cho A= (-3x^2+x+1)(x^2+x-5)+x^2(3x^2+2x-17)
tim x de A=1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( x + 22 ) \(⋮\)( x + 1 )
x + 1 + 21 \(⋮\)( x + 1 )
Mà x + 1 \(⋮\)x + 1 → 21 \(⋮\)x + 1 \(\in\)Ư ( 21 )
( x - 2 ) . ( 2y + 1 ) = 17
Mà 17 là số nguyên tố và bằng 1 . 17
→ Nếu ( x - 2 ) = 1 thì ( 2y + 1 ) = 17
→ Nếu ( 2y + 1 ) = 1 thì ( x - 2 ) = 17
a) \(pt< =>x^3-3.x^2.3+3.x.9-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=4\)
\(< =>x^3-27-x^3+27-9x^2+27x+9x^2+18x+9=4\)
\(< =>45x=4-9=-5< =>x=-\frac{5}{45}=-\frac{1}{9}\)
b) \(pt< =>x\left(x^2-25\right)-\left(x^3+8\right)=17\)
\(< =>x^3-25x-x^3-8=17< =>25x=-8-17=-25< =>x=-1\)
a) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 4
<=> x3 - 9x2 + 27x - 27 - ( x3 - 27 ) + 9( x2 + 2x + 1 ) = 4
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 = 4
<=> 45x + 9 = 4
<=> 45x = -5
<=> x = -5/45 = -1/9
b) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 17
<=> x( x2 - 25 ) - ( x3 + 23 ) = 17
<=> x3 - 25x - x3 - 8 = 17
<=> -25x - 8 = 17
<=> -25x = 25
<=> x = -1
Giải:
a) \(F\left(x\right)+G\left(x\right)-H\left(x\right)\)
\(=4x^2+3x-2+3x^2-2x+5-\left[x\left(5x-2\right)+3\right]\)
\(=4x^2+3x-2+3x^2-2x+5-\left(5x^2-2x+3\right)\)
\(=4x^2+3x-2+3x^2-2x+5-5x^2+2x-3\)
\(=2x^2+3x\)
Để \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)
\(\Leftrightarrow2x^2+3x=0\)
\(\Leftrightarrow x\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(F\left(x\right)-3x+5\)
\(=4x^2+3x-2-3x+5\)
\(=4x^2+3\)
Vì \(x^2\ge0;\forall x\)
\(\Leftrightarrow4x^2\ge0;\forall x\)
\(\Leftrightarrow4x^2+3\ge3>0;\forall x\)
Vậy ...
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
x=-3/2
nếu sai thì sửa nha
thank you