Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(7x\left(16x^2-1\right)=0\)
\(\Leftrightarrow7x\left(4x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
=>2x(x^2-11x+18)=0
=>x(x-2)(x-9)=0
=>\(x\in\left\{0;2;9\right\}\)
\(a,\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow\left(3x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{3}\)
Bài 2:
a: \(x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
a(b+3)-b(3+b)
=(3+b)(a-b)
Thay số, có: (3+1997).(2003-1997)
= 2000.6 =12000
xy(x+y)-2x-2y
xy(x+y)- 2(x+y)
(x+y).(xy-2)
Thay số, co: 7. (8-2)
7.4=28
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
pt <=> (16x^2-24x)+(2x-3) = 0
<=> (2x-3).(8x+1) = 0
<=> 2x-3=0 hoặc 8x+1=0
<=> x=3/2 hoặc x=-1/8
Vậy ..............
Theo bài ra ta có:\(16.x^2+22x-3=0\)
\(\Rightarrow\left(16x+22\right)x=3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)
Tự kẻ bẳng nha