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a, Ta có: \(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\Leftrightarrow\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}\) và \(5x+y-2z=28\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}=\dfrac{5x+y-2z}{50+6-42}=\dfrac{28}{14}=2\)
+) \(\dfrac{5x}{50}=2\Rightarrow5x=100\Rightarrow x=20\)
+) \(\dfrac{y}{6}=2\Rightarrow y=12\)
+) \(\dfrac{2z}{42}=2\Rightarrow2z=84\Rightarrow z=42\)
Vậy ...
b, Ta có:
\(3x=2y\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}\)
\(7y=5z\Leftrightarrow\dfrac{y}{5}=\dfrac{z}{7}\)
Ta lại có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{15}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{15}=\dfrac{z}{21}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\) và \(x-y+z=32\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{32}{16}=2\)
+) \(\dfrac{x}{10}=2\Rightarrow x=20\)
+) \(\dfrac{y}{15}=2\Rightarrow y=30\)
+) \(\dfrac{z}{21}=2\Rightarrow z=42\)
Vậy ...
D = x6 - x4yz + x3 yz2 - x3y2z + x3y2z - z6 + 2018
=> D = -z6 + x3 yz2 + ( - x4) yz + x6 + 2018
=> D = - ( z6 - x3 yz2 + x4yz - x6 - 2018 )
.... :)
b/
\(\dfrac{x+y-6}{z}=\dfrac{x+z+4}{y}=\dfrac{y+z+2}{x}=\dfrac{6}{x+y+z}\)
Đặt 0\(k=\dfrac{x+y-6}{z}=\dfrac{x+z+4}{y}=\dfrac{y+z+2}{x}=\dfrac{6}{x+y+z}\)
\(\Rightarrow k=\dfrac{\left(x+y-6\right)+\left(x+z+4\right)+\left(y+z+2\right)}{z+y+x}\)
\(\Rightarrow k=\dfrac{2x+2y+2z-6+4+2}{z+y+x}\)
\(\Rightarrow k=\dfrac{2\left(x+y+z\right)}{z+y+x}\)
\(\Rightarrow k=2\) (*)
Từ (*)
\(\Rightarrow\dfrac{x+y-6}{z}=2\Rightarrow x+y-6=2z\)
\(\Rightarrow\dfrac{x+z+4}{y}=2\Rightarrow x+z+4=2y\)
\(\Rightarrow\dfrac{y+z+2}{x}=2\Rightarrow y+z+2=2x\)
\(\Rightarrow\dfrac{6}{x+y+z}=2\Rightarrow\dfrac{6}{2}=x+y+z\)
\(\Rightarrow x+y+z=3\)
Thay vào biểu thức x+y+z = 3
\(\Rightarrow\dfrac{3-z-6}{z}=\dfrac{3-y+4}{y}=\dfrac{3-x+2}{x}=2\)
\(\Rightarrow\dfrac{-3-z}{z}=\dfrac{7-y}{y}=\dfrac{5-x}{x}=2\)
\(\text{Ta có :}\dfrac{-3-z}{z}=2\)
\(\Rightarrow-3-z=2z\)
\(\Rightarrow-3=3z\)
\(\Rightarrow z=-1\)
*) \(\dfrac{7-y}{y}=2\)
\(\Rightarrow7-y=2y\)
\(\Rightarrow7=3y\)
\(\Rightarrow y=\dfrac{7}{3}\)
*)\(\dfrac{5-x}{x}=2\)
\(\Rightarrow5-x=2x\)
\(\Rightarrow5=3x\)
\(\Rightarrow x=\dfrac{5}{3}\)
Vậy x = 5/3 ; y = 7/3 ; z = -1
Lời giải:
$A=13,5.\frac{-8}{27}.x^4.x^3.y^9.z^3.z^6$
$=-4x^7y^9z^9$
$B=\frac{-4}{7}.\frac{49}{4}.x^3.x^4.y^5.y^4.z^2.z^7$
$=-7.x^7.y^9.z^9$
a)\(x.x=\frac{y}{-3}.\frac{y}{-3}=\frac{z}{4}.\frac{z}{4}=\frac{x^2+y^2-z^2}{1+9-16}=\frac{6}{-6}=-1\)
không tồn tại vì x.x>=0
b)\(\frac{x}{5}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{6}\)
\(\frac{x}{5}=\frac{y}{2}\Rightarrow\frac{z}{8}=\frac{y}{6}\)
Suy ra \(\frac{x}{15}=\frac{y}{6}=\frac{z}{8}=\frac{x-y+z}{15-6+8}=\frac{10}{17}\)
\(x=15.\frac{10}{17}=\frac{150}{17}\)
\(y=6.\frac{10}{17}=\frac{60}{17}\)
c) \(\frac{x}{5}=\frac{y}{3}=\frac{x-y}{5-3}=\frac{14}{2}=7\)
x=7.5=35; y=3.7=21
d) \(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{2x}{4}=\frac{y}{5}=\frac{2x+y}{4+5}=\frac{18}{9}=2\)
x=2.2=4; y=2.5=10
sao lại chai hết cho 6 ????????
hả????????????????
hả?????????????????????????
TL
HT : ( Sai thì bạn thông cảm )
......... đùa nhau à