`y xx1/2+y xx1/4+y xx1/8=22/24`

`=>y xx (1/2+1/4+1/8)=11/12`

`=> y xx (4/8+2/8+1/8)=11/12`

`=> y xx 7/8=11/12`

`=>y=11/12:7/8`

`=>y=11/12xx8/7`

`=>y=22/21`

`y xx 1/2 + y xx 1/4 + y xx 1/8 = 22/24`

`=> y xx (1/2 + 1/4 + 1/8) = 11/12`

`=> y xx ( 4/8 + 2/8+ 1/8) = 11/12`

`=> y xx 7/8 = 11/12`

`=> y= 11/12 : 7/8`

`=> y=11/12 xx 8/7`

`=> y=22/21`

Có \(\left(x+1\right)^{24}\ge0\forall x\)

\(\left(y-1\right)^{28}\ge0\forall y\)

Nên \(\left(x+1\right)^{24}+\left(y-1\right)^{28}\ge0\forall x,y\)

Dấu "=" xảy ra khi \(x=-1,y=1\)

Ta có:

(x + 1)²⁴ \(\ge\) 0 với mọi x \(\in\) R

(y - 1)²⁸ \(\ge\) 0 với mọi y \(\in\) R

\(\Rightarrow\) (x + 1)²⁴ + (y - 1)²⁸ \(\ge\) 0

\(\Rightarrow\) (x + 1)²⁴ + (y - 1)²⁸ = 0 \(\Leftrightarrow\) (x + 1)²⁴ = 0 và (y - 1)²⁸ = 0

*) (x + 1)²⁴ = 0

x + 1 = 0

x = -1

*) (y - 1)²⁸ = 0

y - 1 = 0

y = 1

Vậy x = -1; y = 1

Đặt: \(\frac{1}{x}=a,\frac{1}{y}=b\left(a,b\ne0\right)\)

HPT trở thành: \(\hept{\begin{cases}a+b=\frac{1}{24}\\4a+6b=\frac{1}{20}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=-\frac{7}{120}\\a=\frac{1}{10}\end{cases}}\)nên \(\hept{\begin{cases}x=10\\y=\frac{120}{-7}\end{cases}}\)

Lời giải

Ta có:

\(\left\{{}\begin{matrix}x\left(x+y\right)=\dfrac{1}{48}\\y\left(x+y\right)=\dfrac{1}{24}\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y\right)+y\left(x+y\right)=\dfrac{1}{48}+\dfrac{1}{24}\)

\(\Leftrightarrow\left(x+y\right)^2=\dfrac{1}{48}+\dfrac{2}{48}=\dfrac{3}{48}=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=\dfrac{1}{4}\\x+y=-\dfrac{1}{4}\end{matrix}\right.\)

Với \(x+y=\dfrac{1}{4}\) thì:

\(\left\{{}\begin{matrix}x=\dfrac{1}{48}:\dfrac{1}{4}=\dfrac{1}{12}\\y=\dfrac{1}{24}:\dfrac{1}{4}=\dfrac{1}{6}\end{matrix}\right.\)

Với \(x+y=-\dfrac{1}{4}\) thì:

\(\left\{{}\begin{matrix}x=\dfrac{1}{48}:-\dfrac{1}{4}=-\dfrac{1}{12}\\y=\dfrac{1}{24}:-\dfrac{1}{4}=-\dfrac{1}{6}\end{matrix}\right.\)

\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2=\left(124-24\right)^2=100^2=10000\)

\(\left(x-3\right)^2-\left(x+1\right)^3+12x\left(x-1\right)=\frac{49}{4}-\frac{1}{8}+\frac{\left(-6\right).\left(-3\right)}{2}\)

\(=\frac{97}{8}+9=\frac{169}{8}\)

X(X-Y)+Y(Y-X)=X² -XY +Y² -XY=(X-Y)² =(124-24)² =100² =10000

(x-3)² -(x+1)³ +12x(x-1)=x² -6x+9-x³ -3x² -3x-1+12x² -12x=-x³ +10x² -9x+8

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