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a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)
c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)
a) A=x^3 + 3x^2*5 + 3x*5^2 + 5^3
=(x+5)^3
Thay x = -10 vào biểu thức A ta được:
A = (-10+5)^3
=(-5)^3
=-75
Làm tương tự nhé
a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
1. \(x^6-2x^3+1=0\Leftrightarrow\left(x^3-1\right)^2=0\Leftrightarrow x=1\)
2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-5x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(x-5\right)^2=0\Leftrightarrow x=5\)
5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)
7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)
\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)
\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Trả lời:
j, ( x + 1 )2 - ( 2x - 1 )2 = 0
<=> ( x + 1 - 2x + 1 ) ( x + 1 + 2x - 1 ) = 0
<=> ( 2 - x ) 3x = 0
<=> 2 - x = 0 hoặc 3x = 0
<=> x = 2 hoặc x = 0
Vậy x = 2; x = 0 là nghiệm của pt.
k, Sửa đề: 8x3 + 12x - 1 = 6x2
<=> 8x3 + 12x - 1 - 6x2 = 0
<=> ( 2x )2 - 3.x2.2 + 3.x.22 - 13 = 0
<=> ( 2x - 1 )3 = 0
<=> 2x - 1 = 0
<=> 2x = 1
<=> x = 1/2
Vậy x = 1/2 là nghiệm của pt.
l, x3 + 15x2 + 75x + 125 = 0
<=> x3 + 3.x2.5 + 3.x.52 + 53 = 0
<=> ( x + 5 )3 = 0
<=> x + 5 = 0
<=> x = - 5
Vậy x = - 5 là nghiệm của pt.