NT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
D
2
4 tháng 2 2021
Câu 1:
a) Ta có: 7x+21=0
\(\Leftrightarrow7x=-21\)
hay x=-3
Vậy: S={-3}
b) Ta có: 3x-2=2x-3
\(\Leftrightarrow3x-2-2x+3=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Vậy: S={-1}
c) Ta có: 5x-2x-24=0
\(\Leftrightarrow3x=24\)
hay x=8
Vậy: S={8}
Câu 2:
a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)
b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)
c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+3-3\right)\left(x+3+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)
Vậy: S={0;-3;-6}
K2
6 tháng 12 2020
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow2+\frac{x+4}{2000}+\frac{x+3}{2001}=2+\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2001}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
Suy ra x+2004=0
\(\Leftrightarrow x=-2004\)
D
1
18 tháng 1 2022
\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)
\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)
\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)
\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)
14 tháng 10 2021
\(9x^2-6x+20\)
\(=9x^2-6x+1+19\)
\(=\left(3x-1\right)^2+19>0\forall x\)
TV
3
27 tháng 1 2022
\(\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}=\frac{12}{1-9x^2}\left(ĐKXĐ:x\ne\pm\frac{1}{3}\right)\)
<=> \(\frac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\frac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}=\frac{12}{\left(1-3x\right)\left(1+3x\right)}\)
=> \(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)
<=> \(\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)=12\)
<=> \(-12x=12\)
<=> \(x=-1\left(TMĐK\right)\)
Vậy: ...
TT
27 tháng 1 2022
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\)\(\frac{12}{\left(1-3x\right)\left(1+3x\right)}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}\)
\(\Rightarrow\)\(12=\left(1-3x\right)^2-\left(1+3x\right)^2\)
\(\Leftrightarrow\)\(12=\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow\)\(12=\left(-6x\right).2\)
\(\Leftrightarrow\)\(12=-12x\)
\(\Leftrightarrow\)\(x=-1\)
TL
1
\(x^3+9x^2+11x-21=0\)
\(\Leftrightarrow x^3+7x^2+2x^2+11x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+2x^2+11x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+x^2+7x+x^2+4x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+x\left(x+7\right)+x^2+4x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+4x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+7x-3x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x\left(7+x\right)-3\left(7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-3\\x=1\end{matrix}\right.\)
cách khác: "định hướng HĐT"
\(\left(x^3+3.3.x^2+3.3^2x+3^3\right)+\left[\left(-27x+11x\right)-27-21\right]=\left(x+3\right)^3-16\left(x+3\right)=0\)\(\left(x+3\right)\left[\left(x+3\right)^2-16\right]=\left(x+3\right)\left[\left(x+3\right)-4\right]\left[\left(x+3\right)+4\right]\)
\(\left(x+3\right)\left(x-1\right)\left(x+7\right)=0\)
\(\left[{}\begin{matrix}x=-3\\x=1\\x=-7\end{matrix}\right.\)