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1/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
2/ \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))
\(\Leftrightarrow x=-1\).
3/ \(x^3-6x^2+8x=0\)
\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)
4/ \(x^4-8x^3-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)
\(x^3-9x+7x^2-63=0\)
\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)
\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)
Vậy ...
x3−9x+7x2−63=0x3−9x+7x2−63=0
⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0
⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0
⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0
⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7
Vậy ...
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
a) 5x + 6 = 0
<=> 5x = -6
<=> x = \(-\frac{6}{5}\)
Vậy phương trình có tập nghiệm là: S = {\(-\frac{6}{5}\)}
b) 9x - 3 = 6x + 21
<=> 3x = 24
<=> x = 8
Vậy phương trình có tập nghiệm là: S = {8}
c) x3 - 9x = 0
<=> x(x2 - 9) = 0
<=> x(x - 3)(x + 3) = 0
<=> \(\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: S = {0; 3; -3}
d) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x-2}-\frac{x^2-4}{4-x^2}=0\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{x^2-4}{x^2-4}=0\)
\(\Rightarrow x+2+x^2-4=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: S ={1}
a) Ta có: 5x+6=0
⇔5x=-6
hay \(x=-\frac{6}{5}\)
Vậy: \(S=\left\{-\frac{6}{5}\right\}\)
b) Ta có: 9x-3=6x+21
⇔9x-6x=21+3
⇔3x=24
hay x=8
Vậy: S={8}
c) Ta có: \(x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={-3;0;3}
d) ĐKXĐ: x∉{2;-2}
Ta có: \(\frac{1}{x-2}-\frac{x^2-4}{4-x^2}=0\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{4-x^2}{4-x^2}=0\)
\(\Leftrightarrow\frac{1}{x-2}+1=0\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{x-2}{x-2}=0\)
Suy ra: \(1+x-2=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(tm)
Vậy: S={1}
d: Ta có: \(9x^2+6x-8=0\)
\(\Leftrightarrow9x^2+12x-6x-8=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
e: Ta có: \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f: Ta có: \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
Ta có: \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
hay \(x\in\left\{0;-3\right\}\)
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
\(x^3-6x^2+9x=0\)
\(\Rightarrow x\left(x^2-6x+9\right)=0\)
\(\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x=0\) hoặc \(x=3\)
Chúc bạn học tốt.
x3 - 6x2 + 9x = 0
=> x3 - 3x2 - 3x2 + 9x =0
=> x2 ( x - 3x ) - 3x ( x - 3x ) = 0
=> ( x2 - 3x ) ( x - 3x ) = 0
=> - 2x ( x2 - 3x ) = 0
\(=>\orbr{\begin{cases}x^2-3x=0\\-2x=0\end{cases}}\)
\(=>\orbr{\begin{cases}x^2=3x\\x=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=3\\x=0\end{cases}}\)
\(=>\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=3\end{cases}}\\x=0\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=3\end{cases}}\\x=0\end{cases}}\)Vậy x = 0 ; 3