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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. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)
\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)
\(D=4x.0+2007\)
\(D=2007\)
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1)\(x^2-x=x\left(x-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
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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) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)
=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)
=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)
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\(-\dfrac{15}{23}:\dfrac{22x}{7}=-\dfrac{14x}{11}:\left(13+\dfrac{4}{5}\right)\)
=>\(-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(-\dfrac{105}{23\cdot22x}=\dfrac{-70x}{11\cdot69}\)
=>\(\dfrac{-3}{2x}=\dfrac{-2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
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\(\dfrac{-15}{23}:\dfrac{22x}{7}=\dfrac{-14x}{11}:\left(13+\dfrac{4}{5}\right)\)
\(\Leftrightarrow-\dfrac{15}{23}\cdot\dfrac{7}{22x}=\dfrac{-14x}{11}:\dfrac{69}{5}\)
=>\(\dfrac{-15\cdot7}{23\cdot22x}=\dfrac{-14x}{11}\cdot\dfrac{5}{69}\)
=>\(\dfrac{5\cdot3\cdot7}{23\cdot2\cdot11x}=\dfrac{2\cdot7x}{11}\cdot\dfrac{5}{3\cdot23}\)
=>\(\dfrac{3}{2x}=\dfrac{2x}{3}\)
=>\(4x^2=9\)
=>\(x^2=\dfrac{9}{4}\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
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a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
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x^2+16x+15=0
=>x^2+x+15x+15=0
=>(x^2+x)+(15x+15)=0
=>x(x+1)+15(x+1)=0
=>(x+1)(x+15)=0
=>x+1=0 hoặc x+15=0
=>x=-1 hoặc x=-15
2|x+5|+x=22
=>x=-32 hoặc 2
x^2+16x+15=0
C1:đl vi-ét
tổng các nghiệm
x1+x2\(=-\frac{b}{a}=-16\)
tích các nghiệm
x1+x2=\(\frac{c}{a}=15\)
=>x=-15;-1
C2:denta
<=>162-4(1*15)=196
\(x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-16\pm\sqrt{196}}{2}\)
=>x=-15;-1