" 22.) "(1)/(x)-(1)/(x-2)=3,x!=0,2

Solve forx :(x+1)/(x-1)+(x-2)/(x+2)=4-(2x+3)/(x-2);x!=1,-2,2

On the interval I=[-2,2] , the function f(x)={{:(,(x+1)e^(-((1)/(|x|)+(1)/(x))),x ne 0),(,0,x=0):}

On the interval I=[-2,2] , the function f(x)={{:(,(x+1)e^(-((1)/([x])+(1)/(x))),x ne 0),(,0,x=0):}

On the interval I=[-2,2] , the function f(x)={{:(,(x+1)e^(-((1)/(|x|)+(1)/(x))),x ne 0),(,0,x=0):}

On the interval I=[-2,2] , the function f(x)={{:(,(x+1)e^(-((1)/(|x|)+(1)/(x))),x ne 0),(,0,x=0):}

If x^(2)+3x+1=0 then find x^(3)+(1)/(x^(3)),x^(4)+(1)/(x^(4)),x^(2)-(1)/(x^(2)),x^(2)+(1)/(x^(2))

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)

Solve |((x+1),0,0)) , ((2x+1),(x-1),0)) , ((3x+1),(2x-1),(x-2))| =0