![](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, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)
\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)
\(\Leftrightarrow x^3+3x^2+2x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
1c/
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt có nghiệm duy nhất \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)
\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)
\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)
\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)
b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)
c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)
d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)
a)
/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy x = 1 không phải nghiệm của phương trình nên nhân 2 vế của phương trình với x - 1 ta có:
⇔ x = 1(KTM)
Vậy phương trình đã cho vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Cách 1: Khai triển HĐT rút gọn được 3 x 2 + 6x + 7 = 0
Vì (3( x 2 + 2x + 1) + 4 < 0 với mọi x nên giải được x ∈ ∅
Cách 2. Chuyển vế đưa về ( x + 3 ) 3 = ( x - 1 ) 3 Û x + 3 = x - 1
Từ đó tìm được x ∈ ∅
b) Đặt x 2 = t với t ≥ 0 ta được t 2 + t - 2 = 0
Giải ra ta được t = 1 (TM) hoặc t = -2 (KTM)
Từ đó tìm được x = ± 1
c) Biến đổi được
d) Biến đổi về dạng x(x - 2) (x - 4) = 0. Tìm được x ∈ {0; 2; 4}
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
vô nghiệm bạn nha