tìm x biết: (x2-2x)2 + 2(x2-2x) = 15
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=7\)
hay \(x=-\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .
b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
a. 3x(12x - 4) - 9x(4x - 3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 36x2 - 36x2 - 12x + 27x = 30
<=> 15x = 30
<=> x = 2
b. x(5 - 2x) + 2x(x - 1) = 15
<=> 5x - 2x2 + 2x2 - 2x = 15
<=> -2x2 + 2x2 + 5x - 2x = 15
<=> 3x = 15
<=> x = 5
a) x2 ( 5x3 - x - 1212)= 5x5-x3-1212x
b) ( 3xy - x2 + y ) 2323x2y= 6969x3y2- 2323x4y+ 2323x2y2
c) x2 ( 4x3 - 5xy + 2x ) ( -1212 xy )=(4x5-5x3y+2x3).(-1212xy)
= -4848x6y +6060x4y2-2424x4y
2/ Tìm x, biết
a) 3x( 12x - 4 ) - 9x (4x - 3 ) = 30
=> 36x2-12x-36x2+27x=30
=> -12x +27x=30
=> 15x = 30
=>x =2
b ) x( 5 - 2x ) + 2x ( x - 1 )= 15
=> 5x-2x2+2x2-2x=15
=> 3x=15
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 2)(x2 + 2x + 4) + 2(x2 - 4) - 5(x - 2) = 0
(x - 2)(x + 2)2 + 2(x - 2)(x+2) - 5(x - 2) = 0
(x - 2)[(x+2)2 + 2(x+2) - 5]= 0
(x - 2)[(x + 2)2 + 2(x + 2) + 1 - 6] = 0
( x - 2)[(x + 2 + 1)2 - 6] = 0
(x - 2)[(x + 3)2 - 6] = 0
(x - 2)(x + 3 - \(\sqrt{6}\))(x + 3 + \(\sqrt{6}\)) = 0
TH1. x - 2 = 0 <=> x = 2
TH2. x + 3 - \(\sqrt{6}\) = 0 <=> x = \(\sqrt{6}-3\)
TH3. x + 3 + \(\sqrt{6}\) = 0 <=> x = \(-\sqrt{6}-3\)
S = {2; \(\sqrt{6}-3\); \(-\sqrt{6}-3\)}
Đặt \(x^2+2x=a\), pt trở thành:
\(a^2+2a=15\Leftrightarrow a^2+2a-15=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\\a=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\\x^2-2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\x\in\varnothing\left[x^2-2x+5=\left(x-1\right)^2+4>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)