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a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Bài 2:
a: \(x^2-16-\left(x+4\right)=0\)
=>(x+4)(x-4)-(x+4)=0
=>(x+4)(x-5)=0
=>x=5 hoặc x=-4
b: \(\left(3x-1\right)^2-\left(9x^2-1\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+1=0\)
=>-6x+2=0
=>-6x=-2
hay x=1/3
c: \(4x^2+9=-12x^2\)
\(\Leftrightarrow4x^2+12x^2=-9\)
\(\Leftrightarrow16x^2=-9\)(vô lý)
Do đó: \(x\in\varnothing\)
d: \(4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
=>x=1 hoặc x=1/4
e: \(4x^2-4x+3=0\)
\(\Leftrightarrow4x^2-4x+1+2=0\)
\(\Leftrightarrow\left(2x-1\right)^2=-2\)(vô lý)
Do đó: \(x\in\varnothing\)
\(f\left(x\right)=x^3+5x^2+ax+b\)
\(f\left(-2\right)=0\Leftrightarrow12-2a+b=0\left(1\right)\)
\(f\left(3\right)=0\Leftrightarrow72+3a+b=0\left(2\right)\)
\(\left(2\right)-\left(1\right)=0\Leftrightarrow\left(72+3a+b\right)-\left(12-2a+b\right)=0\Leftrightarrow60+5a=0\Leftrightarrow5a=-60\Leftrightarrow a=-12\)
\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)
\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)
\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)
\(A=-12x\)
\(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)
\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)
\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)
\(B=-3x-12\)
Câu C tương tự.
Chúc bạn học tốt!!!
A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)
A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)
A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)
A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)
A = \(-4x^2-12x-8+4x^2+8=-12x\)
b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)
B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)
B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)
B = \(-6x^2-3x-6\)
1) \(x^3+2x^2+2x+4=0\)
\(\Rightarrow x^2\left(x+2\right)+2\left(x+2\right)=0\)
\(\Rightarrow\left(x^2+2\right)\left(x+2\right)=0\)
\(\Rightarrow x+2=0\) (x2 +2 loại)
\(\Rightarrow x=-2\)
2) \(x^3+4x^2-2x-8=0\)
\(\Rightarrow x^2\left(x+4\right)-2\left(x+4\right)=0\)
\(\Rightarrow\left(x^2-2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\\x=-4\end{matrix}\right.\)
3) \(x^3+3x-4=0\)
\(\Rightarrow x^2\left(x-1\right)+x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x^2+x+4\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+4=0\\x-1=0\end{matrix}\right.\Rightarrow x=1\)
4) \(x^3+x-30=0\)
\(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+10\left(x-3\right)=0\)
\(\Rightarrow\left(x^2+3x+10\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+3x+10=0\\x-3=0\end{matrix}\right.\Rightarrow x=3.\)
P/S: mấy bạn đừng giải lại nếu như có cách làm khác.
Câu 1+ 2 :
Câu 3:
Câu 4: