=>3x-2x^2-3+2x+2x^2+6x-x-3=4

=>10x=10

=>x=1

a) \(\left|2x+1\right|=\left|x+4\right|\Rightarrow\left[{}\begin{matrix}2x+1=x+4\\2x+1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\left|2x-1\right|=x+4\Rightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

\(B=\dfrac{x^2+x}{x^2+x+1}=\dfrac{3x^2+3x}{3\left(x^2+x+1\right)}=\dfrac{-\left(x^2+x+1\right)+4x^2+4x+1}{3\left(x^2+x+1\right)}\)

\(=-\dfrac{1}{3}+\dfrac{\left(2x+1\right)^2}{3\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge-\dfrac{1}{3}\)

\(B_{min}=-\dfrac{1}{3}\) khi \(x=-\dfrac{1}{2}\)

a: \(\left(2x-1\right)^5-\left(2x-1\right)^8=0\)

=>\(\left(2x-1\right)^5\cdot\left\lbrack1-\left(2x-1\right)^3\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(2x-1\right)^5=0\\ 1-\left(2x-1\right)^3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(2x-1\right)^5=0\\ \left(2x-1\right)^3=1\end{array}\right.\)

=>\(\left[\begin{array}{l}2x-1=0\\ 2x-1=1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=1\\ 2x=2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=1\end{array}\right.\)

b: (2x+1)(2x-3)<0

TH1: \(\begin{cases}2x+1\ge0\\ 2x-3\le0\end{cases}\Rightarrow\begin{cases}x\ge-\frac12\\ x\le\frac32\end{cases}\)

=>\(-\frac12\le x\le\frac32\)

TH2: \(\begin{cases}2x+1\le0\\ 2x-3\ge0\end{cases}\Rightarrow\begin{cases}2x\le-1\\ 2x\ge3\end{cases}\Rightarrow\begin{cases}x\le-\frac12\\ x\ge\frac32\end{cases}\)

=>x∈∅

c: (x-1)(2x+3)>0

TH1: \(\begin{cases}x-1>0\\ 2x+3>0\end{cases}\Rightarrow\begin{cases}x>1\\ x>-\frac32\end{cases}\)

=>x>1

TH2: \(\begin{cases}x-1<0\\ 2x+3<0\end{cases}\Rightarrow\begin{cases}x<1\\ x<-\frac32\end{cases}\)

=>\(x<-\frac32\)

a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

\(\Rightarrow5^x=5^5\)

\(\Rightarrow x=5\)

Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a,`

`x^2 + 2x + 1 = 9`

`=> x^2 + 2x + 1 - 9 = 0`

`=> x^2 + 2x - 8 = 0`

`=> x^2 + 4x - 2x - 8 = 0`

`=> (x^2 + 4x) - (2x + 8) = 0`

`=> x(x + 4) - 2(x + 4) = 0`

`=> (x-2)(x+4) = 0`

`=>`\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

Vậy, `x \in {2; 4}`

`b,`

`x^2 - 1 = 15`

`=> x^2 = 15 + 1`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = +-4`

Vậy, `x \in {4; -4}`

`c)`

`19 - 2x^2 = 1`

`=> 2x^2 = 19 - 1`

`=> 2x^2 = 18`

`=> x^2 = 18 \div 2`

`=> x^2 = 9`

`=> x^2 = (+-3)^2`

`=> x = +-3`

Vậy, `x \in {3; -3}.`