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A(x)+B(x)-C(x)
=x^3+2x^2+3x+1-x^3+x+1-2x^2+1=0
=>4x+3=0
=>x=-3/4
\(\frac{2}{5}-\frac{1}{2}\left(x+\frac{1}{3}\right)=\frac{7}{5}\)
\(\frac{1}{2}\left(x+\frac{1}{3}\right)=\frac{2}{5}-\frac{7}{5}\)
\(\frac{1}{2}\left(x+\frac{1}{3}\right)=-1\)
\(x+\frac{1}{3}=-1:\frac{1}{2}\)
\(x+\frac{1}{3}=-2\)
\(x=-2-\frac{1}{3}\)
\(x=-\frac{7}{3}\)
\(\frac{2}{5}-\frac{1}{2}.\left(x+\frac{1}{3}\right)=\frac{7}{5}\)
\(\Rightarrow\frac{1}{2}.\left(x+\frac{1}{3}\right)=\frac{2}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{1}{2}.\left(x+\frac{1}{3}\right)=-1\)
\(\Rightarrow x+\frac{1}{3}=-2\)
\(\Rightarrow x=-\frac{7}{3}\)
Bài 1:
\(f\left(x\right)=-x^{15}+8x^{14}-8x^{13}+...-8x-5\)
Ta xét \(x=7\Leftrightarrow x+1=8\)
Khi đó :
\(f\left(7\right)=-x^{15}+x^{14}\left(x+1\right)-x^{13}\left(x+1\right)+...-x\left(x+1\right)-5\)
\(f\left(7\right)=-x^{15}+x^{15}+x^{14}-x^{14}-x^{13}+...-x^2-x-5\)
\(f\left(7\right)=-x-5\)
\(f\left(7\right)=-7-5\)
\(f\left(7\right)=-12\)
Vậy...
\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)
\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)
\(=\dfrac{44}{75}\)
a) \(P\left(x\right)+Q\left(x\right)=16x^6-3x^4+5\)
\(\Rightarrow\text{}\)\(x^5-2x^4-7+x+Q\left(x\right)=16x^6-3x^4+5\)
\(\Rightarrow Q\left(x\right)=x^5-2x^4-7+x-\left(16x^6-3x^4+5\right)\)
\(\Rightarrow Q\left(x\right)=x^5-2x^4-7+x-16x^6+3x^4-5\)
\(\Rightarrow Q\left(x\right)=-16x^6+x^5+x^4+x-12\)
b) \(P\left(x\right)-R\left(x\right)=x^4\)
\(\Rightarrow x^5-2x^4-7+x-R\left(x\right)=x^4\)
\(\Rightarrow R\left(x\right)=x^5-2x^4-7+x-x^4\)
\(\Rightarrow R\left(x\right)=x^5-3x^4+x-7\)
a) P(x)+Q(x)=16x6−3x4+5
P(x)+Q(x)=16x6−3x4+5
⇒x5−2x4−7+x+Q(x)=16x6−3x4+5x5−2x4−7+x+Q(x)=16x6−3x4+5
⇒Q(x)=x5−2x4−7+x−(16x6−3x4+5)⇒Q(x)=x5−2x4−7+x−(16x6−3x4+5)
⇒Q(x)=x5−2x4−7+x−16x6+3x4−5⇒Q(x)=x5−2x4−7+x−16x6+3x4−5
⇒Q(x)=−16x6+x5+x4+x−12⇒Q(x)=−16x6+x5+x4+x−12
b) P(x)−R(x)=x4P(x)−R(x)=x4
⇒x5−2x4−7+x−R(x)=x4⇒x5−2x4−7+x−R(x)=x4
⇒R(x)=x5−2x4−7+x−x4⇒R(x)=x5−2x4−7+x−x4
⇒R(x)=x5−3x4+x−7
a: =>7(x-5)>0
=>x-5>0
=>x>5
b: =>x-1 thuộc {1;-1;11;-11}
=>x thuộc {2;0;12;-10}
c: =>x+1+7 chia hết cho x+1
=>x+1 thuộc {1;-1;7;-7}
=>x thuộc {0;-2;6;-8}
d: =>(x+2)(x-5)<0
=>-2<x<5
a:(- 7) . ( 5 – x) < 0
=>7(x-5)>0
=>x-5>0
=>x>5
b:11 ⁝ x – 1
=>x-1 thuộc {1;-1;11;-11}
=>x thuộc {2;0;12;-10}
c: x + 8 ⁝ x + 1
=>x+1+7 chia hết cho x+1
=>x+1 thuộc {1;-1;7;-7}
=>x thuộc {0;-2;6;-8}
d: (x + 2) . (5 – x) > 0
=>(x+2)(x-5)<0
=>-2<x<5