3y大概多大 (y+1)(y-3)=-y(3-3y)的解
(y+1)(y-3)=-y(3-3y)的解
(y+1)(y-3)=-y(3-3y)的解
(y+1)(y-3)=-y(3-3y) (y平方)-2y-3=-3y+3(y平方)移项并合并同类项得(移到右边) 2(y平方)+5y+3=0 (因式分解得) (2y+3)(y+1)=0 解得x1=-3/2 x2=-1 能看明白吗?
记得采纳啊
解方程:(1)3(x-2)+2=x-2(x-1); (2)4(y+1)-5(y-3)=6(y+1)-3(y-3)
(1)去括号,得3x-6+2=x-2x+2,
移项,得3x-x+2x=2+6-2,
合并同类项,得4x=6,
系数化为1,得x=
;
(2)去括号,得4y+4-5y+15=6y+6-3y+9,
移项,得4y-5y-6y+3y=6+9-4-15,
合并同类项,得-4y=-4,
系数化为1,得y=1.
3-2(y+1)=2(y-3)
3-2(y+1)=2(y-3)
3-2y-2=2y-6
1-2y=2y-6
2y+2y=1+6
4y=7
y=7/4
解方程:①4(y+1)-5(y-3)=6(y+1)-3(y-3);②2x?13-10x+112=2x+14-1
①去括号得,4y+4-5y+15=6y+6-3y+9,
移项得,4y-5y-6y+3y=6+9-4-15,
合并同类项得,-4y=-4,
系数化为1得,y=1;
②去分母得,4(2x-1)-(10x+1)=3(2x+1)-12,
去括号得,8x-4-10x-1=6x+3-12,
移项得,8x-10x-6x=3-12+4+1,
合并同类项得,-8x=-4,
系数化为1得,x=
.
(y-3)(y-2)+18=(y+9)(y+1)
(y-3)(y-2)+18=(y+9)(y+1);
y²-5y+6+18=y²+10y+9;
15y=15;
y=1;
-1<y<3,化简|y+1|+|y-3|
因为-1<y<3
所以y+1>0,y-3<0
所以原式=y+1-(y-3)
=y+1-y+3
=4

(y-3)/(y+1)<4
移项:(y-3)/(y+1)-4<0
通分:(y-3)/(y+1)-4(y+1)/(y+1)<0
(y-3-4y-4)/(y+1)<0
(-3y-7)/(y+1)<0
(3y+7)/(y+1)>0 相除大于0,则两式同号,所以,相乘大于0
(3y+7)(y+1)>0
所以:y<-7/3或y>-1
祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O
已知y*y+3y-1=0求y*y*y*y/(y*y*y*y*y*y*y*y-3*y*y*y*y+1)的值
因为
y*y+3y-1=0
所以
y-1/y=-3
(y-1/y)²=9
y²+1/y²-2=9
y²+1/y²=11
再平方得
(y²+1/y²)²=121
y的4次方+1÷y的4次方=119
所以
y*y*y*y/(y*y*y*y*y*y*y*y-3*y*y*y*y+1)
=1/(y*y*y*y-3+1/y*y*y*y)
=1/(119-3)
=1/116
求:y(3y-3)=y+1 的解
y(3y-3)=y+1
3y^2-3y=y+1
3y^2 -4y-1=0
即
a=3,b=-4,c=-1
△=b^2-4ac=16+12=28>0
方程有两了不等的实数解
x1=[-b-根号(b^2-4ac)]/(2a)=(4-根号28)/6=(2-根号7)/3
x2=[-b+根号(b^2-4ac)]/(2a)=(4+根号28)/6=(2+根号7)/3