Ta có: \(\left(2x+1\right)\left(-2x+1\right)+\left(1-2x\right)^2=18\)
\(\Leftrightarrow1-4x^2+4x^2-4x+1=0\)
\(\Leftrightarrow-4x=-2\)
hay \(x=\dfrac{1}{2}\)
tìm x biết \(\left(2x+1\right)\cdot\left(x+1\right)^2\cdot\left(2x+3\right)=18\)
Tìm x
a)\(3x\left(2x+1\right)=5\left(2x+1\right)\)
b)\(\left(3x-8\right)^2=\left(2x-7\right)^2\)
c)\(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
d)\(\left(9x^2-16\right)^2-4\left(3x+4\right)^2\)
e)\(\left(2x-1\right)\left(4x^2+2x+1\right)=x\left(x-8\right)\)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
Tìm x
a, \(\dfrac{\left(x+2\right)^2}{2}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) + \(\dfrac{\left(1-2x\right)^2}{8}\) – (1 + x)2 = 0
b, \(\dfrac{\left(x+1\right)^2}{2}\) - \(\dfrac{\left(1-2x\right)^2}{3}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) - \(\dfrac{\left(5-x\right)^2}{6}\)= 0
c, (3 + x)3 – 3x2(x + 4) + (x + 2)3 = (1 – x)3 – 8
1.Tìm x , biết :
a ) \(2x\left(x-3\right)-x\left(2x+3\right)=18\)
b ) \(x\left(5x^2-2\right)+5x\left(1-x^2\right)=3^4\)
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
Tìm x
\(2x.\left(x-5\right)-x.\left(3+2x\right)=26\)
\(\left(x-7\right).\left(x-5\right)-12.\left(3x-7\right)=15\)
\(4.\left(18-5x\right)-12.\left(3x-7\right)=15.\left(2x-16\right)-6.\left(x+14\right)\)
\(\left(x-1\right).\left(x^2+x+1\right)=x^3-2x\)
