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(2x+3)²=4(2x+3)

2x²+3=7x, 2x²-7x+3=0, 2[x²-2×(7/4)×x+(7/4)²]=2×(7/4)²-3, 2(x-7/4)²=25/8, (x-7/4)²=25/16, x-7/4=5/4,或x-7/4=-5/4, x=3,或x=1/2

x/(2x-3)+5/(3x-2)=4 x(3x-2)+5(2x-3)=4(2x-3)(3x-2) 3x²-2x+10x-15=24x²-52x+24 21x²-60x+39=0 7x²-20x+13=0 (7x-13)(x-1)=0 x1=1 x2=13/7

x=[(1+√5)±√(30-6√5)]/2 或 x=[(1-√5)±√(30+6√5)]/2 解析: //“一元四次方程求根公式”,很复杂!! x⁴-2x³-16x²+32x+16=0 x⁴-2x³=16x²-32x-16 x⁴-2x³+x²=(16x²-32x-16)+x² (x²-...

已知函数f(x)=x²+2x-3,x∈[t,t+2]。求函数的值域 解析: f(x) =x²+2x-3 =(x+1)²-4 (1) t+2≤-1时,f(x)在[t,t+2]上单调递增减 fmax=t²+2t-3 fmin=t²+6t+5 值域:[t²+6t+5,t²+2t-3] (2) t≥-1时,f(x)在[t,...

cos2x=cos²x-sin²x=1-2sin²x 所以2sin²x=1-cos2x f(x)=sin(2x-π/4)-2√2sin²x =sin2xcosπ/4-cos2xsinπ/4-√2(1-cos2x) =√2/2 sin2x- √2/2 cos2x +√2cos2x-√2 =√2/2 sin2x+√2/2 cos2x-√2 =sin(2x+π/4)-√2 最小正周期是2π/...

题目有误, 如果是y=-x²+2x+3, 可以写成 y=-x²+2x+3=-(x²-2x-3)=-(x²-2x+1-1-3)=-(x²-2x+1)+4=-(x-1)²+4

1/2x+﹙-3/2x+1/3y²﹚-﹙2x-2/3y²﹚ =(1/2)x -(3/2)x + (1/3)y² -2x +(2/3)y² =-3x+y²

3x2+2x=3? 应该是3x²+2x=3吧? 解: 3x²+2x=3 x²+(2/3)x-1=0 x²+2×(1/3)x+(1/3)²-(1/3)²-1=0 (x+1/3)²-(1/3)²-1=0 (x+1/3)²=10/9 x+1/3=±(√10)/3 x=(-1±√10)/3 所求方程的根为:x=(-1+√10)/3、x=...

2x²+7x-4=0 x²+7/2x=2 x²+7/2x+(7/4)²=2+49/16 (x+7/4)²=81/16 x+7/4=±√81/16 x1=1/2 x2=-4

1×2+2×3+3×4+……+59×60 =1²+1+2²+2+3²+3+……+59²+59 =1/6×59×(59+1)×(2×59+1)+(1+59)×59÷2 =1/6×59×60×119+60×59÷2 =59×1190+59×30 =59×1220 =71980 希望帮到你 望采纳 谢谢 加油

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