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a) x=\(\dfrac{5.6}{-10}=-3.\)
b) y=\(\dfrac{3.77}{-33}=-7.\)
a) x.21=6.7
x.21=42
x=42:21
x = 2
b) y . 20 = -5.28
y.20 = -140
y = (-140) : 20
y = -7
a)=>x*21=7*6
=>x*21=42
=>x=42/21
x=2
b)=>y*20=(-5)*28
=>y*20=-140
=>y=-140/20
y=-7
a, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)
=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}
vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3
xét bảng
| 5-2y | 1 | -1 | 3 | -3 |
| y | 2 | 3 | 1 | 4 |
| x | 18 | -18 | 6 | -6 |
vậy giá trị x,y cần tìm là: {x=18.y=2}
{x=-18.y=3}
{x=6, y=1}Ư
{x=-6,y=4}
a) \(x\)=1 \(y\)= 12
b)\(x\)=4 \(y\)= 14
hoặc \(x\)= 6 \(y \)=21
...
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
\(\dfrac{x}{6}-\dfrac{5}{2y+1}=\dfrac{2}{3}\)
\(\dfrac{x}{6}-\dfrac{5.2}{2y.2+1.2}=\dfrac{4}{6}\)(vì 2y + 1 là số lẻ)
\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)
Để \(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)thì y = 1 để cùng mẫu số
Khi đó ta có\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{4+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{6}=\dfrac{4}{6}\)
Vì 4+10 = 14 => x = 14
Vậy y = 1; x = 14