Tính : (\(\frac{7y}{3}+\frac{5y}{2}\))2
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ta co 3x-2/5 =5y+8 /9 (1)
ap dung tinh chat day ti so bang nhau ta co
3x-2/5=5y+8/ 9=3x-2+5y+8 /9 =3x+5y+6 /14 =3x+5y+6 /7y
=> 14=7y=>y= 2
thay y=2 vao (1) ta co 3x-2 /5 = 5.2+8 /9= 18/9=2 =>3x-2 /5 =2
=>3x-2 =10 =>3x =12 =>x =4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Ta có xy=-6
Với x=-6 => y=1
x=-3 => y=2
x= -2 => y=3
x=-1 => y=6
2/ Ta có x=y+4
Thay x=y+4 vào bt, ta được
<=> y+4-3/y-2 =3/2
<=> y+1/y-2=3/2
<=> 2(y+1)=3(y-2)
<=> 2y +2 = 3y - 6
<=> 3y - 2y= 2+ 6
<=> y= 8 <=> x= 12
3/ -4/8 = x/-10 <=> x= (-4)*(-10)/8=5
-4/8 = -7/y <=> y=(-7)*8/(-4) =14
-4/8 = z/-24 <=> z= (-4)*(-24)/8=12
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\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{\left(1+5y\right)-\left(1+7y\right)}{5x-4x}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{-2y}{x}\)
\(\Rightarrow\frac{1+3y}{12}=\frac{-10y}{5x}\)
\(\Rightarrow\frac{1+5y}{5x}=-\frac{10y}{5x}\)
\(\Rightarrow1+5y=-10y\)
\(\Rightarrow-15y=1\)
\(\Rightarrow y=\frac{1}{-15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e) Ta có:
\(\left\{{}\begin{matrix}2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Leftrightarrow\frac{1}{7}.\frac{x}{3}=\frac{1}{7}.\frac{y}{2}\Leftrightarrow\frac{x}{21}=\frac{y}{14}\\7z=5y\Leftrightarrow\frac{z}{5}=\frac{y}{7}\Leftrightarrow\frac{1}{2}.\frac{z}{5}=\frac{1}{2}.\frac{y}{7}\Leftrightarrow\frac{z}{10}=\frac{y}{14}\end{matrix}\right.\)
\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=20\end{matrix}\right.\)
f)Ta có:
\(\frac{x}{4}=\frac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)
\(\Rightarrow xy=4k5k=20k^2=80\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
TH1: \(k=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
TH2: \(k=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
g)Ta có:
\(\frac{x+3}{5}=\frac{y-2}{3}=\frac{z-1}{7}=\frac{3\left(x+3\right)}{15}=\frac{5\left(y-2\right)}{15}=\frac{7\left(z-1\right)}{49}=\frac{3x+9}{15}=\frac{5y-10}{15}=\frac{7z-7}{49}=\frac{3x+9+5y-10-\left(7z-7\right)}{15+15-49}=\frac{3x+5y-7z+\left(9-10+7\right)}{-19}=\frac{38}{-19}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-13\\y=-4\\z=-13\end{matrix}\right.\) h)Ta có: \(\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x^2}{4^2}=\frac{y^2}{3^2}=\frac{x^2-y^2}{16-9}=\frac{63}{7}=9\) \(\Rightarrow\left\{{}\begin{matrix}x^2=144\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\\y^2=81\Leftrightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\end{matrix}\right.\) Vậy \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=9\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-9\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
t =1/3x-1
ta có phương trình
\(\hept{\begin{cases}2t+5y=7\\-3t+7y=4\end{cases}}\)
=> t= 1
y =1
ta có 1/3x-1=1
=> x= 2/3
vậy hệ phương trình có nghiệm x y lần lượt là 2/3 và 1
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ta có tích của ba đơn thức trên là :
\(-\frac{2}{11}.x^2y^{41}.\frac{11}{7}x^5y^6.\frac{-49}{3}x^7y=\frac{14}{3}.x^{14}.y^{62}\ge0\)
Do đó ba đơn thức không thể cùng âm được.
\(\left(\frac{7y}{3}+\frac{5y}{2}\right)^2\)
\(=\left(\frac{14y+15y}{6}\right)^2\)
\(=\left(\frac{29y}{6}\right)^2\)
\(=\frac{841y}{36}\)