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\(2x^2+10x+15=0\)
\(\Leftrightarrow2.\left(x^2+5x+\frac{15}{2}\right)=0\Leftrightarrow x^2+5x+\frac{15}{2}=0\)
\(\Leftrightarrow x^2+5x+\frac{25}{4}+\frac{6}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=-\frac{6}{4}\)
Vậy...
\(f\left(x\right)=x^2+x^2+4x+6x+4+9+2\)
\(=\left(x^2+4x+4\right)+\left(x^2+6x+9\right)+2\)
\(=\left(x+2\right)^2+\left(x+3\right)^2+2>0\)
Vậy đa thức trên ko có ngiệm
Câu 1:
a) \(P\left(x\right)=x^5+7x^4-9x^3+\left(-3x^2+x^2\right)-\frac{1}{4}x\)
\(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)
\(Q\left(x\right)=-x^5+5x^4-2x^3+\left(x^2+3x^2\right)-\frac{1}{4}\)
\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)
b) \(P\left(x\right)+Q\left(x\right)=\left(x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\right)+\left(-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\right)\)
\(P\left(x\right)+Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)
\(P\left(x\right)+Q\left(x\right)=\left(x^5-x^5\right)+\left(7x^4+5x^4\right)-\left(9x^3+2x^3\right)+\left(-2x^2+4x^2\right)-\frac{1}{4}x-\frac{1}{4}\)
\(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}-\frac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\right)-\left(-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x+x^5-5x^4+2x^3-4x^2+\frac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(x^5+x^5\right)+\left(7x^4-5x^4\right)+\left(-9x^3+2x^3\right)-\left(2x^2+4x^2\right)-\frac{1}{4}x+\frac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=2x^5+2x^4-7x^3-6x^2-\frac{1}{4}x+\frac{1}{4}\)
c) \(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)
\(P\left(0\right)=0^5+7\cdot0^4-9\cdot0^3-2\cdot0^2-\frac{1}{4}\cdot0\)
\(P\left(0\right)=0\)
\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)
\(Q\left(0\right)=0^5+5\cdot0^4-2\cdot0^3+4\cdot0^2-\frac{1}{4}\)
\(Q\left(0\right)=-\frac{1}{4}\)
Vậy \(x=0\) là nghiệm của đa thức P(x) nhưng không là nghiệm của đa thức Q(x)
1.a) Theo đề bài,ta có: \(f\left(-1\right)=1\Rightarrow-a+b=1\)
và \(f\left(1\right)=-1\Rightarrow a+b=-1\)
Cộng theo vế suy ra: \(2b=0\Rightarrow b=0\)
Khi đó: \(f\left(-1\right)=1=-a\Rightarrow a=-1\)
Suy ra \(ax+b=-x+b\)
Vậy ...
\(f\left(x\right)=ax^2+bx+c\)
Ta có : \(f\left(-2\right)=4a-2b+c\)
\(f\left(3\right)=9a+3b+c\)
\(\Rightarrow\) \(f\left(-2\right)+f\left(3\right)=4a-2b+c+9a+3b+c\)
\(=13a+b+c\)
\(=0\)
\(\Rightarrow\) \(-f\left(-2\right)=f\left(3\right)\)
\(\Rightarrow\) \(f\left(-2\right).f\left(3\right)=f\left(-2\right).-f\left(-2\right)=-\left[f\left(-4\right)\right]^2\le0\)
\(\Rightarrow\) \(đpcm\)
Study well ! >_<
Giải:
a) \(\dfrac{1}{4}+x-\dfrac{1}{4}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}+\dfrac{3}{4}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3}{4}x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy ...
b) \(\left|x^2-2x\right|+\left|x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x^2-2x\right|=0\\\left|x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow x=0\)
Vậy ...
c) \(\left|3x^2-2x\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=x\\3x^2-2x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2=3x\\3x^2=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-3x=0\\3x^2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x\left(x-1\right)=0\\x\left(3x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
f(1)=\(1^4-987.1^2+1\)=-985
f(-1)=\(\left(-1\right)^4-987.\left(-1\right)^2+1\)=-985
f(1)-f(-1)=-985-(-985)=-985+985=0
Thay x=1 vào f(x),ta có:
f(1)=\(^{ }1^4-987.1^2+1\)
= 1-987+1
= - 985 ( 1)
Thay x= -1 vào f(x) , ta có:
f(-1) = \(\left(-1\right)^4-987.\left(-1\right)^2+1\)
= 1-987+1
= - 985 (2)
Từ (1) (2) => f(1) - f(-1) = -985 - (-985) = 0 ( đccm)
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