Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
8x^3 - 8x^2 + 20x^2 - 20x + 26x - 26 =0
<=> 8x^2 ( x-1) + 20x(x-1) + 26(x-1)=0
<=>( 8x^2 + 20x + 26)(x-1)=0
<=> ( x-1)= 0 ( Vì 8x^2 + 20x + 26 >=13,5)
<=> x=1
a) \(x^3-6x^2+12x-9=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-1=0\)
\(\Leftrightarrow\left(x-2\right)^3=1\)
\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)
b) \(8x^3+12x^2+6x-26=0\)
\(\Leftrightarrow8x^3+12x^2+6x+1-27=0\)
\(\Leftrightarrow\left(2x+1\right)^3=27\)
\(\Leftrightarrow2x+1=3\Leftrightarrow x=1\)
A= x^3-3x^2+3x5
=x2(3x3+x-3)
Để giá trị của A nhỏ nhất
=>x=2.Thay x=2 vào ta đc:
A=22(3*23+2-3)=4(3*8+2-3)
=4(24+2-3)=4*23=92
B=x^3 + 6x^2+12x-1
=x3+6x2+12x+8-9
=(x+2)3-9
Để giá trị của B nhỏ nhất
=>x=-1.Thay x=-1 vào ta được:
B=[(-1)+2]3-9=[1]3-9=-8
\(8x^3+12x^2+6x-26=0\)
<=> \(4x^3+6x^2+3x-13=0\)
<=> \(4x^3-4x^2+10x^2-10x+13x-13=0\)
<=> \(4x^2\left(x-1\right)+10x\left(x-1\right)+13\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(4x^2+10x+13\right)=0\)
<=> \(x-1=0\)
<=> \(x=1\)
Vậy...
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
x^3-6^2+12x-8=1
(x-2)^3=1
=>x-2=1
=>x=3
Câu b tương tự nha