Tìm x
3x^3-7x^2+17x-5=0
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1.\(\left|9-7x\right|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=-5x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-9-3\\-7x+5x=-9-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-12:\left(-12\right)\\x=-12:\left(-2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
2.\(8x-\left|4x+1\right|=x+2\)
\(\Rightarrow\left|4x+1\right|=8x-x+2\)
\(\Rightarrow\left|4x+1\right|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-7x=2-1\\4x+7x=2-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1:\left(-3\right)\\x=1:11\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{1}{11}\end{cases}}\)
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)
b: =>|17x-5|=|17x+5|
=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5
=>34x=0
hay x=0
c: =>|3x+4|=|4x-18|
=>4x-18=3x+4 hoặc 4x-18=-3x-4
=>x=22 hoặc 7x=14
=>x=22 hoặc x=2
chào các bạn,có 2 tốt bụng thì tk mik nhé,cần lắm những người như thế
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
Bài 2
\(a,x^3+5x^2+3x-9\)
\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)
\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)
b,\(x^3-7x-6\)
\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)
\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
Bài 1:
a)\(28x^3+15x^2+75x+125=0\)
\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)
Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)
\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)
b)\(4x^2-x-5=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)
\(\Rightarrow x=-1;x=\frac{5}{4}\)
Bài 2:
a)\(x^3+5x^2+3x-9\)
\(=\left(x-1\right)\left(x+3\right)^2\)
b)\(x^3-7x-6\)
\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)