\(x^2-9x+18-4=0\Leftrightarrow x^2-9x+14=0\Leftrightarrow x^2-2x-7x+14=0\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)
=> x=2 hoặc x=7
Bài 1 : Tìm x ,y,z biết:
a, 3/x-1 = 4/y-2 = 5/z-3 và x+y+z = 18
b, 3/x-1 = 4/y-2 = 5/z-3 và x.y.z = 192
Bài 2 : Tìm x,y,z biết : x^3+y^3/6 = x^3-2y^3/4 và x^6.y^6 = 64
Bài 3 : Tìm x,y,z biết :x+4/6 = 3y-1/8 = 3y-x-5/x
Bài 4 :Tìm x,y,z biết : x+y+2005/z = y+z-2006 = z+x+1/y = 2/x+y+z
tìm x biết
1) x=-1/2 + 3/4 2) x - 1/5 = 2/11 3) x - 5/6 = 16/42 + -8/56
4) x/5 = 5/6 + -19/30 5) |x| - 1/4 = 6/18 6)x = -1/2 +3/4
7) x/15 = 3/5 + -2/3 8)11/8 + 13/6 = 85/x 9) x - 7/8 = 13/12
10) x - -6/15 = 4/27 11) -\(\dfrac{-6}{12}\)+ x = 9/48 12) x - 4/6 = 5/25 + -7/15
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
6/3+5/x+6/x=3/4
tìm x
a)tìm x,y biết:x-1/y+2 và x+y=23
b)tìm x biết: 4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=8^2x-6
Tìm x biết:
a) 6* 8^x-1 + 8^x+1 = 6* 8^19 + 8^21
b) 4* 3^x-1 + 2* 3^x+2 = 4* 3^6 +2* 3^9
(x+2)(x+3)(x+4)-x^2(x-5)=6 Tìm x:
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
Tìm x a) |5/3.x|=|-1/6|
B) |3/4.x-3/4|-3/4=|-3/4|
C)|x+3/5|-|x-7/3|=0
tìm x biết x-1/7 + x-2/3 + x-3/5 + x-4/2 = 6
