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a) \(\left(x-3\right)^2.\left(x+3\right).\left(x-3\right)\)
\(=\left(x-3\right).\left(x-3\right).\left(x+3\right).\left(x-3\right)\)
\(=\left(x-3\right)^3.\left(x+3\right)\)
\(=\left(3x-9\right).\left(x+3\right)\)
Phần b tương tự
\(\left|2x-\frac{1}{2}\right|+1=3x\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)
a) IaI+a =\(\orbr{\begin{cases}a+a=2a\left(a\ge0\right)\\-a+a=0\left(a< 0\right)\end{cases}}\)
b) IaI-a =\(\orbr{\begin{cases}a-a=0\left(a\ge0\right)\\-a-a=-2a\left(a< 0\right)\end{cases}}\)
c)IaI.a=\(\orbr{\begin{cases}a.a=a^2\left(a\ge0\right)\\-a.a=-a^2\left(a< 0\right)\end{cases}}\)
d)IaI:a =\(\orbr{\begin{cases}a:a=1\left(a>0\right)\\-a:a=-1\left(a< 0\right)\end{cases}}\)(a là số chia=>a khác 0)
e)3(x-1)-I2x+3I =\(\orbr{\begin{cases}3x-1-\left(2x+3\right)=x-4\left(x\ge-\frac{3}{2}\right)\\3x-1+\left(2x+3\right)=5x+2\left(x,-\frac{3}{2}\right)\end{cases}}\)
g)2Ix-3I-I4x-1I
xét các th
TH1)\(x< \frac{1}{4}=>=2\left(3-x\right)+4x-1=5+2x\)
TH2)\(\frac{1}{4}\le x\le3=>=2\left(3-x\right)-\left(4x-1\right)=7-6x\)
TH3)\(x>3=>=2\left(x-3\right)-\left(4x-1\right)=-2x-5\)
\(a,\left|a\right|+a=\orbr{\begin{cases}2a\left(a\ge0\right)\\0\left(a< 0\right)\end{cases}}\)
\(b,\left|a\right|-a=\orbr{\begin{cases}0\left(a\ge0\right)\\-2a\left(a< 0\right)\end{cases}}\)
\(c,\left|a\right|.a=\orbr{\begin{cases}a^2\left(a\ge0\right)\\-a^2\left(a< 0\right)\end{cases}}\)
\(d,\left|a\right|:a=\orbr{\begin{cases}1\left(a\ge0\right)\\-1\left(a< 0\right)\end{cases}}\)
\(e,3\left(x-1\right)-2\left|x+3\right|=\orbr{\begin{cases}3x-3-2x-6=x-9\left(x\ge-3\right)\\3x-3+2x+6=5x+3\left(x< -3\right)\end{cases}}\)
\(g,2\left|x-3\right|-\left|4x-1\right|=\orbr{\begin{cases}4-5x\left(x\le\frac{1}{4}\right)\\2+3x\left(\frac{1}{4}< x\le3\right)\end{cases}}\)hoặc \(=5x-4\left(x>3\right)\)
\(A=3\left(2x-1\right)-\left|x-5\right|\)
\(=6x-3-\left|x-5\right|\)
TH1 : \(x-5\ge0\Rightarrow x\ge5\Rightarrow\left|x-5\right|=x-5\)
\(A=6x-3-x+5\)
\(=5x+2\)
TH2 : \(x-5< 0\Rightarrow x< 5\Rightarrow\left|x-5\right|=5-x\)
\(A=6x-3-5+x\)
\(=7x-8\)
Vậy ....
a) \(x^3-2x^2+x=0\)
\(\Leftrightarrow x\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy....
b) \(-x^4-x^2-3=0\)
\(\Leftrightarrow x^4+x^2+3=0\)
\(\Leftrightarrow\left(x^2\right)^2+2\cdot x^2\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2=\frac{-11}{4}\)( vô lý )
Đa thức vô nghiệm
a) 3(x-1)-2|x+3| = 3x-3-2(x+3) = 3x-3-2x+6 = (3x-2x)+(6-3)=x+3
b) 2|x-3|-|4x-1| = 2(x-3)-(4x-1) = 2x-6-4x+1 = (2x-4x)-(6-1) = -2x-5