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Bài 1:
\(a,\dfrac{x}{3}=\dfrac{y}{7}\) và \(x+y=20\)
\(=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\Rightarrow x=2.3=6\)
\(y=2.7=14\)
Vậy \(x=6\) và \(y=14\)
\(b,\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=6\)
\(=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\Rightarrow x=2.5=10\)
\(y=2.2=4\)
Vậy \(x=10\) và \(y=4\)
\(c,\dfrac{x}{7}=\dfrac{18}{14}\)
Từ tỉ lệ thức trên ta có:
\(14x=7.18\)
\(x=\dfrac{7.18}{14}\)
\(x=9\)
Vậy \(x=9\)
\(d,6:x=1\dfrac{3}{4}:5\)
\(6:x=\dfrac{7}{20}\)
\(x=6:\dfrac{7}{20}\)
\(x=\dfrac{120}{7}\)
Vậy \(x=\dfrac{120}{7}\)
\(e,\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(x-y+z=8\)
\(=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)
\(\Rightarrow x=2.2=4\)
\(y=2.4=8\)
\(z=2.6=12\)
Vậy \(x=4;y=8;z=12\)
a, \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{1}{2}\)
Từ đó suy ra x=1,5; y=3,5
b,\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{1}{2}\)
Từ đó suy ra x=2,5; y=1
c,\(\dfrac{x}{7}=\dfrac{18}{14}\Leftrightarrow\dfrac{x}{7}=\dfrac{9}{7}\Rightarrow x=9\)
d,\(\dfrac{6}{x}=\dfrac{\dfrac{7}{4}}{5}\Leftrightarrow\dfrac{6}{x}=\dfrac{24}{7}\left(\dfrac{\dfrac{7}{4}}{5}\right)\Leftrightarrow\dfrac{6}{x}=\dfrac{6}{\dfrac{120}{7}}\Rightarrow x=\dfrac{120}{7}\)
e,\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{8}=\dfrac{x-y+z}{2-4+8}=\dfrac{4}{3}\)
Từ đó suy ra x=\(\dfrac{8}{3}\); y=\(\dfrac{16}{3}\); z=\(\dfrac{32}{3}\)
b: Ta có: x/y=7/9
nên x/7=y/9
=>x/49=y/63
Ta có: y/z=7/3
nên y/7=z/3
=>y/63=z/27
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{49}=\dfrac{y}{63}=\dfrac{z}{27}=\dfrac{x-y+z}{49-63+27}=\dfrac{-15}{13}\)
Do đó: x=-735/13; y=-945/13; z=-405/13
c: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}=\dfrac{2x+5y-2z}{2\cdot7+5\cdot20-2\cdot32}=\dfrac{100}{50}=2\)
Do đó: x=14; y=40; z=64
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)
Do đó: x=24; y=15; z=6
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\) và \(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
\(\Rightarrow\dfrac{x+y-z}{20+24-21}=\dfrac{138}{23}\)=6
\(\Rightarrow\dfrac{x}{20}=6\Rightarrow x=120\)
\(\Rightarrow\dfrac{y}{24}=6\Rightarrow y=144\)
\(\Rightarrow\dfrac{z}{21}=6\Rightarrow z=126\)
Vậy : x = 120
y = 144
z = 126
a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)
Theo đề bài, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))
\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)
\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)
\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)
Vậy x=60; y=72; z=63
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
a; \(\dfrac{x}{5}\) = \(\dfrac{y}{6}\); \(\dfrac{7}{8}\) = \(\dfrac{z}{7}\); \(x\) + y - z = 69
z = \(\dfrac{7}{8}\). 7 = \(\dfrac{49}{8}\); Thay z = \(\dfrac{49}{8}\) vào biểu thức \(x\) + y - z = 69 ta có:
\(x\) + y - \(\dfrac{49}{8}\) = 69 ⇒ \(x\) + y = 69 + \(\dfrac{49}{8}\) = \(\dfrac{601}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}\) = \(\dfrac{y}{6}\) = \(\dfrac{x+y}{5+6}\) = \(\dfrac{\dfrac{601}{8}}{11}\) = \(\dfrac{601}{88}\)
\(x\) = \(\dfrac{601}{88}\) x 5 = \(\dfrac{3005}{88}\); y = \(\dfrac{601}{88}\) x 6 = \(\dfrac{1803}{44}\)
Vậy (\(x\); y; z) = (\(\dfrac{3005}{88}\);\(\dfrac{1803}{44}\);\(\dfrac{49}{8}\))
b; 2\(x\) = 3y; 5y = 7z; 3\(x\) + 5z + 7y = 30
2\(x\) = 3y ⇒ \(x\) = \(\dfrac{3}{2}\)y;5y = 7z ⇒ z = \(\dfrac{5}{7}\)y
thay \(x\) = \(\dfrac{3}{2}\)y; z = \(\dfrac{5}{7}\)y vào biểu thức 3\(x\) + 5z + 7y = 30 ta có:
3.\(\dfrac{3}{2}\)y + 5.\(\dfrac{5}{7}\)y + 7y = 30
y.(3.\(\dfrac{3}{2}\) + 5.\(\dfrac{5}{7}\) + 7) = 30
y.(\(\dfrac{9}{2}\) + \(\dfrac{25}{7}\) + 7) = 30
y.\(\dfrac{211}{14}\) = 30
y = 30 : \(\dfrac{211}{14}\)
y = \(\dfrac{420}{211}\); \(x\) = \(\dfrac{3}{2}\).\(\dfrac{420}{211}\) = \(\dfrac{630}{211}\); z = \(\dfrac{420}{211}\). \(\dfrac{5}{7}\) = \(\dfrac{300}{211}\)
Vậy...
Ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{6}{23}\)(do \(x+y-z=6\))
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{6}{23}.20=\dfrac{120}{23}\\y=\dfrac{6}{23}.24=\dfrac{144}{23}\\z=\dfrac{6}{23}.21=\dfrac{126}{23}\end{matrix}\right.\)
Chúc bạn học tốt!!!
Theo đề bài ta có: \(\dfrac{x}{5}=\dfrac{y}{6}\); \(\dfrac{y}{8}=\dfrac{z}{7}\)
\(\Rightarrow\)\(\dfrac{x}{20}=\dfrac{y}{24}\); \(\dfrac{y}{24}=\dfrac{z}{21}\)
\(\Rightarrow\)\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Theo t/chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)\(=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Do đó:
\(\dfrac{x}{20}=3\Rightarrow x=60\)
\(\dfrac{y}{24}=3\Rightarrow y=72\)
\(\dfrac{z}{21}=3\Rightarrow z=63\)
Vậy................