Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

`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`

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: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

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: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow x=2\)

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

\(\Leftrightarrow x^2-x+6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

a) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

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

a) (x+2)²+(x-3)² = 2x(x+7)

⇒x²+4x+4+x²-6x+9=2x²+14x

⇒x²+4x+4+x²-6x+9-2x²-14x=0

⇒ -16x+13=0

⇒ x=\(\dfrac{13}{16}\)

b) (x+3)(x²-3x+9) = x(x²+4)-1

⇒x(x²-3x+9)+3(x²-3x+9)=x³+4x-1

⇒x³-3x²+9x+3x²-9x+27-x³-4x+1=0

⇒-4x+28=0

⇒x=7

a: Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2=2x\left(x+7\right)\)

\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(\Leftrightarrow-16x=-13\)

hay \(x=\dfrac{13}{16}\)

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)=x\left(x^2+4\right)-1\)

\(\Leftrightarrow x^3+27-x^3-4x=-1\)

\(\Leftrightarrow-4x=-28\)

hay x=7