Bạn sao chép câu này nhé.

Tớ làm trên đó rồi.

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3( 3x - 1 ) = 3x + 5

<=> 9x - 3x = 5 + 3

<=> 6x = 8

<=> x = 4/3

Vậy S = { 4/3 }

kết bn với mik nha! >.< mik chỉ cách lm cho. Xin sđt

\(P=3x+2y+\frac{6}{x}+\frac{8}{y}\)

\(2P=6x+4y+\frac{12}{x}+\frac{16}{y}\)

\(=\left(3x+\frac{12}{x}\right)+\left(y+\frac{16}{y}\right)+3\left(x+y\right)\)

\(\ge2\sqrt{3x\cdot\frac{12}{x}}+2\sqrt{y\cdot\frac{16}{y}}+3\cdot6=12+8+18=38\)( bđt AM-GM và giả thiết x + y ≥ 6 )

=> P ≥ 19

Đẳng thức xảy ra <=> \(\hept{\begin{cases}3x=\frac{12}{x}\\y=\frac{16}{y}\\x+y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy MinP = 19

Ta có: \(P=3x+2y+\frac{6}{x}+\frac{8}{y}=\left(\frac{3}{2}x+\frac{3}{2}y\right)+\left(\frac{3}{2}x+\frac{6}{x}\right)+\left(\frac{y}{2}+\frac{8}{y}\right)\)

Vì \(\frac{3}{2}x+\frac{3}{2}y=\frac{3}{2}\left(x+y\right)\ge\frac{3}{2}.6=9\)

\(\frac{3x}{2}+\frac{6}{x}\ge2\sqrt{\frac{3x}{2}.\frac{6}{x}}=6;\frac{y}{2}+\frac{8}{y}\ge2\sqrt{\frac{y}{2}.\frac{8}{y}}=4\)

\(\Rightarrow P\ge9+6+4=19\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}x+y=6\\\frac{3x}{2}=\frac{6}{x}\\\frac{y}{2}=\frac{8}{y}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy GTNN của P là 19

a/ \(9x^2+y^2=18x+6y-18\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

a) \(9x^2+y^2=18x+6y-18\)

\(\Rightarrow9x^2+y^2-18x-6y+9=0\)

\(\Rightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)

\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)

Mà \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)

Vậy ....................

Câu b để mik nghĩ  tiếp

\(P=\frac{n^3+2n^2-1}{n^3+2n^2+2n+1}\)

ĐKXĐ : \(n\ne-1\)

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

\(=\frac{\left(n+1\right)\left(n^2+n-1\right)}{\left(n+1\right)\left(n^2-n+1\right)+2n\left(n+1\right)}=\frac{\left(n+1\right)\left(n^2+n-1\right)}{\left(n+1\right)\left(n^2+n+1\right)}=\frac{n^2+n-1}{n^2+n+1}\)

Với n nguyên, đặt ƯC( n² + n - 1 ; n² + n + 1 ) = d

=> n² + n - 1 ⋮ d và n² + n + 1 ⋮ d

=> ( n² + n + 1 ) - ( n² + n - 1 ) ⋮ d

=> n² + n + 1 - n² - n + 1 ⋮ d

=> 2 ⋮ d => d = 1 hoặc d = 2

Dễ thấy n² + n + 1 ⋮/ 2 ∀ n ∈ Z ( bạn tự chứng minh )

=> loại d = 2

=> d = 1

=> ƯCLN( n² + n - 1 ; n² + n + 1 ) = 1

hay P tối giản ( đpcm )

Ta có:

\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^3+y^3+z^3+xyz\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x^3+y^3+z^3+xyz=0\end{cases}}\)

Xét \(x+y+z=0\)

\(\Rightarrow\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}}\)

\(\Rightarrow F=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)

\(=\frac{x}{-x}+\frac{y}{-y}+\frac{z}{-z}=-1-1-1=-3\)

Xét \(x^3+y^3+z^3+xyz=0\)

\(\Rightarrow F-1=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-1\)

\(=\frac{x^3+y^3+z^3+xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)

\(\Rightarrow F=1\)

Đặt \(a=x;2b=y;3c=z\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow xy+yz+zx=0\)

\(\Rightarrow Q=\frac{\frac{y}{2}.\frac{z}{3}}{x^2}+\frac{x.\frac{c}{3}}{2y^2}+\frac{x.\frac{y}{2}}{3z^2}\)

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

\(=\frac{x^3y^3+y^3z^3+z^3x^3-3x^2y^2z^2+3x^2y^2z^2}{6x^2y^2z^2}\)

\(=\frac{\left(xy+yz+zx\right)\left(x^2y^2+y^2z^2+z^2x^2-x^2yz-y^2zx+z^2xy\right)+3x^2y^2z^2}{6x^2y^2z^2}\)

\(=\frac{3x^2y^2z^2}{6x^2y^2z^2}=\frac{1}{2}\)