Cho \(\dfrac{1}{x}=\dfrac{3}{y}=\dfrac{4}{z}\). Cmr (x+3y+4z)2=26(x2+9y2+16z2)
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MH
17 tháng 9 2021
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
17 tháng 9 2021
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
4 tháng 10 2021
\(\left(x+4\right)\left(x^2-4x+16\right)=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
1 tháng 12 2021
\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)
\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)
23 tháng 8 2021
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(3)=(y)/(4)=(x+y)/(3+4)=(90)/(7)`
`->` $\begin{cases}x=\dfrac{90}{7}.3=\dfrac{30}{7} \\ y=\dfrac{90}{7}.4=\dfrac{360}{7} \end{cases}$
23 tháng 8 2021
1)\(\dfrac{x}{5}=\dfrac{y}{3}\) áp dụng...ta đc:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{20}{2}=10\)
x=50
y=30
5 tháng 10 2021
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
S
3 tháng 1 2018
a) \(\dfrac{3cy-4bz}{2x}=\dfrac{4az-2cx}{3y}=\dfrac{2bx-3ay}{4z}\)
=> \(\dfrac{3cy-4bz}{2x}.\dfrac{2x}{2x}=\dfrac{4az-2cx}{3y}.\dfrac{3y}{3y}=\dfrac{2bx-3ay}{4z}.\dfrac{4z}{4z}\)
=> \(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}\)
Áp dụng t/c ...
\(\dfrac{6cxy-8bzx}{4x^2}=\dfrac{12azy-6cxy}{9y^2}=\dfrac{8bxz-12ayz}{16z^2}=\dfrac{6cxy-8bzx+12azy-6cxy+8bxz-12ayz}{4x^2+9y^2+16z^2}=\dfrac{0}{4x^2+9y^2+16z^2}=0\)
Ta có : 6cxy - 8bzx = 0
=> 6cxy = 8bzx
=>3cx = 4bz
=>\(\dfrac{c}{4z}=\dfrac{b}{3y}\) (1)
Ta có : 12azy - 6cxy = 0
=> 12azy = 6cxy
=> 4az = 2cx
=> \(\dfrac{a}{2x}=\dfrac{c}{4z}\) (2)
Từ (1),(2) => \(\dfrac{a}{2x}=\dfrac{b}{3y}=\dfrac{c}{4z}\) (ĐPCM)