\)