y x 4 + y x 3 - y x 2 = 45
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\(\frac{2}{3}x\frac{4}{y}=\frac{4}{45}:\frac{1}{5}\)\(=\frac{4}{15}\)
\(\frac{4}{y}=\frac{4}{15}:\frac{2}{3}\)\(=\frac{2}{5}\)
y=4:\(\frac{2}{5}\) y=10
\(\frac{3}{4}x\frac{y}{5}=\frac{15}{4}\)
\(\frac{y}{5}=\frac{15}{4}:\frac{3}{4}=5\)
y=5x5=25
Hệ đã cho \(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-4\left(x+y\right)+4=49\\\left(x-y\right)^2-2\left(x-y\right)+1=4\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+y-2\right)^2=49\\\left(x-y-1\right)^2=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=9\\x-y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=3\end{cases}}}\)
Vậy nghiệm của hệ là (x;y) = (6;3)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\)
\(\frac{x}{4}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\frac{x+y+z}{4+5+6}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\) mà x + y + z = 45
\(\Rightarrow\frac{45}{15}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow3=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\hept{\begin{cases}x=3\cdot4=12\\y=3\cdot6=18\\z=3\cdot5=15\end{cases}}\)
Ta có : Vì \(\left(x+y\right)^2-4\left(x+y\right)=45\)
\(\Rightarrow\left(x+y\right)^2-2.\left(x+y\right).2+2^2=49\)
\(\Rightarrow\left(x+y-2\right)^2=49\)\(\Rightarrow\orbr{\begin{cases}x+y-2=7\\x+y-2=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x+y=9\\x+y=-5\end{cases}}\)
Vì \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(x-y\right)^2-2.\left(x-y\right)+1^2=4\)
\(\Rightarrow\left(x-y+1\right)^2=4\)
\(\Rightarrow\orbr{\begin{cases}x-y-1=2\\x-y-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x-y=3\\x-y=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=y+3\\x=y-1\end{cases}}\)
* \(x+y=9\)\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=3;x=6\\x=y-1\Rightarrow y=5;x=4\end{cases}}\)
* \(x+y=-5\)\(\Rightarrow\orbr{\begin{cases}x=y+3\Rightarrow y=-4;x=-1\\x=y-1\Rightarrow y=-2;x=-3\end{cases}}\)
Vậy cặp số x;y là (6;3) , (4;5) , (-1;-4) , (-3;-2)
Kb với tớ nhé, mn!
a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)
a: Ta có: \(x^2-2xy+y^2-4z^2\)
\(=\left(x-y\right)^2-\left(2z\right)^2\)
\(=\left(x-y-2z\right)\left(x-y+2z\right)\)
\(=\left(6+4-2\cdot45\right)\left(6+4+2\cdot45\right)\)
\(=-8000\)
b: Ta có: \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48\)
\(=3\left(x^2+4x-21\right)+\left(x-4\right)^2+48\)
\(=3x^2+12x-63+x^2-8x+16+48\)
\(=2x^2+4x+1\)
\(=2\cdot\dfrac{1}{4}+4\cdot\dfrac{1}{2}+1\)
\(=\dfrac{7}{2}\)
Áp dụng BĐT AM-GM ta có: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\)
Suy ra: \(P=6\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+8\left[\left(x^2+y^2\right)^2-2\left(xy\right)^2\right]+\frac{5}{xy}\)
\(\ge6\left(1-\frac{3}{4}\right)+8\left(\frac{1}{4}-\frac{1}{8}\right)+\frac{5}{\frac{1}{4}}\) (Do x+y=1) \(\Rightarrow P\ge6-\frac{9}{2}+2-1+20=\frac{45}{2}\)(đpcm).
Dấu "=" xảy ra <=> x=y=1/2.
y x 4 + y x 3 - y x 2 = 45
y x (4+3-2)=45
y x 5=45
y=45:5
y=9
y x 4 + y x 3 - y x 2 = 45 Tick mik nhá !!!!!
y x ( 4 + 3 - 2 ) = 45
y x 5 = 45
y= 9