![](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)
\(\left(2x+3\right)\left(x+2\right)^2\left(2x+5\right)=3\)
\(\Leftrightarrow4x^4+32x^3+95x^2+124x+57=0\)
\(\Leftrightarrow4x^4+4x^3+28^3+28x^2+67x^2+67x+57x+57=0\)
\(\Leftrightarrow4x^3\left(x+1\right)+28x^2\left(x+1\right)+67x\left(x+1\right)+57\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x^3+28x^2+67x+57\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x^3+12x^2+16x^2+48x+19x+57\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x^2\left(x+3\right)+16x\left(x+3\right)+19\left(x+3\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(4x^2+16x+19\right)=0\)
\(\forall x\)ta có: \(4x^2+16x+19=4x^2+16x+16+3=\left(2x+4\right)^2+3\ge3>0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy tập nghiệm của pt là S={-1;-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-2\right)+4=5x-2\left(x-1\right)\\ \Leftrightarrow3x-6+4=5x-2x+2\\ \Leftrightarrow0x=4\left(vôlý\right)\)
Vậy pt vô nghiệm
\(2\left(x-2\right)-3\left(1-2x\right)=5\\ \Leftrightarrow2x-4-3+6x=5\\ \Leftrightarrow8x=12\\ \Leftrightarrow x=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(x^2+4x\right)^2+31\left(x^2+4x\right)+60=3\)
\(t=x^2+4x\)
\(4t^2+31t+57=0\)
\(\orbr{\begin{cases}t=\frac{-31-7}{8}=\frac{-19}{4}\\t=\frac{-31+7}{8}=-3\end{cases}}\)
\(x^2+4x+\frac{19}{4}=0\Rightarrow vn\)
\(x^2+4x+3=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>x^4+4x^2+9-4x^3-6x^2+12x<x^4-4x^3-2x^2+15x-3
=>-2x^2+12x+9<-2x^2+15x-3
=>-3x<-12
=>x>4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\)
\(ĐKXĐ:x^2+2x-3=\left(x+1\right)\left(x-3\right)\\ \Rightarrow x\ne-1;x\ne3\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x^2\)
\(\Leftrightarrow x^2+x+x^2-3x=4x^2\)
\(\Leftrightarrow2x^2-2x=4x^2\)
\(\Leftrightarrow2x^2-4x^2-2x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow2x\left(-x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\-x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(N\right)\\x=-1\left(L\right)\end{cases}}\)
Tự kết luận tập nghiệm bạn nhé!
x2+2x-3 = (x+1)(x-3)
vậy MSC = 2(X+1(X-3) qui đồng mẫu số r làm dc r, đk x khác 1; -3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2x(x + 5) - (x - 3)2 = x2 + 6
<=> 2x2 + 10x - (x2 - 6x + 9) = x2 + 6
<=> 2x2 + 10x - x2 + 6x - 9 - x2 = 6
<=> 16x = 6 + 9
<=> 16x = 15
<=> x = 15/16
Vậy...
b, (4x + 7)(x - 5) - 3x2 = x(x - 1)
<=> 4x2 - 20x + 7x - 35 - 3x2 = x2 - x
<=> 4x2 - 20x + 7x - 3x2 - x2 + x = 35
<=> -12x = 35
<=> x = -35/12
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
dễ mà ban,chuyển vế sang r nhân tung ra
tung cai dau mj ra