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a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)
Bậc của P(x) là 3
\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)
Bậc của Q(x) là 3
b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
|2x+6|+ |x+2| + |2x+2| =4
=>2x+6+ x+2 + 2x+2 =4 hoặc 2x+6+ x+2 + 2x+2 =-4
=>2x+6+ x+2 + 2x+2 =4 2x+6+ x+2 + 2x+2 =-4
=>(2x+x+2x)+(6+2+2)=4 (2x+x+2x)+(6+2+2)=-4
=> 5x+10=4 5x+10=-4
=>5x=4-10 5x=(-4)-10
=>5x=-6 5x=-14
=>x=(-6):5 x= (-14):5
=>x=-6/5 x=-14/5
vậy x∈{-6,5;-14/5}
\(\left|2x+6\right|+\left|x+2\right|+\left|2x+2\right|=4\)
\(\Leftrightarrow2\left|x+3\right|+\left|x+2\right|+2\left|x+1\right|=4\) (1)
+, Với \(x< -3\) thì (1) trờ thành:
\(2(-x-3)+(-x-2)+2(-x-1)=4\)
\(\Rightarrow-2x-6-x-2-2x-2=4\)
\(\Rightarrow-5x-10=4\)
\(\Rightarrow-5x=14\)
\(\Rightarrow x=-\dfrac{14}{5}\left(ktm\right)\)
+, Với \(-3\le x< -2\), thì (1) trở thành:
\(2\left(x+3\right)+\left(-x-2\right)+2\left(-x-1\right)=4\)
\(\Rightarrow2x+6-x-2-2x-2=4\)
\(\Rightarrow-x+2=4\)
\(\Rightarrow x=-2\left(ktm\right)\)
+, Với \(-2\le x< -1\), thì (1) trở thành:
\(2\left(x+3\right)+\left(x+2\right)+2\left(-x-1\right)=4\)
\(\Rightarrow2x+6+x+2-2x-2=4\)
\(\Rightarrow x+6=4\)
\(\Rightarrow x=-2\left(tm\right)\)
+, Với \(x\ge-1\), thì (1) trở thành:
\(2\left(x+3\right)+\left(x+2\right)+2\left(x+1\right)=4\)
\(\Rightarrow2x+6+x+2+2x+2=4\)
\(\Rightarrow5x+10=4\)
\(\Rightarrow5x=-6\)
\(\Rightarrow x=-\dfrac{6}{5}\left(ktm\right)\)
Vậy \(x=-2\)
#Urushi☕
a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)
Biết nội quy r ko cần nói âu -_-
2k5???=))