a) -15(x-2)+7(3-x) =7
b) (x-5).(x-7) < 0
c) -3 /3x - 1/= - 6 .
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2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
2:
a: =>-2x=10
=>x=-5
b: =>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c)(x-4).(2x+6)=0
=>(x-4)=0 hoặc (2x+6)=0
với x-4 = 0
x =0+4
x =4
với 2x+6=0
2x =0-6
2x =-6
x =-6:2
x =-3
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
\(a,2.\left(x+3\right)-3x=-2x+7\)
\(\Leftrightarrow2x+6-3x=-2x-7\)
\(\Leftrightarrow-x+6=-2x+7\)
\(\Leftrightarrow x=-6+7\)
\(\Leftrightarrow x=1\)
\(b,5.\left(x-3\right)-29=-2.\left(7-x\right)-23\)
\(\Leftrightarrow5x-15-29+23=14-2x\)
\(\Leftrightarrow5x-21=14-2x\)
\(\Leftrightarrow5x+2x=35\)
\(\Leftrightarrow x=5\)
\(c,-17+\left|5-x\right|=-2.7\)
\(\Leftrightarrow-17+\left|5-x\right|=-14\)
\(\Leftrightarrow\left|5-x\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=8\end{cases}}\)
\(d,21-\left|x+7\right|=-42:\left(-2\right)\)
\(\Leftrightarrow21-\left|x+7\right|=21\)
\(\Leftrightarrow\left|x+7\right|=0\)
\(\Leftrightarrow x+7=0\)
\(\Leftrightarrow x=7\)
\(f,\left|3x-5\right|-\left(-15\right)=5\)
\(\Leftrightarrow\left|3x-5\right|+15=5\)
\(\Leftrightarrow\left|3x-5\right|=-10\)
Vì \(\left|a\right|\ge0\)mà \(-10< 0\)nên \(x\in\Phi\)
Đã giải quyết xong!!!
giúp mình với: so sánh -37/56 với -377/567 ai giải đc mình cho
a) \(-15\left(x-2\right)+7\left(3-x\right)=7\)
\(-15x+30+21-7x=7\)
\(-22x+51=7\)
\(-22x=-44\)
\(x=2\)
vậy \(x=2\)
b) \(\left(x-5\right)\left(x-7\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-5< 0\\x-7>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-5>0\\x-7< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 5\\x>7\end{cases}}\) hoặc \(\hept{\begin{cases}x>5\\x< 7\end{cases}}\)
\(\Rightarrow5< x< 7\)
\(\Rightarrow x=6\)