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Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
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a) \(\dfrac{x}{6}=\dfrac{y}{-5}=\dfrac{z}{4}\rightarrow\dfrac{z}{4}=\dfrac{2x}{12}=\dfrac{y}{-5}=\dfrac{2x+y}{12-5}=\dfrac{21}{7}=3\)
-Suy ra: x=6.3=18; y=(-5).3=-15; z=4.3=12
b) \(\dfrac{x}{y}=\dfrac{9}{10}\rightarrow\dfrac{x}{9}=\dfrac{y}{10}\rightarrow\dfrac{x}{27}=\dfrac{y}{30}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\rightarrow\dfrac{z}{4}=\dfrac{y}{3}\rightarrow\dfrac{z}{40}=\dfrac{y}{30}\)
\(\rightarrow\dfrac{x}{27}=\dfrac{y}{30}=\dfrac{z}{40}=\dfrac{x-y-z}{27-30-40}=\dfrac{43}{43}=1\)
\(\rightarrow x=27;y=30;z=40\)
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\(\frac{20-x}{x+7}=\frac{2}{5}\)
=> \(5\left(20-x\right)=2\left(x+7\right)\)
<=> 100 - 5x = 2x + 14
=> 2x + 5x = 100 - 14
=> 7x = 86
=> x = 86/7
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a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-1}{30+60-28}=\dfrac{186}{62}=3\)
\(\dfrac{x}{15}=3\Rightarrow x=45\\ \dfrac{y}{20}=3\Rightarrow y=60\\ \dfrac{z}{28}=3\Rightarrow x=84\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\dfrac{x}{2}=5\Rightarrow x=10\\ \dfrac{y}{3}=5\Rightarrow y=15\\ \dfrac{z}{4}=5\Rightarrow z=20\)
c) x : y :z : t = 3 : 4 : 5 :6\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{t}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{t}{6}=\dfrac{x+y+z+t}{3+4+5+6}=\dfrac{3,6}{18}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}\Rightarrow x=\dfrac{3}{5}\\ \dfrac{y}{4}=\dfrac{1}{5}\Rightarrow y=\dfrac{4}{5}\\ \dfrac{z}{5}=\dfrac{1}{5}\Rightarrow z=1\\ \dfrac{t}{6}=\dfrac{1}{5}\Rightarrow t=\dfrac{6}{5}\)
d) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=-\dfrac{49}{7}=-7\)
\(\dfrac{x}{10}=-7\Rightarrow x=-70\\ \dfrac{y}{15}=-7\Rightarrow y=-105\\ \dfrac{z}{12}=-7\Rightarrow z=-84\)
e) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\dfrac{x}{2}=4\Rightarrow x=8\\ \dfrac{y}{3}=4\Rightarrow y=12\\ \dfrac{z}{4}=4\Rightarrow z=16\)
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Câu 3:
\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)
\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)
Câu b:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)
\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)
Câu c:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)
\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)
Câu d:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)
\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)
Câu e:
\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)
\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)
3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)
4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)
\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)
5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)
6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)
7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)
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Ta có: \(\frac{x}{y}=\frac{9}{10}\Rightarrow\frac{x}{9}=\frac{y}{10}\Rightarrow\frac{x}{27}=\frac{y}{30}\left(1\right)\)
\(\frac{y}{z}=\frac{3}{4}\Rightarrow\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{y}{30}=\frac{z}{40}\)(2)
- Từ (1) và (2) \(\Rightarrow\)\(\frac{x}{27}=\frac{y}{30}=\frac{z}{40}\)
- Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
....................
Đến đây bạn tự làm nhé
Chúc bạn học tốt💯
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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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a: 3x=7y
=>x/7=y/3=(x-y)/(7-3)=-16/4=-4
=>x=-28; y=-12
b: x/6=y/5
=>x/6=2y/10=(x+2y)/(6+10)=20/16=5/4
=>x=30/4=15/2; y=25/4
c: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot2+3\cdot\left(-3\right)+5\cdot5}=\dfrac{6}{20}=\dfrac{3}{10}\)
=>x=3/5; y=-9/10; z=3/2
d: x/2=y/3
=>x/8=y/12
y/4=z/5
=>y/12=z/15
=>x/8=y/12=z/15
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
=>x=16; y=24; z=30
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a) ADTCDTSBN
có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)
=> x/2 = 3 => x = 6
y/3 = 3 => y = 9
z/4 = 3 => z = 12
KL:...
b,c làm tương tự nha
d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)
ADTCDTSBN
có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)
=>...
e) ADTCDTSBN
có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)
\(=\frac{21+6}{9}=\frac{27}{9}=3\)
=>...
g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)
mà xy = 12 => 4k.3k = 12
12.k2 = 12
k2 = 1
=> k = 1 hoặc k = -1
=> x = 4.1 = 4
y = 3.1 = 3
x=4.(-1) = -4
y=3.(-1) = -3
KL:...
h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
=>...
\(\frac{x}{3}=\frac{y}{4}\Leftrightarrow\frac{x}{12}=\frac{y}{16},\frac{x}{4}=\frac{z}{5}\Leftrightarrow\frac{x}{12}=\frac{z}{15}\)
suy ra \(\frac{x}{12}=\frac{y}{16}=\frac{z}{15}\).
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{12}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{12+16+15}=\frac{43}{43}=1\)
\(\Leftrightarrow\hept{\begin{cases}x=1.12=12\\y=1.16=16\\z=1.15=15\end{cases}}\)