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Ta có: \(5x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{5}=k\left(k\inℝ\right)\)
Thay vào \(3\left(3k\right)^2-\left(5k\right)^2=-28\)
\(\Leftrightarrow27k^2-25k^2=-28\)
\(\Leftrightarrow2k^2=-28\)
\(\Rightarrow k^2=-14\)
=> vô lý
=> không tồn tại x,y thỏa mãn
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\(\frac{x}{5}=\frac{y}{3},\frac{y}{6}=\frac{z}{21}\)
\(\frac{x}{30}=\frac{y}{18},\frac{y}{18}=\frac{z}{63}\)
\(\Rightarrow\frac{x}{30}=\frac{y}{18}=\frac{z}{63}=\frac{5x}{150}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{30}=\frac{y}{18}=\frac{z}{63}=\frac{5x}{150}=\frac{5x+y-z}{150+18-63}=\frac{28}{105}=\frac{4}{15}\)
đến đây tự tính tiếp
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a) \(\frac{x}{5}=\frac{y}{3};\frac{y}{2}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)
Theo tính chất dãy tỉ số bằng nhau
Ta có: \(\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
\(\Rightarrow x=2\times10=20\)
\(y=2\times6=12\)
\(z=2\times21=42\)
Vậy x = 20; y = 12 ; z = 42
b) \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{z-y}{5-4}=\frac{-6}{1}=-6\)
\(\Rightarrow x=\left(-6\right)\times3=-18\)
\(y=\left(-6\right)\times4=-24\)
\(z=\left(-6\right)\times5=-30\)
Vậy x = -18; y = -24; z = -30
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Đặt \(\frac{x}{5}=\frac{y}{6}=\frac{z}{3}=k\)
\(\Rightarrow\hept{\begin{cases}x=5k\\y=6k\\z=3k\end{cases}}\)
Thay các x , y , z vào biểu thức , ta có :
5x + y - 2z = 28
=> 25k + 6k - 6k = 28
=> 25k = 28
=> k = \(\frac{28}{25}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{5.28}{25}=\frac{28}{5}\\y=\frac{6.28}{25}=\frac{168}{25}\\z=\frac{3.28}{25}=\frac{84}{25}\end{cases}}\)
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Gọi 5x = 5y (1)
x + y = 28 (2)
từ (2) có x = 28 - y
Thế (2) vào (1) có:
5( 28 - y)= 2y
140 - 5y = 2y
140 = 2y + 5y
140 = 7y
y = 140 : 7
y = 20
thay y = 20 vào (2) có: x = 28 - 20 = 8
Vậy (x,y) = (8,20)
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\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
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(5x + 3)28 = 167
(5x+3)28 = (24)7
(5x+3)28 = 228
5x + 3 = 2
5x = 2 - 3
5x = -1
x = -1/5
(5x+3)28 = (24)7 = -(24)7
\(\Rightarrow\) 5x+3 = 2 = -(2)
\(\left[{}\begin{matrix}5x+3=2\\5x+3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-1\\5x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=-1\end{matrix}\right.\)
Vậy x = \(\dfrac{-1}{5}\) hoặc x = -5