giải pt: (2x+1)(x+1)2(2x+3)=18
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)
(2x+1)(x+1)2(2x+3)=18
<=> (2x+2-1)(x+1)2(2x+2+1)=18
Đặt y=x+1, ta có:
(2y-1)y2(2y+1)=18
Ta có
(2x+1)(x+1)2(2x+3)=18
=> (x+1)2(4x2+8x+3)-18=0
=> (x2+2x+1)(4x2+8x+3)-18=0
Đặt x2+2x+1=a ta có
a.(4a-1)-18=0
=> 4a2-a-18=0
=> 4a2 +8a-9a-18=0
=> 4a(a+2)-9(a+2)=0
=> (a+2)(4a-9)=0
Với a=x2+2x+1biểu thức trên trở thành
(x2+2x+3)(4x2+8x-5)=0
=> x2+2x+3=0 hoặc 4x2+8x-5=0
• x2+2x+3=0 => phương trình vô nghiệm
• 4x2+8x-5=0 => x=1/2 hoặc x=-5/2
Vậy x=1/2 và x=-5/2 là nghiệm của phương trình
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)
\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)
\(4x^2+12x+18=-2x-5x^2+5\)
\(4x^2+12x+18+2x+5x^2-5=0\)
\(9x^2-14x+13=0\)
=> vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\-4+\sqrt{7}\le x\le-1\end{matrix}\right.\)
Khi x thỏa ĐKXĐ, vế phải luôn dương, bình phương 2 vế ta được:
\(\Leftrightarrow3x^2+16x+17+2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=4x^2+16x+16\)
\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=x^2-1\)
\(\Leftrightarrow4\left(x^2-1\right)\left(2x^2+16x+18\right)=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\4\left(2x^2+16x+18\right)=x^2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\7x^2+64x+73=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-32+3\sqrt{57}}{7}\\x=\dfrac{-32-3\sqrt{57}}{7}\left(loại\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow4\left(x^2+2x+\frac{3}{4}\right)\left(x^2+2x+1\right)-18=0\)
Đặt \(a=x^2+2x+\frac{3}{4}\) \(a=x^2+2x+\frac{3}{4}\)
\(\Rightarrow4a\left(a+\frac{1}{4}\right)-18=0\)
\(\Leftrightarrow4a^2+a-18=0\)
\(\Leftrightarrow4a^2-8a+9a-18=0\)
\(\Leftrightarrow\left(4a+9\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4a+9=0\\a-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}a=-\frac{9}{4}\\a=2\end{cases}}\)
\(\left(+\right)a=-\frac{9}{4}\Rightarrow x^2+2x+\frac{3}{4}=-\frac{9}{4}\)
\(\Leftrightarrow x^2+2x+\frac{3}{4}+\frac{9}{4}=0\)\(\Leftrightarrow x^2+2x+3=0\)
\(\Leftrightarrow\left(x+1\right)^2+2=0\)
( vô lí )
\(\left(+\right)a=2\Rightarrow x^2+2x+\frac{3}{4}=2\)
\(\Leftrightarrow x^2+2x-\frac{5}{4}=0\)
\(\Leftrightarrow x^2+2x+1-\frac{9}{4}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(\frac{3}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+1-\frac{3}{2}\right)\left(x+1+\frac{3}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=0\\x-\frac{1}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{1}{2}\end{cases}}}\)
=> (2x+1)(2x+3)(x+1)2=18
=> (2x+2-1)(2x+2+1)(x+1)2=18
=> ((2x+2)2-1)(x+1)2=18
=>(2x+2)2(x+1)2 _ (x+1)2 - 18 =0
=> (2(x+1))2(x+1)2_(x+1)2 - 18=0
=> 4(x+1)4 - (x+1)2 -18 =0
đặt (x+1)2=a
phương trình <=> 4a2 - a-18=0
=> 4a2 + 8a - 9a -18=0
=> 4a(a+2)-9(a+2)=0
=> (a+2)(4a-9)=0
từ đó tìm ra a xong tìm ra x mình nghĩ bạn giải đc :D
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>4x-3x=1-2
=>x=-1
b: =>3x=12
=>x=4
c: =>2(x^2-6)=x(x+3)
=>2x^2-12-x^2-3x=0
=>x^2-3x-12=0
=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)
\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)
Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)
pt \(\Leftrightarrow4t^2-t-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)
\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)