- Timf x biết:
- (x+2)(x2-2x+4)-x(x2+2)=0
- (x-1)3-(x+3)(x2-3x+9)+3(x2-4)=2
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`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)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (x+2)2+(x-3)2=2x(x+7)
x.2+2.2+x.2+(-3).2-2x=8
2x+4+2x-6-2x=8
(2x+2x)+(4-6)=8
4x-2=8
4x=8+2
4x=10
X=10:4
X=5/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
Trước khi xem lời giải bài toán này bạn nên xem qua video để hiểu cách biến đổi biểu thức 1 cách nhanh,gọn:Khai triển, rút gọn đa thức bằng máy tính casio . Bài này nhìn rồi mắt chứ rút gọn thì easy
a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\Leftrightarrow-\left(2x-8\right)\) ( Dùng máy tính casio để biến đổi cho nhanh nha =))
\(\Leftrightarrow-2x+8=0\Leftrightarrow8-2x=0\Leftrightarrow2x=8\Leftrightarrow x=4\)
b) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)-2=0\)
\(\Leftrightarrow3x-42=0\Leftrightarrow3x=42\Leftrightarrow x=14\)