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3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
VM
25 tháng 10 2019
\(2x^2-6x=0\)
\(\Rightarrow2x.\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}.\)
\(2x.\left(x+2\right)-3.\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right).\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-2\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-2;\frac{3}{2}\right\}.\)
\(x^3-16x=0\)
\(\Rightarrow x.\left(x^2-16\right)=0\)
\(\Rightarrow x.\left(x^2-4^2\right)=0\)
\(\Rightarrow x.\left(x-4\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0+4\\x=0-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4;-4\right\}.\)
Chúc bạn học tốt!
14 tháng 1 2018
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{2}{3};-1;\dfrac{1}{2}\right\}\)
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x^2\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)\left(1+x\right)-\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1-x-x-3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(-3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1\right\}\)
\(c,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\-5x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{7}{5}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-2;\dfrac{7}{5}\right\}\)
\(d,x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm duy nhất x = -1
\(e,x^3-7x+6=0\)
\(\Leftrightarrow x^3-4x-3x+6=0\)
\(\Leftrightarrow x\left(x^2-4\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;2;-3\right\}\)
\(f,x^4-4x^3+12x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)-\left(4x^3-12x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)-4x\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3>0\forall x\\x^2-4x-3>0\forall x\end{matrix}\right.\)
Vậy phương trình vô nghiệm
\(g,x^5-5x^3+4x=0\)
\(\Leftrightarrow x\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^4-4x^2-x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\) hoặc x = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\\x=-1\end{matrix}\right.\) hoặc x =0
Vậy tập nghiệm của pt \(S=\left\{0;1;-1;2;-2\right\}\)
\(h,x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^4-x^2\right)-\left(4x^3-4x\right)+\left(4x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4x\left(x^2-1\right)+4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{1;-1;2\right\}\)
HP
14 tháng 8 2016
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
T
1
19 tháng 2 2020
1) Sửa đề: \(x^3-x^2+2=0\)
\(\Leftrightarrow x^3+x^2-2x^2-2x+2x+2=0\)
\(\Leftrightarrow x^2\left(x+1\right)-2x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2x+2\right)=0\)(1)
Ta có: \(x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+1\ge1\ne0\forall x\)(2)
Từ (1) và (2) suy ra \(x+1=0\)
hay x=-1
Vậy: x=-1
2) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-2x-10x+5=0\)
\(\Leftrightarrow2x\left(2x-1\right)-5\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
3) Ta có: \(x^4+6x^2+8=0\)
\(\Leftrightarrow x^4+4x^2+2x^2+8=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)+2\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x^2+2\right)=0\)(3)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+4\ge4\ne0\forall x\)(4)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+2\ge2\ne0\forall x\)(5)
Từ (3), (4) và (5) suy ra phương trình \(x^4+6x^2+8=0\) vô nghiệm
Vậy: x∈∅
4) Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3+5x^2-6x^2-30x+9x+45=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
Vậy: x∈{-5;3}
31 tháng 5 2018
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
10 tháng 1 2018
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........