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\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
a)
x (x - 1) + x - 1 = 0
x2 - x + x - 1 = 0
x2 - 1 = 0
x2 = 1
\(\Rightarrow\) x = \(\pm\)1
b)
3( x - 3) - 4x - 12 = 0
3x - 9 - 4x - 12 = 0
-x - 21 = 0
-x = 21
\(\Rightarrow\)x = -21
c)
x3 - 5x = 0
x( x2 - 5) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x2-5=0\Rightarrow x2=5\Rightarrow x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow\)x = 0
d)
(3x - 2)2 - (x + 2)2 = 0
9x2 - 12x + 4 - x2 - 4x - 4 = 0
8x2 - 16x = 0
(làm tương tự như c)
e)
x2 - 9 - 4(x + 3) = 0
x2 - 9 - 4x - 12 = 0
(x2 - 4x + 4) - 13 -12 = 0
(x - 2)2 - 25 = 0
(x - 2)2 = 25
\(\Rightarrow\) x - 2 = 5
\(\Rightarrow\)x = 7
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
a ) \(\left(5x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=7\end{matrix}\right.\)
b ) \(\left(x^2-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)
c )\(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d ) \(x^2+x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e ) \(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
g ) \(\left(x^2+1\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-2\end{matrix}\right.\)
i ) \(x^4+2x^3-2x^2+2x-3=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^3-x^2+x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-3\end{matrix}\right.\)
h) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+3x+2x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
a)<=>\(\left(x^3+x^2-2x\right)+\left(3x^2+3x-6\right)=0\)
<=>\(x\left(x^2+x-2\right)+3\left(x^2+x-2\right)=0\)
<=>\(\left(x^2+x-2\right)\left(x+3\right)=0\)
Phương trình trên bạn tự bấm máy tính nha
<=>\(\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
Đến đây tự làm đc rồi
Vậy x=1 hoặc -2 hoặc -3
b)<=>\(\left(x^3-4x^2+4x\right)+\left(x^2-4x+4\right)=0\)
<=>\(x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)
<=>\(\left(x+1\right)\left(x^2-4x+4\right)=0\)
<=>\(\left(x+1\right)\left(x-2\right)^2=0\)
<=>\(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
c)Câu c mik chưa làm đc
Đáp án câu C:
\(x^3-4x^2+5x=0\)
\(\Leftrightarrow x\left(x^2-4x^2+5x\right)=0\)
\(Tacó:x^2-4x+5=x^2-4x+2^2+1\)
\(=\left(x-2\right)^2+1\)
\(Mà\left(x-2\right)^2\ge0\)
\(Nên\left(x-2\right)^2+1\ge1\)
\(Khiđó:x\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x=0\)
Tìm x,biết:
a/ x + 5x2 =0
⇔x ( 1 + 5x ) = 0
\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0
1) x = 0
2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)
b/x+1=(x+1)2
\(\Leftrightarrow\) (x+1) - (x+1)2 = 0
\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0
\(\Leftrightarrow\) (x+1).(-x) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x = 0
\(\Leftrightarrow\) x= -1 ; 0
Vậy: S=\(\left\{-1;0\right\}\)
c/ x3+x=0
\(\Leftrightarrow\) x(x2 + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0
Ta có : x2 + 1 \(\ge\) 0 vs mọi x
Vậy: S = \(\left\{0\right\}\)
d/5x(x−2)−(2−x)=0
\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(5x+1) = 0
\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0
\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)
g/ x(x−4)+(x−4)2=0
⇔ (x - 4)( x+x-4) = 0
\(\Leftrightarrow\) x - 4 = 0 hoặc 2x-4=0
\(\Leftrightarrow\) x = 4 hoặc x = 2
Vậy: S= \(\left\{2;4\right\}\)
h/ x2−3x=0
⇔x (x-3) = 0
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy: S = \(\left\{0;3\right\}\)
Vậy: S= \(\left\{0;3\right\}\)
i/4x(x+1)=8(x+1)
⇔4x(x+1)-8(x+1) = 0
\(\Leftrightarrow\) 4(x+1) (x - 2) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x= -1 hoặc x = 2
Vậy: S=\(\left\{-1;2\right\}\)
a) \(5x\left(x-4\right)-x^2+16=0\)⇔\(5x^2-20x-x^2+16=0\)
⇔\(4x^2-20x+16=0\)⇔\(\left(2x-5\right)^2-9=0\)
⇔\(\left(2x-8\right)\left(2x-2\right)=0\)⇔\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b) \(x^2-4x+3=0\)⇔\(x^2-x-3x+3=0\)
⇔\(x\left(x-1\right)-3\left(x-1\right)=0\)⇔\(\left(x-1\right)\left(x-3\right)=0\)
⇔\(\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)