![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt C(x)=0
\(\Leftrightarrow-2x\left(2x-3\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow-4x^2+6x-2x+2=0\)
\(\Leftrightarrow-4x^2+4x+2=0\)
\(\Leftrightarrow4x^2-4x-2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}\\2x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3}+1\\2x=-\sqrt{3}+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{2}\\x=\dfrac{-\sqrt{3}+1}{2}\end{matrix}\right.\)
Đặt Q(x)=0
\(\Leftrightarrow2\left(x-3\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x-6-x+1=0\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có\(\frac{2^x+2^{x+1}+2^{x+2}}{7}=\frac{3^x+3^{x+1}+3^{x+2}}{13}\)
\(\Rightarrow\frac{2^x\left(1+2+2^2\right)}{7}=\frac{3^x\left(1+3+3^2\right)}{13}\)
\(\Rightarrow\frac{2^x\left(1+2+4\right)}{7}=\frac{3^x\left(1+3+9\right)}{13}\)
\(\Rightarrow\frac{2^x.7}{7}=\frac{3^x.13}{13}\)
\(\Rightarrow2^x=3^x\)
\(\Rightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
I3.(x+1)I - I2(2+x)I + I 1-xI =4
I3x+3I - I4+2xI + I1+xI =4
Lập bảng xét dấu:
Đến đây bạn tự lmf nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)\times\left(2+x\right)=\frac{1}{2}-\frac{1}{3}\)
\(\left(x+2\right)\times\left(2+x\right)=\frac{1}{6}\)
\(\left(x+x\right)\times\left(2+2\right)=\frac{1}{6}\)
\(x\times2\times4=\frac{1}{6}\)
\(x\times2=\frac{1}{6}:4\)
\(x\times2=\frac{1}{24}\)
\(x=\frac{1}{24}:2\)
\(x=\frac{1}{48}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\).
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(=-\frac{1}{x+3}=\frac{1}{2010}\)
\(x=2010-\left(-3\right)=2013\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
=> (x-1)(x+3) = (x-2)(x+2)
=> x2 + 3x - x - 3 = x2 + 2x - 2x - 4
=> 2x - 3 = -4 (bớt 2 vế đí x2)
=> 2x = -1
=> x = - 0,5