Tìm x :
\(\frac{5}{-x}=\frac{x}{-20}\)
Tick nhé :33
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
ta co 6/11.x=20
x=20.11/6
x=110/3
ta co 9/2y=20
y=20.2/9=40/9
ta co 18/5 z=20
z= 20.5/18
z=50/9
\(x=20:\frac{6}{11}=\frac{110}{3}\)
\(y=20:\frac{9}{2}=\frac{40}{9}\)
\(z=20:\frac{18}{5}=\frac{50}{9}\)
\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{y}{4}\)
\(\Rightarrow\frac{5}{x}=\frac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
tu xet bang
tớ có cách khác:))
\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Rightarrow\frac{20+xy}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{40+2xy}{8x}=\frac{x}{8x}\)
\(\Rightarrow40+2xy=x\)
\(\Rightarrow40=x\left(1-2y\right)\)
Cách này xem cho vui nha.dài hơn cách của Phương Uyên.
a ) x/4 = 16/128
Áp dụng tính chất phân số bằng nhau ta có 4 . 16 = x . 128
64 = x . 128
=> x = 0,5
b ) \(1\frac{5}{6}=\frac{-x}{5}\)
11/6 = -x/5
Áp dụng tính chất phân số bằng nhau ta có 11 . 5 = 6 . -x
=> không tồn tại x ( vì 11 . 5 kết quả là dương , 6 . -x sẽ có kết quả âm , dương không thể bằng âm
c ) 4,25 : 8 = -3,5 : x
4,25 = -3.5 . 17/-14
=> 8 = x . 17/-14
=> x = -112/17
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1
\(\frac{5}{-x}=\frac{x}{-20}\)
\(\Leftrightarrow-100=-x^2\)
\(\Leftrightarrow100=x^2\)
\(\Leftrightarrow x=\pm\sqrt{100}=\pm10\)
5/-x = x/-20
<=> -5/x = -x/20
<=> (-5).20 = (-x).x
<=>-100 = x2
<=> x = 10; -10
=> x = 10; -10