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![](https://rs.olm.vn/images/avt/0.png?1311)
\(-8x-10=\left|-8x-10\right|\)
\(\left|-8x-10\right|=-8x-10\) khi \(-8x-10\ge0\Leftrightarrow-8x\ge10\Leftrightarrow x\le-\dfrac{5}{4}\)
PT trở thành :
\(-8x-10=-8x-10\)
\(\Leftrightarrow-8x+8x=-10+10\)
\(\Leftrightarrow0x=0\)
PT có vô số nghiệm
Vậy \(S=\){\(x\in R\)| x\(\le-\dfrac{5}{4}\)}
\(\left|-8x-10\right|=-\left(-8x-10\right)=8x+10\) khi \(-8x-10< 0\Leftrightarrow-8x< 10\Leftrightarrow x>-\dfrac{5}{4}\)
PT trở thành :
\(-8x-10=8x+10\)
\(\Leftrightarrow-8x-8x=10+10\)
\(\Leftrightarrow-16x=20\)
\(\Leftrightarrow4x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{4}\)(KTMĐK)
Vậy \(S=\){\(x\in R\)|x \(\le-\dfrac{5}{4}\)}
=> Đáp án B đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(\Leftrightarrow2x-8>=2x+1\)
=>-8>=1(vô lý)
d: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>10x-30>3x+5
=>7x>35
hay x>5
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
![](https://rs.olm.vn/images/avt/0.png?1311)
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời giùm đi mọi người , mình đang cần gấp. Cảm ơn nhiều!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(đkxđ:x\ne1;2;3;4;5\\ \Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}=\dfrac{1}{15}\\ \Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}=\dfrac{1}{15}\\ \Leftrightarrow\dfrac{1}{x-5}-\dfrac{1}{x-1}=\dfrac{1}{15}\\ \Leftrightarrow60=x^2-6x+5\\ \)
\(\Leftrightarrow60=x^2-6x+5\\ \Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\\ \Rightarrow D\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)
\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)
\(\Leftrightarrow x^2-x-72=0\)
=>(x-9)(x+8)=0
=>x=9 hoặc x=-8
6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0 hoặc x=5
5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x
<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8
6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9
<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5
7, <=> (x-1)^2 = (3x+3)^2
<=> (x-1-3x-3)(x-1+3x+3) = 0
<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2
8, = (x^2-10x-15)(x^2-10x+25)
`B.S={x|x<-8/5}`
`-5x-8=|5x+8|`
`<=>-(5x+8)=|5x+8|`
`<=>5x+8<=0`
`<=>x<=-8/5`
`B.S={x|x<=-8/5}`
`-5x-8=|5x+8|`
`<=>-(5x+8)=|5x+8|`
`<=>5x+8<=0`
`<=>x<=-8/5`