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cách 2: 2x + 1 = 3y + 3= 5z + 3 => \(2.\left(x+\frac{1}{2}\right)=3.\left(y+1\right)=5.\left(z+\frac{3}{5}\right)\)
=> \(\frac{2.\left(x+\frac{1}{2}\right)}{30}=\frac{3\left(y+1\right)}{30}=\frac{5\left(z+\frac{3}{5}\right)}{30}\) => \(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z+\frac{3}{5}}{6}\)
Áp dụng tc dãy tỉ số bằng nhau => \(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z+\frac{3}{5}}{6}=\frac{x+\frac{1}{2}-y-1+z+\frac{3}{5}}{15-10+6}=\frac{1,1+\frac{1}{10}}{11}=\frac{6}{55}\)
=> \(x+\frac{1}{2}=\frac{6}{55}.15=\frac{18}{11}\Rightarrow x=\frac{25}{11}\)
tương tự, y = ...; z ...
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Đặt 2x + 1 = 3y + 3 = 5z + 3 = k
=> x = \(\frac{k-1}{2}\); y = \(\frac{k-3}{3}\); z = \(\frac{k-3}{5}\)
=> x - y + z = \(\frac{k-1}{2}\)- \(\frac{k-3}{3}\) + \(\frac{k-3}{5}\) = \(\frac{15\left(k-1\right)-10\left(k-3\right)+6\left(k-3\right)}{30}=\frac{11k-3}{30}=1,1\)
=> 11k - 3 = 33 => k = \(\frac{36}{11}\)
=> x = \(\frac{25}{22}\); y = \(\frac{1}{11}\); z = \(\frac{3}{55}\)
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Gợi ý nhá
Bài 3: câu 1: làm tương tự như câu hỏi lần trước bạn gửi.
b) Bạn chỉ cần cho tử và mẫu mũ 3 lên. theé là dễ r
\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow=\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow=\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)
tự tính tiếp =)
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1)a)
x/3=y/4=>x/15=y/20
y/5=z/7=>y/20=z/28
=>x/15=y/20=z/18
Áp dụng tính chất dãy tỉ số bằng nhau ta có
x/15=y/20=z/28=2x+3y-z/30+60-28=372/62=6
=>x=90
y=120
z=168
b)
2x=3y=5z
2x=3y=>x/3=y/2=>x/15=y/10
3y=5z=>y/5=z/3=>y/10=z/6
Áp dụng tính chất dãy tỉ số bằng nhau ta có
x/15=y/10=z/6=x+y-z/15+10-6=95/19=5
=>x=75
y=50
z=30
a) Ta co :x/3=y/4 suy ra x/15=y/20 (1)
y/5=z/7 suy ra y/20=z/28 (2)
Tu (1) va (2) suy ra y/20=x/15=z/28
còn lại tự làm nhé dễ rùi
b)Ta co : 2x=3y=5z suy ra x phan 1/2=y phan 1/3 = z phan 1/5
de rui tu lam nha
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1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
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b) 3x = 2y
=> x/2 = y/3 (1)
7y = 5z
=> y/5 = z/7 (2)
Từ (1) và (2), có:
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\)\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
áp dụng tính chất của dãy tỉ số bằng nhau, có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)
x/10 = 2 => x = 2 x 10 =20
y/15 = 2 => y = 2 x 15 = 30
z/21 = 2 => z = 2 x 21 = 42
Ta có: \(2x+1=3y+3=5z-3.\)
\(\Rightarrow2.\left(x+\frac{1}{2}\right)=3.\left(y+1\right)=5.\left(z-\frac{3}{5}\right)\)
\(\Rightarrow\frac{2.\left(x+\frac{1}{2}\right)}{30}=\frac{3.\left(y+1\right)}{30}=\frac{5.\left(z-\frac{3}{5}\right)}{30}.\)
\(\Rightarrow\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z-\frac{3}{5}}{6}\) và \(x-y+z=1,1.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z-\frac{3}{5}}{6}=\frac{x+\frac{1}{2}-y-1+z-\frac{3}{5}}{15-10+6}=\frac{\left(x-y+z\right)+\left(\frac{1}{2}-1-\frac{3}{5}\right)}{11}=\frac{1,1-\frac{11}{10}}{11}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x+\frac{1}{2}}{15}=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\\\frac{y+1}{10}=0\Rightarrow y+1=0\Rightarrow y=-1\\\frac{z-\frac{3}{5}}{6}=0\Rightarrow z-\frac{3}{5}=0\Rightarrow z=\frac{3}{5}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(-\frac{1}{2};-1;\frac{3}{5}\right).\)
Chúc bạn học tốt!