\(tìmx:\)\(\left(-3x+9\right).x^2-7x+21\) giải hộ nghennnnnnn
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a, \(\left|5x\right|=x+2\)
Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)
Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)
b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)
Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )
Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )
Vậy phương trình vô nghiệm
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(x-180=\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)
\(\Rightarrow x-180=\left(\dfrac{104}{21}-5\right):-\dfrac{1}{42}:-\dfrac{1}{64}\)
\(\Rightarrow x-180=-\dfrac{1}{21}:-\dfrac{1}{42}:-\dfrac{1}{64}\)
\(\Rightarrow x-180=-\dfrac{1}{21}\cdot-42:-\dfrac{1}{64}\)
\(\Rightarrow x-180=2\cdot-64\)
\(\Rightarrow x-180=-128\)
\(\Rightarrow x=-128+180\)
\(\Rightarrow x=52\)
\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
ĐK: \(x\ne\dfrac{2}{7}\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8+2-7x\right)=\left(x-5\right)\left(3x+8+2-7x\right)\\ \Leftrightarrow\left(2x+3\right)\left(10-4x\right)=\left(x-5\right)\left(10-4x\right)\\ \Leftrightarrow\left(10-4x\right)\left(2x+3-x+5\right)=0\\ \Leftrightarrow\left(10-4x\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10-4x=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{5}{2};-8\right\}\)
=>(x^2-3x)^2+3(x^2-3x)+2=2
=>(x^2-3x)(x^2-3x+3)=0
=>x^2-3x=0
=>x=0 hoặc x=3
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)