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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 :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)
=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)
\(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)
\(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
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) Ta có :\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)\(=\dfrac{5x+2y+z}{555+444+333}=\dfrac{1100}{1332}=\dfrac{275}{333}\)
Từ đó tìm được x;y;z
b) Từ \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)
Đặt \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4k\\y^2=9k\end{matrix}\right.\)
\(\Rightarrow x^2\cdot y^2=4k\cdot9k=52\)
\(\Rightarrow36k^2=52\)
\(\Rightarrow k^2=\dfrac{13}{9}\) (sai đề)
b: Sửa đề: x^2+y^2=52
Đặt x/2=y/3=k
=>x=2k; y=3k
x^2+y^2=52
=>4k^2+9k^2=52
=>k^2=4
TH1: k=2
=>x=4; y=6
TH2: k=-2
=>x=-4; y=-6
c: Đặt x/5=y/3=k
=>x=5k; y=3k
x^2-y^2=16
=>25k^2-9k^2=16
=>k^2=1
TH1: k=1
=>x=5; y=3
TH2: k=-1
=>x=-5; y=-3
d: Đặt x/2=y/3=k
=>x=2k; y=3k
Ta có: xy=54
=>2k*3k=54
=>6k^2=54
=>k^2=9
TH1: k=3
=>x=6; y=9
TH2: k=-3
=>x=-6; y=-9
e: Đặt x/4=y/3=k
=>x=4k; y=3k
Ta có: xy=12
=>4k*3k=12
=>k^2=1
TH1: k=1
=>x=4; y=3
TH2: k=-1
=>x=-4; y=-3
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) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
Lời giải:
a) Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{3}=t\Rightarrow x=2t; y=5t; z=3t\)
Khi đó:
\(3x+2y-z=13\)
\(\Leftrightarrow 3.2t+2.5t-3t=13\)
\(\Leftrightarrow 13t=13\Rightarrow t=1\)
Do đó: \(\left\{\begin{matrix} x=2t=2\\ y=5t=5\\ z=3t=3\end{matrix}\right.\)
b) Đặt \(\frac{x}{2}=\frac{y}{3}=t\Rightarrow x=2t, y=3t\)
Khi đó: \(x^2+y^2=52\Leftrightarrow (2t)^2+(3t)^2=52\)
\(\Leftrightarrow 13t^2=52\Rightarrow t^2=4\rightarrow t=\pm 2\)
Với \(t=2\Rightarrow x=2t=4; y=3t=6\)
Với \(t=-2\Rightarrow x=2t=-4; y=3t=-6\)