tìm x biết:
-\(\frac{-1}{x}\)=\(\frac{x-1}{6}\)
nhanh, đúng tick 4 tick
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Theo đề ra ,ta có :
- 1 / 12 < x < 1 / 8 mà x có giá trị nguyên
=> x = 0
\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)
\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)
\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)
Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)
=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018
=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0
=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0
=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )
=> x=-2017
Vậy x=-2017
k mk nha
a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)
\(\Leftrightarrow x=66\)
rút 4 ra ngoài nhan bạn 4(2(x+1/x)^2+(x^2+1/x^2)^2-(x^2+1/x^2)(x+1/x)^2=(x+4)^2
mik xét cái này cho dễ nhìn nhan
2(x+1/x)^2-(x^2+1/x^2)(x+1/x)^2
= (x+1/x)^2(2-x^2-1/x^2)
= -(x+1/x)^2(x^2-2+1/x^2)
= -(x+1/x)^2(x-1/x)^2=-(x^2-1/x^2)^2
thế ở trên ta có
4(-(x^2-1/x^2)^2+(x^2+1/x^2)^2)=(x+4)^2
4(-x^4+2-1/x^4+x^4+2+1/x^4)=x^2+8x+16
4.4=x^2+8x+16
suy ra x^2+8x=0
x(x+8)=0
suy ra x=0 hoặc x=-8
mak nhìn để bài thì x=0 ko được nên x=-8
\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{y}{4}\)
\(\Rightarrow\frac{5}{x}=\frac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
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.
\(\frac{x-1}{3}+\frac{1}{y}=\frac{-1}{6}\)
\(\frac{\left(x-1\right)y}{3y}+\frac{3}{3y}=\frac{-1}{6}\)
\(\frac{\left(x-1\right)y+3}{3y}=\frac{-1}{6}\)
\(\frac{\left(x-1\right)y}{y}=\frac{\left(-1\right)-3}{6:3}\)
\(x-1=-2\)
\(x=\left(-2\right)+1\)
\(x=-1\)
\(\frac{x-1}{3}+\frac{1}{y}=\frac{-1}{6}\)
\(\frac{\left(x-1\right)y}{3y}+\frac{3}{3y}=\frac{-1}{6}\)
\(\frac{\left(x-1\right)y+3}{3y}=\frac{-1}{6}\)
\(x-1=\frac{\left(-1\right)-3}{6:3}\)
\(x-1=-2\)
\(x=\left(-2\right)+1\)
\(x=-1\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)
\(\left(2x-1\right)=-\frac{4}{63}\)
2x= -4/63 + 1
2x = 59/63
x = 59/63 : 2
x = 59/126
1/3:(2.x-1)=-5-1/4
1/3:(2.x-1)=-21/4
2.x-1=1/3:-21/4
2.x-1=-4/63
2.x=-4/63+1
2.x=\(3\frac{59}{63}\)
x=\(3\frac{59}{63}\):2
x=\(1\frac{61}{63}\)
\(\frac{-1}{x}=\frac{x-1}{6}\Leftrightarrow x\left(x-1\right)=-6\Leftrightarrow x^2-x=-6\Leftrightarrow x^2-x+\frac{1}{4}=-\frac{23}{4}< 0\)
\(Mà:x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\text{ vô lí}\)
\(Vậy:x\in\varnothing\)
\(-\frac{-1}{x}=\frac{x-1}{6}\left(x\ne0\right)\)
\(\Leftrightarrow\frac{1}{x}=\frac{x-1}{6}\)
\(\Leftrightarrow x\left(x-1\right)=1.6\)
\(\Leftrightarrow x^2-x=6\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)