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解方程(1)x%2/4%2x%3/6=1 (2)x+4/5%(x%5)=x+3/3%...

原式得 6*(x+2)-4*(2x-3)=24 展开 6x+12-4X+12=24 展开应该得到的是6X+12-8X+12=24 是吧! 移项 2X=0 x=0 你的原题应该为(x+2)/2-(x-5)=(x+3)/3-(x-2)/2吧 (x+2)/2-(x-5)=(x+3)/3-(x-2)/2 方程两边同时乘以6 3(x+2)-6(x-5)=2(x+3)-3(x-2) 3x+6-6...

(x+3)/(x+2)+(x+4)/(x+3)=(x+5)/(x+4)+(x+2)/(x+1) 1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1) x+3+x+2/(x+2)(x+3)=x+1+x+4/(x+4)(x+1) 2x+5/(x+2)(x+3)=2x+5/(x+4)(x+1) (1)2x+5不为0 (x+2)(x+3)=(x+4)(x+1) x^2+5x+...

解:根据比例的基本性质解比例: 2/5:(1+X)=4/5:6 4/5(1+x)=2/5*6 1+x=12/5÷4/5 1+x=3 x=3-1 x=2

(x-1/x-2)-(x-2/x-3)=(x-4/x-5)-(x-5/x-6) (x-1/x-2)-(x-4/x-5)=(x-2/x-3)-(x-5/x-6) [(x-1)(x-5)-(x-4)(x-2)]/[(x-2)(x-5)]=[(x-2)(x-6)-(x-5)(x-3)]/[(x-3)(x-6)] -3/[(x-2)(x-5)]=-3/[(x-3)(x-6)] (x-2)(x-5)=(x-3)(x-6) x²-7x+10=x...

int main() { int x=0; while(1){ if(x%2==1&&x%3==2&&x%5==4&&x%6==5&&x%7==0) { printf("%d",x);break; } else x++; } } 楼上错误,少了模7为0,剩多少就是余数是多少的意思,调试通过,结果为119

5-6(5/6x+1/2)=2(X-3/4) 解:去括号,得 5-5x+3=2x-3/2 移项,得 -5x-2x=-3/2-5-3 合并,得 -7x=-19/2 系数化为1,得 x=19/14 5-6(5/6×1/2)=2(X-3/4) 5-5/2=2x-3/2 2x=4 x=2

2(x+1)/3=5(x+1)/6-1 2[2(x+1)]=5(x+1)-6 4x+4=5x+5-6 4-5+6=5x-4x 5=x x=5

1/(x+2)-1/(x+3)-1/(x+4)+1/(x+5)=0 1/(x+2)-1/(x+3)=1/(x+4)-1/(x+5) 1/(x+2)(x+3)=1/(x+4)(x+5) (x+2)(x+3)=(x+4)(x+5) x²+5x+6=x²+9x+20 4x=-14 x=-3.5

说错了。不是方程,更不能是恒等式。

1+1/(x-1)+1+1/(x-4)=1+1/(x-2)+1+1/(x-3) 1/(x-1)+1/(x-4)=1/(x-2)+1/(x-3) (2x-5)/(x-1)(x-4)=(2x-5)/(x-2)(x-3) 因此2x-5=0, 得:x=5/2 或(x-1)(x-4)=(x-2)(x-3)=0, 得:x^2-5x+4=x^2-5x+6, 不符。 故只有一解为x=5/2

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