Giúp mk vài con với : a) 3/x-5=-4/x+2 ; B) x+3/-4=-9/x+3
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\(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{1000}{999}\)
\(=\frac{3\times4\times5\times...\times1000}{2\times3\times4\times...\times999}=\frac{1000}{2}=500\)
b ) -2(3x+2)-3 =28
-6x-4-3=28
-6x = 28 +4+3
-6x = 35
x =\(\dfrac{35}{-6}\)= \(\dfrac{-35}{6}\)= -5,8
vậy x= -5,8
c ) -(x+3)-5(x-7)=10x-20
-(x+3)-5x+35=10x-20
-x+3-5x+35=10x-20
-x-5x-10x=-20-3-35
-16x=-58
x=\(\dfrac{-58}{-16}\)=\(\dfrac{58}{16}\)=3,625
vậy x =3,625
d)2x+37=-6(3x-8)
2x+37=-18x+48
2x+18x=48-37
20x=11
x=\(\dfrac{11}{20}\)=0,55
vậy x=0,55
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\\ 4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ 2x+255=0\\ 2x=-255\\ x=-\dfrac{255}{2}\)
(x-5)^2+(x+3)^2 = x^2 -10x + 25 + x^2 + 6x +9= 2(x^2 - 16) -5x +7 = 2(x-4)(x+4) - 5x + 7
a) <=> \(3x^4-9x^3+9x^2-27x=0\)
<=>\(3x\left(x^3-3x^2+3x-9\right)=0\)
<=>\(3x\left(x-3\right)\left(x^2+3\right)\)=0
<=>x=0 hoặc x=3
b) \(\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
<=>\(\left(x+3\right)\left(x^2-4x+5\right)=0\)
<=>\(\left(x+3\right)\left(\left(x-2\right)^2+1\right)=0\)
=> x=-3
a) 3x4 - 9x3 = -9x2 + 27x
3x4 - 9x3 + 9x2 - 27x = 0
3x(x3 - 3x2 + 3x - 9) = 0
3x[x2(x - 3) + 3(x - 3)] = 0
3x(x - 3)(x2 + 3) = 0
vì x2 + 3 > 0 nên:
3x = 0 hoặc x - 3 = 0
x = 0 : 3 x = 0 + 3
x = 0 x = 3
=> x = 0 hoặc x = 3
b) (x + 3)(x2 - 3x + 5) = x2 + 3x
x3 - 3x2 + 5x + 3x2 - 9x = x2 + 3x
x3 - 4x + 15 = x2 + 3x
x3 - 4x + 15 - x2 - 3x = 0
x3 - 7x + 15 - x2 = 0
(x2 - 4x + 5)(x + 3) = 0
vì x2 - 4x + 5 > 0 nên
x + 3 = 0
=> x = -3
a) \(\frac{3}{x-5}=\frac{-4}{x+2}\) ĐKXĐ : \(x\ne5;-2\)
\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+4x=20-6\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )
b) \(\frac{x+3}{-4}=\frac{-9}{x+3}\) ĐKXĐ : \(x\ne-3\)
\(\Rightarrow\left(x+3\right)\left(x+3\right)=\left(-9\right)\left(-4\right)\)
\(\Leftrightarrow\left(x+3\right)^2=36\)
\(\Leftrightarrow\left(x+3\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}}\)( thỏa mãn ĐKXĐ )
Vậy....
a, \(\frac{3}{x-5}=\frac{-4}{x+2}\Rightarrow\frac{3x+6}{\left(x-5\right)\left(x+2\right)}=\frac{-4x+20}{\left(x-5\right)\left(x+2\right)}\)
\(\Rightarrow3x+6=-4+20\Rightarrow3x=-4+20-6=10\Rightarrow x=\frac{10}{3}\)