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a: Vì \(2.8\cdot0.4=1.4\cdot0.8\)
nên 2,8/0,8=1,4/0,4; 2,8/1,4=0,8/0,4; 0,8/2,8=0,4/1,4; 1,4/2,8=0,4/0,8
b: Vì x,y,z tỉ lệ với 3;5;6 nên x/3=y/5=z/6=k
=>x=3k; y=5k; z=6k
\(M=\dfrac{2x-3y+4z}{x-11y-4z}=\dfrac{6k-15k+24k}{3k-55k-24k}=\dfrac{-15}{76}\)
a) Có x:y:z=3:5:6
\(\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{6}\)
Đặt \(k=\frac{x}{3}=\frac{y}{5}=\frac{z}{6}\)
\(\Rightarrow x=3k\)
\(\Rightarrow y=5k\)
\(\Rightarrow z=6k\)
Thay vào \(\frac{2x-3y+4z}{x-11y-4z}=\frac{2.3k-3.5k+4.6k}{3k-11.5k-4.6k}\)\(=\frac{k.\left(2.3-3.5+4.6\right)}{k.\left(3-11.5-4.6\right)}=\frac{k.15}{k.\left(-76\right)}=\frac{15}{-76}\)
b) Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{1+2y}{18}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}\)\(=\frac{2+8y}{18+6x}=\frac{2.\left(1+4y\right)}{2.\left(9+3x\right)}=\frac{1+4y}{9+3x}\)
\(\Rightarrow\frac{1+4y}{9+3x}=\frac{1+4y}{24}\Rightarrow9+3x=24\Rightarrow x=5\)
Theo đề ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)
Suy ra: \(x=3k;y=4k;z=5k\) Thay vào biểu thức P ta có:
\(P=\frac{3k+8k+15k}{6k+12k+20k}+\frac{6k+12k+20k}{9k+16k+25k}+\frac{9k+16k+25k}{12k+20k+30k}\)
\(P=\frac{26k}{38k}+\frac{38k}{50k}+\frac{50k}{62k}=\frac{13}{19}+\frac{19}{25}+\frac{25}{31}=\frac{33141}{14725}\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{2y}{\dfrac{8}{3}}=\dfrac{4z}{5}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{2y}{\dfrac{8}{3}}=\dfrac{4z}{5}=\dfrac{x+2y+4z}{\dfrac{3}{2}+\dfrac{8}{3}+5}=\dfrac{220}{\dfrac{55}{6}}=24\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{\dfrac{3}{2}}=24\\\dfrac{2y}{\dfrac{8}{3}}=24\\\dfrac{4z}{5}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=32\\z=30\end{matrix}\right.\)
Vậy ...
Đặt \(\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{6}=t\)
=> \(\left[{}\begin{matrix}x=2k-1\\y=4k-3\\z=6k-5\end{matrix}\right.\)
Ta có: 2x + 3y + 4z = 9
=> 2(2k - 1) + 3(4k - 3) + 4(6k - 5) = 9
=> 4k - 2 + 12k - 9 + 24k - 20 = 9
=> (4k + 12k + 24k) - 31 = 9
=> 40k - 31 = 9
=> 40k = 9 + 31
=> 40k = 40
=> k = 1
*Với k = 1 ta có:
x = 2.1 - 1 = 1
y = 4.1 - 3 = 1
z = 6.1 - 5 = 1
Giải:
Theo đề ra, ta có:
\(2x+3y+4z=10\) và \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{4z}{20}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{5}{19}\\\dfrac{y}{4}=\dfrac{5}{19}\\\dfrac{z}{5}=\dfrac{5}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{19}.3\\y=\dfrac{5}{19}.4\\z=\dfrac{5}{19}.5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{19}\\y=\dfrac{20}{19}\\z=\dfrac{25}{19}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{4z}{20}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
\(\dfrac{x}{3}=\dfrac{5}{19}\Rightarrow x=\dfrac{15}{19}\)
\(\dfrac{y}{4}=\dfrac{5}{19}\Rightarrow y=\dfrac{20}{19}\)
\(\dfrac{z}{5}=\dfrac{5}{19}\Rightarrow z=\dfrac{25}{19}\)
Xét \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)
Thay (1) vào P
=> P = \(\dfrac{3k+2.4k+3.5k}{2.5k+3.4k+4.5k}+\dfrac{2.5k+3.4k+4.5k}{3.3k+4.4k+5.5k}\) + \(\dfrac{3.3k+4.4k+5.5k}{4.3k+5.4k+6.5k}\)
=> P = \(\dfrac{26k}{42k}+\dfrac{42k}{50k}\) + \(\dfrac{50k}{62k}\)
=> P = \(\dfrac{13}{21}+\dfrac{21}{25}+\dfrac{25}{31}\approx2,265499232\)
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\Rightarrow x=15k;y=20k;z=24k\)
\(M=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186}{245}\)
Theo đề bài, ta có:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{2x}{6}\)=\(\dfrac{3y}{15}\)=\(\dfrac{4z}{24}\)= \(\dfrac{2x-3y+4z}{6-15+24}\)=\(\dfrac{2x-3y+4z}{15}\)(*)
\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{6}\)=\(\dfrac{x}{3}\)=\(\dfrac{11y}{55}\)=\(\dfrac{4z}{24}\)=\(\dfrac{x-11y-4z}{3-55-24}\)=\(\dfrac{x-11y-4z}{-76}\)(**)
Từ (*) và (**) suy ra:
\(\dfrac{2x-3y+4z}{15}\)=\(\dfrac{x-11y-4z}{-76}\)=\(\dfrac{2x-3y+4z}{x-11y-4z}\)=\(\dfrac{15}{-76}\)
=> m=\(\dfrac{15}{-76}\)
Vậy m=\(\dfrac{15}{-76}\)