Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(a,\dfrac{x}{3}=\dfrac{y}{7}\) và \(x+y=20\)
\(=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\Rightarrow x=2.3=6\)
\(y=2.7=14\)
Vậy \(x=6\) và \(y=14\)
\(b,\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=6\)
\(=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\Rightarrow x=2.5=10\)
\(y=2.2=4\)
Vậy \(x=10\) và \(y=4\)
\(c,\dfrac{x}{7}=\dfrac{18}{14}\)
Từ tỉ lệ thức trên ta có:
\(14x=7.18\)
\(x=\dfrac{7.18}{14}\)
\(x=9\)
Vậy \(x=9\)
\(d,6:x=1\dfrac{3}{4}:5\)
\(6:x=\dfrac{7}{20}\)
\(x=6:\dfrac{7}{20}\)
\(x=\dfrac{120}{7}\)
Vậy \(x=\dfrac{120}{7}\)
\(e,\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(x-y+z=8\)
\(=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)
\(\Rightarrow x=2.2=4\)
\(y=2.4=8\)
\(z=2.6=12\)
Vậy \(x=4;y=8;z=12\)
a, \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{1}{2}\)
Từ đó suy ra x=1,5; y=3,5
b,\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{1}{2}\)
Từ đó suy ra x=2,5; y=1
c,\(\dfrac{x}{7}=\dfrac{18}{14}\Leftrightarrow\dfrac{x}{7}=\dfrac{9}{7}\Rightarrow x=9\)
d,\(\dfrac{6}{x}=\dfrac{\dfrac{7}{4}}{5}\Leftrightarrow\dfrac{6}{x}=\dfrac{24}{7}\left(\dfrac{\dfrac{7}{4}}{5}\right)\Leftrightarrow\dfrac{6}{x}=\dfrac{6}{\dfrac{120}{7}}\Rightarrow x=\dfrac{120}{7}\)
e,\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{8}=\dfrac{x-y+z}{2-4+8}=\dfrac{4}{3}\)
Từ đó suy ra x=\(\dfrac{8}{3}\); y=\(\dfrac{16}{3}\); z=\(\dfrac{32}{3}\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
=> x = 12
a) \(\frac{4}{x+5}=\frac{3}{x-4}\)
=> 4.(x - 4) = 3.(x + 5)
=> 4x - 16 = 3x + 15
=> 4x - 3x = 15 + 16
=> 1x = 31
=> x = 31 : 1
=> x = 31
Vậy x = 31.
b) \(5-\frac{2}{x}=\frac{3}{-7}\)
=> \(\frac{2}{x}=5-\frac{-3}{7}\)
=> \(\frac{2}{x}=\frac{38}{7}\)
=> 2 . 7 = 38 . x
=> 14 = 38 . x
=> x = 14 : 38
=> x = \(\frac{14}{38}=\frac{7}{19}\)
Vậy x = \(\frac{7}{19}\).
e) \(\frac{x}{7}=-\frac{15}{14}\)
=> x . 14 = (-15) . 7
=> x . 14 = -105
=> x = (-105) : 14
=> x = \(-7,5=-\frac{15}{2}\)
Vậy x = \(-\frac{15}{2}\).
f) 2 - (2x + 3) = 7
=> 2x + 3 = 2 - 7
=> 2x + 3 = -5
=> 2x = (-5) - 3
=> 2x = -8
=> x = (-8) : 2
=> x = -4
Vậy x = -4.
Chúc bạn học tốt!
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
Ta có:
(A): x14 : x = x14 - 1 = x13
(B): x7 .x2 = x7+2 = x9
(C): x8. x6 = x8+6 = x14
(D): x14.x = x14+1 = x15
Chọn (C)x8. x6.