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

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

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

\(D=2007\)

mấy cái kia cũng làm giống vậy

1)\(x^2-x=x\left(x-1\right)=0\)

\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Thay \(x=15\) vào và tính thôi -.-

a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)

=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)

=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)

b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)

=> \(-\frac{7}{8}x-\frac{23}{4}=3\)

=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)

=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)

c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)

=> \(2x-\frac{2}{7}-7=-11\)

=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)

=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)

d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)

=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)

=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)

\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)

\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)

=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)

=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)

=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)

=>\(4x^2=9\)

=>\(x^2=\dfrac{9}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)

=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)

=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)

=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)

=>\(4x^2=9\)

=>\(x^2=\dfrac{9}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)

giải chi tiết được không ạ

nhớ điền đúng minh mới giải cho bạn được

 nhớ nha

a) (x² - 4).(x + 5) = 0

=> x² - 4 = 0 hoặc x + 5 = 0

+) x² - 4 = 0 => x² = 4 = 2²

=> x \(\in\){-2; 2}

+) x + 5 = 0=> x = -5

Vậy x \(\in\){-2; -5; 2}

b)