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\(\Leftrightarrow2x^2+10x-x^2+6x-9=x^2+6\)
=>16x-9=6
=>16x=15
hay x=15/16
\(PT\Leftrightarrow2x^2+10x-x^2+6x-9-x^2-6=0.\)
\(\Leftrightarrow16x-15=0.\\ \Leftrightarrow x=\dfrac{15}{16}.\)
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)
\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)
\(\Leftrightarrow x^2+x-x^2+3x=4x\)
\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)
\(\Leftrightarrow0x=0\)
Phương trình có vô số nghiệm , trừ x = -1,x = 3
Vậy ...
\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)
\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)
\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)
\(\Leftrightarrow12x+1< 36x+4-24x-3\)
\(\Leftrightarrow12x+1< 12x+1\)
\(\Leftrightarrow12x-12x< 1-1\)
\(\Leftrightarrow0x< 0\)
Vậy S = {x | x \(\in R\)}
\(\Leftrightarrow36\left(x+6\right)+36\left(x-6\right)=\dfrac{9}{2}\left(x^2-36\right)\)
\(\Leftrightarrow x^2\cdot\dfrac{9}{2}-162-72x=0\)
\(\Leftrightarrow9x^2-144x-324=0\)
\(\Leftrightarrow x^2-16x-36=0\)
=>(x-18)(x+2)=0
=>x=18 hoặc x=-2
ĐKXĐ:\(x\ne\pm6\)
\(\dfrac{36}{x-6}+\dfrac{36}{x+6}=4,5\\ \Leftrightarrow36\left(\dfrac{1}{x-6}+\dfrac{1}{x+6}\right)=4,5\\ \Leftrightarrow\dfrac{x+6}{\left(x-6\right)\left(x+6\right)}+\dfrac{x-6}{\left(x-6\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x+6+x-6}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{2x}{x^2-36}=\dfrac{1}{8}\\ \Leftrightarrow x^2-36=16x\\ \Leftrightarrow x^2-16x-36=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(18x+36\right)=0\\ \Leftrightarrow x\left(x+2\right)-18\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-18\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=18\left(tm\right)\end{matrix}\right.\)
\(\dfrac{36}{x+6}+\dfrac{36}{x-6}=4,5\)
\(\Leftrightarrow36\left(x-6\right)+36\left(x+6\right)=4,5\left(x^2-36\right)\)
\(\Leftrightarrow36x-216+36x+216=4,5x^2-162\)
\(\Leftrightarrow-4,5x^2+72x+162=0\)
\(\Leftrightarrow\left(x-18\right)\left(-4,5x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-2\end{matrix}\right.\)
bạn làm rõ hơn ở chỗ này đc ko, mk ko hiểu
⇔−4,5x2+72x+162=0⇔−4,5x2+72x+162=0
⇔(x−18)(−4,5x−9)=0
`|5x| = - 3x + 2`
Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :
`5x =-3x+2`
`<=> 5x +3x=2`
`<=> 8x=2`
`<=> x= 2/8=1/4` ( thỏa mãn )
Nếu `5x<0<=>x<0` thì phương trình trên trở thành :
`-5x = -3x+2`
`<=>-5x+3x=2`
`<=> 2x=2`
`<=>x=1` ( không thỏa mãn )
Vậy pt đã cho có nghiệm `x=1/4`
__
`6x-2<5x+3`
`<=> 6x-5x<3+2`
`<=>x<5`
Vậy bpt đã cho có tập nghiệm `x<5`
\(\left|3x-6\right|=x^2-4x+6\)
\(3x-6=x^2-4x+6\)
\(-x^2+3x+4x-6-6=0\)
\(-x^2+7x-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=4\\x_2=3\end{matrix}\right.\)
Đặt \(\left|x\right|=t\left(t\ge0\right)\). Ta có phương trình \(t^2-t=6\)
\(\Rightarrow t^2-t-6=0\Rightarrow t^2-3t+2t-6=0\)
\(\Rightarrow\left(t-3\right)\left(t+2\right)=0\Rightarrow\left[{}\begin{matrix}t=3\left(TM\right)\\t=-2\left(L\right)\end{matrix}\right.\)
\(\Rightarrow\left|x\right|=3\Rightarrow x=\pm3\)