搜狗做题网7.已知∣x-3∣-∣x 2∣的最大值为p,最小值为q,求pq的值.
来源：学生作业帮助网 编辑：作业帮 时间：2024/06/05 17:19:50
1/10x4+3x2+1=x4-x3+(x3+3x2+x)-x+1=x4-x3+x(x2+3x+1)-x+1=x4-x3-x+1=x4-(x3+3x2+x)+3x2+1=x4-x(x2+3x+1)+3
∵x2+x-1=0，∴x2+x=1，2x3+3x2-x=2x（x2+x）+x2-x=2x+x2-x=x2+x=1．
x^2+3x-2＝0,x(x+3)=2,∴(2x-4)╱(x^2+6x+9)÷（x-3）×(x^2-9)╱(2x-x^2)=2(x-2)/(x+3)^*1/(x-3)*(x+3)(x-3)/[x(2-
∵x=2是方程3a-x=x2+3的解，∴3a-2=1+3解得：a=2，∴原式=a2-2a+1=22-2×2+1=1．
x^2-3x+2=0(x-2)(x-1)=0x=2或x=1当x=2时x^2+1/x^2=2^2+1/2^2=4+1/4=17/4当x=1时x^2+1/x^2=1^2+1/1^2=1+1=2
x+1/x=3(x+1/x)^2=3^2=9x^2+1/x^2+2=9x^2+1/x^2=7
解析：由集合A={x∣x2
x^2+3x+1=0方程两边同除以xx+3+1/x=0x+1/x=-3x^2+1/x^2=(x+1/x)^2-2=(-3)^2-2=9-2=7
x3-3x2-11x+8=x（x2-3x)-11x+8=x-11x+8=-10x+8
4x^2-（3m-5）x-6m^2=0△=b^2-4ac=(3m-5)^2+96m^2>0|x1/x2|=3/2.x1=(3/2)*x2或x1=(-3/2)*x2根据根与系数的关系得：x1+x2=(3
x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x
2x2-3x
x^2+2x+5=7x^2+2x=23x^2+6x+3=3(x^2+2x)+3=3*2+3=9
∵x^2+3x+5=7∴x^2+3x=23x^2+9x+2=3(x^2+3x)+2=3×2+2=8
x/(x²-3x+1)=2(x²-3x+1)/x=1/2翻过来x+1/x=7/2(x^4+x2+1)/x²翻过来=x²+2+1/x²=(x+1/x)&
M={y|y=(x-2)^2-1,x∈Z｝={y|y=k^2-1,且k∈Z｝,这个说明,x是整数,则y也是整数了,相当于M=｛-1,0,3,8,15.｝N=｛y∣y=-(x+1)^2+1,x∈Z｝=｛
因为A=｛2｝,因此方程x^2+2ax+b=0有重要x=2,所以x^2+2ax+b=(x-2)^2=x^2-4x+4,所以a=-2,b=4,不等式化为-2x^2+4x+6>0,因此x^2-2x-3再问
对任意定义域内的xf（x×x）=f（x）+f（x）=2f（x）f（-x×-x）=f（-x）+f（-x）=2f（-x）f（-x×-x）=f（x×x）即2f（x）=2f（-x）即f（x）=f（-x）所以得
你好x²+3x-1=0两边同除以xx+3-1/x=0x-1/x=-3两边平方得x²+1/x²-2=9x²+1/x²=11x²+1/x
X2-3X-1=0则X-3-1/X=0则X-1/X=3则(X-1/X)²=3²=9则X²-2+1/X²=9则X²+1/X²=9+2=11