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Đặt \(\frac{x}{-5}=\frac{y}{6}=\frac{z}{-2}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow x=-5k;y=6k;z=-2k\)
\(\Rightarrow A=\frac{3.k.\left(-5\right)+6.k-2.\left(-2\right).k}{-3.\left(-5\right).k-5.6.k+6.\left(-2\right).k}=\frac{-15k+6k+4k}{15k-30k-12k}=\frac{-5k}{-27k}=\frac{5}{27}\)
Vậy \(A=\frac{5}{27}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
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Ta có: 3x = 4y => \(\frac{x}{4}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{6}\)
5y = 6z => \(\frac{y}{6}=\frac{z}{5}\)
=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y-z}{8+6-5}=\frac{18}{9}=2\)
=> \(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{6}=2\\\frac{z}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.8=16\\y=2.6=12\\z=2.5=10\end{cases}}\)
Vậy ....
![](https://rs.olm.vn/images/avt/0.png?1311)
3x=4y
=>x/4=y/3
=>x/8=y/6
5y=6z
=>y/6=z/5
=>x/8=y/6=z/5
Đặt x/8=y/6=z/5=k
=>x=8k; y=6k; z=5k
xyz=30
=>8k*6k*5k=30
=>240k^3=30
=>k^3=1/8
=>k=1/2
=>x=8*1/2=4; y=6*1/2=3; z=5*1/2=5/2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=k\)\(\Rightarrow\begin{cases}x=5k+1\\y=3k+2\\z=2k+2\end{cases}\)
Theo đề bài: 3x-5y+6z <=> 3(5k+1)-5(3k+2)+6(2k+2)=9
<=>15k+3-15k-10+12k+12=9
<=>12k+5=9
<=>12k=4
<=>k=\(\frac{4}{12}=\frac{1}{3}\)
=>\(\Rightarrow\begin{cases}x=5.\frac{1}{3}+1=\frac{8}{3}\\y=3.\frac{1}{3}+2=3\\z=2.\frac{1}{3}+2=\frac{8}{3}\end{cases}\)
Vậy ............
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}\) = \(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\)
= \(\frac{3x-3-\left(5y-10\right)+6z-12}{15-15+12}\) = \(\frac{3x-3-5y+10+6x-12}{12}\)
= \(\frac{9-5}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\)
=> \(\left[\begin{array}{nghiempt}x-1=\frac{5}{3}\\y-2=1\\z-2=\frac{2}{3}\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x=\frac{8}{3}\\y=3\\z=\frac{8}{3}\end{array}\right.\)
Vậy x = \(\frac{8}{3}\) ; y = 3 ; z = \(\frac{8}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
khó quá
k nhé tớ k lại cho
hihihiihih ^_^ ~ hihihihihih
Vì \(\left(3x-2y\right)^{100}\ge0\forall x,y\inℤ\)
\(|5y-6z|\ge0\forall y,z\inℤ\Rightarrow|5y-6z|^{153}\ge0\forall y,z\inℤ\)
Nên \(\Rightarrow\hept{\begin{cases}(3x-2y)^{100}=0\\|5y-6z|^{153}=0\end{cases}}\Rightarrow\hept{\begin{cases}3x-2y=0\\5y-6z=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=6z\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{6}=\frac{z}{5}\end{cases}}}\)
Từ \(\frac{x}{2}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)suy ra\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
Ta có
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{2x}{8}=\frac{5y}{30}=\frac{3z}{15}=\frac{2x-5y+3z}{8-30+15}=\frac{56}{-7}=-8\)
Do đó
\(\frac{x}{4}=-8\Rightarrow x=-32\)
\(\frac{y}{6}=-8\Rightarrow y=-48\)
\(\frac{z}{5}=-8\Rightarrow z=-40\)
Vậy \(x=-32;y=-48;z=-40\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho
Vì 3x = 5y = 6z
=> \(\frac{x}{5}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)
\(=>\frac{x}{30}=\frac{y}{18};\frac{y}{18}=\frac{z}{15}\)
\(hay\)\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{30}=\frac{y}{18}=\frac{z}{15}=\frac{x+y-z}{30+18-15}=\frac{22}{33}=\frac{2}{3}\)
Do đó suy ra:
\(3x=\frac{2}{3}=>x=\frac{2}{9}\)
\(5y=\frac{2}{3}=>y=\frac{2}{15}\)
\(6z=\frac{2}{3}=>x=\frac{1}{9}\)
Vậy \(\left(x;y;z\right)\in\left\{\frac{2}{9};\frac{2}{15};\frac{1}{9}\right\}\)
toán lớp 7 ak