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* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a/Ta có: M(x)+N(x) = (2x5 - 4x3 + 2x2 + 10x - 1) + (-2x5 + 2x4 + 4x3 + x2 + x - 10)
= 2x5 - 2x5 - 4x3 + 4x3 + 2x4 + 2x2 + x2 + 10x + x -1 - 10
= 2x4 + 3x2 + 11x - 11
b/ Ta có: A(x) = N(x)-M(x) = (-2x5 + 2x4 + 4x3 + x2 + x - 10) - (2x5 - 4x3 + 2x2 + 10x - 1)
= -2x5 - 2x5 + 2x4 + 4x3 + 4x3 + x2 - 2x2 + x - 10x -10 + 1
= -2x5 + 2x4 + 8x3 - x2 - 9x -9
1: Trường hợp 1: x<-2
Pt sẽ là -x-2+5-x=7
=>-2x+3=7
=>-2x=4
hay x=-2(loại)
Trường hợp 2: -2<=x<5
Pt sẽlà x+2+5-x=7
=>7=7(luôn đúng)
Trường hợp 3: x>=5
Pt sẽ là x+2+x-5=7
=>2x-3=7
=>x=5(nhận)
4: \(\left|x^2-2x\right|=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(x^2-2x\right)^2=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-2x-x\right)\left(x^2-2x+x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x^2-3x\right)\left(x^2-x\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;1;3\right\}\)
5: Ta có: \(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(2x+3+x+2\right)\left(2x+3-x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+5\right)\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};-1\right\}\)
6: |5x-4|=|x+2|
=>5x-4=x+2 hoặc 5x-4=-x-2
=>4x=6 hoặc 6x=2
=>x=3/2 hoặc x=1/3
a/ \(M=\left(-2x^4+x^2+5\right)-\left(5x^2-x^3+4x\right)\)
\(=-2x^4+x^2+5-5x^2+x^3-4x\)
\(=-2x^4+x^3-4x^2-4x+5\)
Vậy...
b/ \(M=-2x^4+x^2+5+5x^2-x^3+4x\)
\(=-2x^4-x^4+6x^2+4x+5\)
Vậy...
c/ \(M=\left(5x^2-x^3+4x\right)-\left(-2x^4+x^2+5\right)\)
\(=5x^2-x^3+4x+2x^4-x^2-5\)
\(=2x^4-x^3+4x^2-5\)
Vậy...
d/ \(M=-\left(5x^2-x^3+4x\right)\)
\(=x^4-5x^2-4x\)
Vậy..