1. \(A=x^{15}+3x^{14}+5=x^{14}\left(x+3\right)+5\)

Thay \(x+3=0\)vào đa thức ta được:\(A=x^{14}.0+5=5\)

2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

Thay \(x=-3\)vào đa thức ta được: \(B=\left[x^{2006}\left(-3+3\right)+1\right]^{2017}=\left(x^{2006}.0+1\right)^{2017}=1^{2017}=1\)

3. \(C=21x^4+12x^3-3x^2+24x+15=3x\left(7x^3+4x^2-x+8\right)+15\)

Thay \(7x^3+4x^2-x+8=0\)vào đa thức ta được: \(C=3x.0+15=15\)

4. \(D=-16x^5-28x^4+16x^3-20x^2+32x+2007\)

\(=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)

Thay \(-4x^4-7x^3+4x^2-5x+8=0\)vào đa thức ta được: \(D=4x.0+2007=2007\)

1. \(A=x^{15}+3x^{14}+5\)

\(A=x^{14}\left(x+3\right)+5\)

\(A=x^{14}+5\)

2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(B=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=1^{2007}=1\)

3. \(C=21x^4+12x^3-3x^2+24x+15\)

\(C=3x\left(7x^2+4x^2-x+8+5\right)\)

\(C=3x\left(0+5\right)\)

\(C=15x\)

4. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)

\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)

\(D=4x.0+2007\)

\(D=2007\)

\(Dat:\left\{{}\begin{matrix}4x+3=a\\3x-8=b\end{matrix}\right.\Rightarrow-4x-3+8-3x=5-7x=-a-b\Rightarrow a^3-\left(a+b\right)^3+b^3=0\Leftrightarrow-3ab\left(a+b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\3x-8=0\\5-7x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{4}\\x=\frac{8}{3}\\x=\frac{5}{7}\end{matrix}\right.\)

Áp dụng hằng đẳng thức : 

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)

Ta viết lại phương trình thành :

\(\left(4x+3+5-7x+3x-8\right)\left[\left(4x+3\right)^2+\left(5-7x\right)^2+\left(3x-8\right)^2\right]+3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)

\(\Leftrightarrow3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\5-7x=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{7}\\x=\dfrac{8}{3}\end{matrix}\right.\)

Vậy : \(S=\left\{-\dfrac{3}{4};\dfrac{5}{7};\dfrac{8}{3}\right\}\).

Đặt 4x+3=a; 3x-8=b

=>a^3+b^3-(a+b)^3=0

=>a^3+b^3-a^3-b^3-3ab(a+b)=0

=>ab(a+b)=0

=>(4x+3)(3x-8)(7x-5)=0

=>\(x\in\left\{-\dfrac{3}{4};\dfrac{8}{3};\dfrac{5}{7}\right\}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

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