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Bài 1 :
a) \(\frac{x}{7}=\frac{18}{14}\)
=> x.14 = 7.18
x.14 = 126
x = 126:14
x = 9
b) \(\frac{6}{x}=\frac{7}{4}\)
=> \(x=\frac{6.4}{7}=\frac{24}{7}\)
c) Theo mình đề thế này mới đúng \(\frac{5,7}{0,35}=\frac{\left(-x\right)}{0,45}\)
=> 5,7.0,45 = 0,35.(-x)
2,565 = 0,35.(-x)
(-x) = 2,565:0,35
(-x) = 513/70
=> -x = -513/70
x = 513/70
Bài 2 : Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x-y+z}{2-4+6}=\frac{8}{4}=2\)
\(\frac{x}{2}=2\)
x = 2.2
x = 4
\(\frac{y}{4}=2\)
y = 2.4
y = 8
\(\frac{z}{6}\) = 2
z = 2.6
z = 12
Vậy x=4 ; y=8 và z=12
a)\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\dfrac{x}{3}=2\Rightarrow x=6\)
\(\dfrac{y}{7}=2\Rightarrow y=14\)
b)\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\dfrac{x}{5}=2\Rightarrow x=10\)
\(\dfrac{y}{2}=2\Rightarrow y=4\)
a)\(\dfrac{x}{7}=\dfrac{18}{14}\)
\(\Rightarrow x=\dfrac{7.18}{14}=9\)
b)\(6:x=1\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{6}{x}=\dfrac{7}{4}\)
\(\Rightarrow x=\dfrac{6.4}{7}=\dfrac{24}{7}\)
c)5,7:0,35=(-x):0,45
\(\Leftrightarrow\dfrac{114}{7}=\dfrac{-x}{0,45}\)
\(\Rightarrow\left(-x\right)=\dfrac{114.0,45}{7}=\dfrac{-513}{70}\)
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
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=\(\dfrac{32}{3}\)
\(a,\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{x}{7}\) và \(x+y+z=138\)
\(\dfrac{x}{5}=\dfrac{y}{6}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}\) \(\left(1\right)\)
\(\dfrac{y}{8}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{24}=\dfrac{z}{21}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y+z}{20+24+21}=\dfrac{138}{65}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{138}{65}\\\dfrac{y}{24}=\dfrac{138}{65}\\\dfrac{z}{21}=\dfrac{138}{65}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{553}{13}\\y=\dfrac{3312}{65}\\z=\dfrac{2898}{65}\end{matrix}\right.\)
Vậy.......
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)
\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)
\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)
Đến đây tự làm tiếp nhé
b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)
=> x = 75, y = 50, z = 30
c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)
\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)
\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)
\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)
=> x=... , y=... , z=...
d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)
Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3
Với k = 3 => x = 6, y = 15
Với k = -3 => x = -6, y = -15
Vậy...
e, Tương tự câu d
b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
Câu 2:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{8}=1\)
Do đó: x=2; y=4; z=6