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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
Đặt \(a=x^2+x\)
\(\Leftrightarrow a^2+4a=12\)
\(\Leftrightarrow a^2+4a-12=0\)
\(\Leftrightarrow a^2+6a-2a-12=0\)
\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)
\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+x=-6\\x^2+x=2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+3\right)^2\left(x^2+6x+1\right)=9\)
\(\Leftrightarrow\left(x^2+6x+9\right)\left(x^2+6x+1\right)=9\)
Đặt: \(x^2+6x+5=t\)thì:
\(\left(1\right)\Leftrightarrow\left(t-4\right)\left(t+4\right)=9\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x\left(x+6\right)=0\left(x^2+6x+10=\left(x+3\right)^2+1>0\right)\)
.... bạn tự giả tiếp
Chúc bạn hc tốt :D
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt: \(x^2-6x+9=t\left(t\ge0\right)\)
Khi đó: \(\left(x^2-6x+9\right)^2-15\left(x^2-6x+10\right)=1\)
\(\Leftrightarrow t^2-15\left(t+1\right)=1\Leftrightarrow t^2-15t-15=1\)
\(\Leftrightarrow t^2-15t-16=0\Leftrightarrow\left(t-16\right)\left(t+1\right)=0\Leftrightarrow t=16\left(t\ge0\right)\)
\(\Leftrightarrow x^2-6x+9=16\Leftrightarrow\left(x-3\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=4\\x-3=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
Tập nghiệm của pt: \(S=\left\{7;-1\right\}\)
Đặt \(x^2-6x+9=t\)
\(\Rightarrow\)Phương trình ban đầu trở thành: \(t^2-15\left(t+1\right)=1\)
\(\Leftrightarrow t^2-15t-15=1\)\(\Leftrightarrow t^2-15t-16=0\)
\(\Leftrightarrow\left(t^2+t\right)-\left(16t+16\right)=0\)\(\Leftrightarrow t\left(t+1\right)-16\left(t+1\right)=0\)
\(\Leftrightarrow\left(t+1\right)\left(t-16\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}t+1=0\\t-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=-1\\t=16\end{cases}}\)
Ta thấy: \(x^2-6x+9=\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow t\ge0\)\(\Rightarrow t=16\)\(\Rightarrow x^2-6x+9=16\)
\(\Leftrightarrow x^2-6x-7=0\)\(\Leftrightarrow\left(x^2+x\right)-\left(7x+7\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-7\left(x+1\right)=0\)\(\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=7\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-1;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)
=>18x-6+18x-6=4-12x
=>36x-12=4-12x
=>48x=16
hay x=1/3
2: \(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
=>(2x-1)(3x-4)=0
=>x=1/2 hoặc x=4/3
![](https://rs.olm.vn/images/avt/0.png?1311)
( x - 2 ).( x + 3 )2 - ( x - 2 ).(x - 1)2 = 0
(=) ( x - 2 ).[ ( x + 3 )2 - ( x - 1 )2 ] = 0
(=) ( x - 2).[ x2 + 6x + 9 - x2 + 2x - 1] = 0
(=) ( x - 2 ) .( 8x + 8 ) = 0
(=) \(\orbr{\begin{cases}x-2=0\\8x+8=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy phương trình có nghiệm là : x = 2 , -1
b) 9x2 - 6x + 1 = 4x2
(=) 9x2 - 6x + 1 - 4x2 = 0
(=) 5x2 - 6x + 1 = 0
(=) 5x2 - 5x - x + 1 = 0
(=) 5x.( x - 1 ) - (x - 1) = 0
(=) ( x - 1 ).( 5x - 1) = 0
(=)\(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy phương trình có nghiệm là : x = 1 , \(\frac{1}{5}\)
c) ( x - 3 ) - \(\frac{\left(x-3\right)\left(2x+1\right)}{3}\)= 1
(=) \(\frac{3\left(x-3\right)}{3}\)\(-\)\(\frac{\left(x-3\right)\left(2x+1\right)}{3}\)= \(\frac{3}{3}\)
(=) 3.( x - 3) - ( x - 3 ).( 2x +1 ) = 3
(=) 3x - 9 - 2x2 +5x +3 -3 = 0
(=) -2x2 +8x -9 = 0 (loại )
Vậy phương trình vô nghiệm
d) x2 + 6x - 7 =0
(=) x2 +7x - x - 7 = 0
(=) x.( x + 7 ) - ( x + 7 ) = 0
(=) ( x - 1 ) .( x+7 ) = 0
(=) \(\orbr{\begin{cases}x-1=0\\x+7=0\end{cases}}\)(=) \(\orbr{\begin{cases}x=1\\x=-7\end{cases}}\)
Vậy phương trình có nghiệm là : x = 1 , -7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x^2-6x+10\right)^2+2\left(x^2-6x\right)-65=0\\ \Leftrightarrow3\left(\left(x-3\right)^2+1\right)^2+2\left(x^2-6x+9\right)-83=0\\ \Leftrightarrow3\left(\left(x-3\right)^4+2.\left(x-3\right)^2+1\right)+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+6\left(x-3\right)^2+3+2\left(x-3\right)^2-83=0\\ \Leftrightarrow3\left(x-3\right)^4+8\left(x-3\right)^2-80=0\\ \Leftrightarrow\left(x-3\right)^2\left(3.\left(x-3\right)^2+8\right)=80\\ \Leftrightarrow\left(x-3\right)^2.\left(3x^2-18x+27+8\right)=80\)
Phá ra rồi giải tiếp !!
Áp dụng cái bổ đề nhưu ở lần trước mình CM cho bạn ý
Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) thì ta có:
\(3\left(x^2-6x+9\right)\left(1-x^2\right)\left(6x-10\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(1-x\right)\left(1+x\right)\left(6x-10\right)=0\)
\(\Rightarrow x\in\left\{-1;1;3;\frac{5}{3}\right\}\)
bạn Đoạt Mệnh Thương giải rõ hơn đc ko