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( fx) + g(x) + (f(x) - g(x) = 6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9
= 10x^4 + 4x^2 + 8x - 14
=> 2fx = 2 ( 5x^4 + 2 x^2 + 4x - 7)
=> f(x) = 5x^4 + 2x^2 + 4x - 7
Tính tiếp g(x) nha
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\(P(x)=4x^2-1\)
\(4x^2-1=0\)
\(\Leftrightarrow 4x^2=1\)
\(\Leftrightarrow x^2=\frac{1}{4}\)
\(\Rightarrow\left[\begin{array}{} x=\frac{1}{2}\\ x=\frac{-1}{2} \end{array} \right.\)
Vậy đa thức có nghiệm: \(x=\frac{1}{2};x=\frac{-1}{2}\)
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Câu 1:
b) \(\left|x-\frac{1}{2}\right|-\left(\frac{1}{9}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\left|x-\frac{1}{2}\right|-\frac{1}{81}=\frac{1}{16}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{16}+\frac{1}{81}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{97}{1296}.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{2}=\frac{97}{1296}\\x-\frac{1}{2}=-\frac{97}{1296}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{97}{1296}+\frac{1}{2}\\x=\left(-\frac{97}{1296}\right)+\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{745}{1296}\\x=\frac{551}{1296}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{745}{1296};\frac{551}{1296}\right\}.\)
Chúc bạn học tốt!
a)4x=3y;5y=3z và 2x-3y+z=6
Có : \(4x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{4}\Leftrightarrow\frac{1}{3}\times\frac{x}{3}=\frac{1}{3}\times\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)
\(5y=3z\Leftrightarrow\frac{y}{3}=\frac{z}{5}\Leftrightarrow\frac{1}{4}\times\frac{y}{3}=\frac{1}{4}\times\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\Leftrightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)
-Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)
-Do đó :
\(\frac{2x}{18}=2\Leftrightarrow\frac{x}{9}=2\Rightarrow x=2\times9=18\)
\(\frac{3y}{36}=2\Leftrightarrow\frac{y}{12}=2\Rightarrow y=2\times12=24\)
\(\frac{z}{20}=2\Rightarrow z=2\times20=40\)
Vậy x = 18 , y = 24 , z = 40
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\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)
\(f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\)
suy ra: \(\left[f\left(x\right)+g\left(x\right)\right]+\left[f\left(x\right)-g\left(x\right)\right]=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\)
\(\Leftrightarrow\)\(2f\left(x\right)=10x^4-6x^3+4x^2+8x-14\)
\(\Rightarrow\)\(f\left(x\right)=5x^4-3x^3+2x^2+4x-7\)
\(g\left(x\right)=6x^4-3x^2-5-f\left(x\right)\)
\(=x^4+3x^3-5x^2-4x+2\)
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\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)
Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
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\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(4x^2+4x=2\)
=>\(2x^2+2x=1\)
=>\(2x^2+2x-1=0\)
\(\Delta=2^2-4\cdot2\cdot\left(-1\right)=4+8=12>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-2-\sqrt{12}}{2\cdot2}=\dfrac{-2-2\sqrt{3}}{4}=\dfrac{-1-\sqrt{3}}{2}\\x_2=\dfrac{-1+\sqrt{3}}{2}\end{matrix}\right.\)