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1/√4%9x²Dx

解: ∫(1-x)/√(4-9x²)dx =∫1/√(4-9x²)-∫x/√(4-9x²)dx =1/3 ∫1/√[1-(3x/2)²] d(3x/2)+1/18 ∫1/√(4-9x²) d(4-9x²) =1/3·arcsinx+1/18·2/3·(4-9x²)^(3/2)+C =1/3·arcsinx+1/27·(4-9x²)^(3/2)+C

∫ (1 - x)/√(4 - 9x²) dx Let x = (2/3)sinz,dx = (2/3)cosz dz √(4 - 9x²) = √(4 - 4sin²z) = 2cosz => ∫ (1 - (2/3)sinz)/(2cosz) * (2/3)cosz dz = (1/3)∫ (1 - (2/3)sinz) dz = (1/3)∫ dz - (2/9)∫ sinz dz = (1/3)z - (2/9)...

你是不是要分解因式 9x+1/4+81x²=1/4(81×4x²+9×4x+1) =1/4[(18x)²+2×18x+1] =1/4(18x+1)²

9x² + 16y² = 1 x²/(1/3)² + y²/(1/4)² = 1 长半轴:a = 1/3 短半轴:b=1/4 c=√(a²-b²) = √(1/9-1/16) = √7/12 焦点:F1(-√7/12,0),F2(√7/12,0) 离心率:e = c/a = √7/4 顶点: (-1/3,0),(...

2(3x-1)=3(3x+1)或者2(3X-1)=-3(3x+1) 6x-2=9x+3或者6x-2=-9x-3 x=-5\9或者x=-1/15

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