已知3( 5x 2) 5
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这俩式子相加得A-C=-x-1所以C-A=x+1
5x2-5x-3=7,根据等式的性质1,两边同时+3得:5x2-5x-3+3=7+3,即:5x2-5x=10,根据等式的性质2,两边同时除以5得:5x2−5x5=105,即:x2-x=2.
5x2-3x-5=0x=(3+-根109)/10x2-1/x=(+-6根109+118)/100-10/(3+-根109)=(37-+根109)/25再问:x=(3+-根109)/10是10分之3加根
∵A=5x2+4x-1,B=-x2-3x+3,C=8-7x-6x2,∴A-B+C=(5x2+4x-1)-(-x2-3x+3)+(8-7x-6x2)=5x2+4x-1+x2+3x-3+8-7x-6x2=
∵x2+4x-5=0,∴x2+4x=5,∴3x2+12x-5=3(x2+4x)-5=3×4-5=7.故答案为7.
5x2—3xy—4y2—11xy—7x2+2y2=-2x2-2y2-14xy=-2(x2+y2)-14xy=-14+28=14
x+y=7xy=27^2=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2*2x^2+y^2=45(1)y=7-xx(7-x)=2x^2-7x+2=0x=(7+41^(1/2))/2或x=(
∵x2+3x-1=0,∴x2+3x=1,x3+5x2+5x+18=x(x2+3x)+2x2+5x+18=x+2x2+5x+18=2(x2+3x)+18=2+18=20.
再问:你确定吗?再答:我保证再答:亲,那一步不明白可以问再问:你省略了一步吧!再答:能看明白吗再问:no再答: 再答:是不是这部再问:那么第一步是怎么得来的呢?再答:第一步不是你
根据韦达定理:x1+x2=-b/ax1*x2=c/a代入:x1+x2=-5/3x1*x2=-2/3即:x1+x2+x1*x2=(-5/3)+(-2/3)=-7/3
x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x
答:x²-5x=3(x-1)(2x-1)-x(x+3)+15/(x²-3)=2x²-3x+1-x²-3x+15/(5x)=x²-5x-x+1+3/x=
∵x^2+3x+5=7∴x^2+3x=23x^2+9x+2=3(x^2+3x)+2=3×2+2=8
(2mx2+5x2+3x+1)-(6x2-4y2+3x)=2mx2+5x2+3x+1-6x2+4y2-3x=(2m-1)x2+1+4y2不含x2的项,就是2m-1=0,m=2分之12m3-[3m3-(
1.=2y1-5y'2>=3y1+y'2>=-5y1无限制,y2>=02.
7x2-3xy-2y2-11xy-5x2+4y2=2x²-14xy+2y²=2(x²+y²)-14xy=2×7-14×(-2)=14+28=42
∵x2+y2=7,xy=-2,∴7x2-3xy-2y2-11xy-5x2+4y2,=(7x2-5x2)+(4y2-2y2)-(3xy+11xy),=2x2+2y2-14xy,=2(x2+y2)-14x
x²-2xy+y²/x²-y²=(x-y)²/(x-y)(x+y)=(x-y)/(x+y)因为x=3,y=-5,所以(3-(-5))/(3+(-5))
已知X1X2为方程5X平方-3X-1=0两个根;所以x1+x2=3/5;x1x2=-1/5;x1-x2=√(x1-x2)²=√[(x1+x2)²-4x1x2]=√(9/25+4/5
5x2-2x-15x2−2x−5=x+5-1x,∵5x2-3x=5,两边同除以5x得:x-1x=35,∴原式=x+5-1x=285.