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\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ 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 ~ )
Bài 1:
b: \(x^3-4x^2+7x-6=0\)
\(\Leftrightarrow x^3-2x^2-2x^2+4x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-2x+3\right)=0\)
=>x-2=0
hay x=2
c: \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x+2+7x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+4x+x+2\right)=0\)
=>(x+1)(x+2)(2x+1)=0
hay \(x\in\left\{-1;-2;-\dfrac{1}{2}\right\}\)
d: \(2x^3-9x+2=0\)
\(\Leftrightarrow2x^3-4x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+1-\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1+\dfrac{\sqrt{6}}{2}\right)\left(x+1-\dfrac{\sqrt{6}}{2}\right)=0\)
hay \(x\in\left\{2;-1-\dfrac{\sqrt{6}}{2};-1+\dfrac{\sqrt{6}}{2}\right\}\)
b. sửa đề
\(6x^4+25x^3+12x-25x^2+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-3\\x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy........
Bài 1 : Giải phương trình
a) (x + 3)4 + (x + 5)4 = 16
Đặt : x + 3 = t
=> x + 5 = x + 3 + 2 = t + 2
Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :
t4 + (t + 2)4 = 16
<=> 2t4 + 8t3 + 24t2 + 32t + 16 = 16
<=> 2(t4 + 4t3 + 12t2 + 16t) = 0
<=> t4 + 4t3 + 12t2 + 16t = 0
<=> (t + 2) . t . (t2 + 2y + 4) = 0
TH1 : t = 0
TH2 : t + 2 = 0 <=> t = -2
TH3 : t2 + 2y + 4 = 0 (vô nghiệm => loại)
Nên t = 0 hoặc t = -2
hay x + 3 = -2 hoặc x + 3 = 0
<=> x = -5 hoặc x = -3
\(S=\left\{-5;-3\right\}\)
b) 6x4 + 25x3 + 12x2 - 25x + 6 = 0
<=> 6x4 + 12x3 + 13x3 + 26x2 - 14x2 - 28x + 3x + 6 = 0
<=> 6x3 (x + 2) + 13x2 (x + 2) - 14x (x + 2) + 3(x + 2) = 0
<=> (x + 2)(6x3 + 13x2 - 14x + 3) = 0
<=> (x + 2)(6x3 + 18x2 - 5x2 - 15x + x + 3) = 0
\(\Leftrightarrow\left(x+2\right)[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)]=0\)
<=> (x + 2)(x + 3) (6x2 - 5x + 1) = 0
<=> (x + 2)(x + 3)(2x - 1)(3x - 1) = 0
TH1 : x + 2 = 0 <=> x = -2
TH2 : x + 3 = 0 <=> x = -3
TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)
TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)
\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)
a) \(\left(x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) \(x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
c) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=-1\)
d) \(x^3+10x^2+25x=0\)
\(\Leftrightarrow x\left(x+5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
a) ( x2 - 1 )( x2 - 25 ) = 0
<=> \(\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) x2 - 8x + 16 = 0
<=> ( x - 4 )2 = 0
<=> x - 4 = 0
<=> x = 4
c) x3 + 3x2 + 3x + 1 = 0
<=> ( x + 1 )3 = 0
<=> x + 1 = 0
<=> x = -1
d) x3 + 10x2 + 25x = 0
<=> x( x2 + 10x + 25 ) = 0
<=> x( x + 5 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
a) \(x^2-8\text{ }x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\Leftrightarrow x=4\)
b) \(\left(3x-2\right)^2-49=0\)
\(\Leftrightarrow\left(3x-2\right)^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=\sqrt{49}\\3x-2=-\sqrt{49}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=9\\3x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{5}{3}\end{cases}}\)
c) \(\left(10x-5\right)^2-9x^2=0\)
\(\Leftrightarrow\left(7x-5\right)\left(13x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x-5=0\\13x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{7}\\x=\frac{5}{13}\end{cases}}\)
d) \(\left(7x+1\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(12x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\12x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{1}{12}\end{cases}}\)