1+2/1x=10/9

谁能帮下我把下面30道一元一次方程解了1.2(x-2)-3(4x-1)=9(1-x) 2.11x+64-2x=100-9x 3.15-(8-5x)=7x+(4-3x) 4.3(x-7)-2[9-4(2-x)]=22 5.3/2[2/3(1/4x-1)-2]-x=2 6.2(x-2)+2=x+1 7.0.4(x-0.2)+1.5=0.7x-0.38 8.30x-10(10-x)=100 9.4(x+2)=5(x-2) 10.120-4(x+5)=25 11.15x+863-65x=54 12.12.3(x-2)+1=x-(2x-1) //13.11x+64-2x=100-9x 14.14.59+x-25.31=0 15.x-48.32+78.51=80 16.820-16x=45.5×8 17.(x-6)×7=2x 18.3x+x=18 19.0.8+3.2=7.2 20.12.5-3x=6.5 21.1.2(x-0.64)=0.54 22.x+12.5=3.5x 23.8x-22.8=1.2 24.1\ 50x+10=60 25.2\ 60x-30=20 26.3\ 3^20x+50=110 27.4\ 2x=5x-3 28.5\ 90=10+x 29.6\ 90+20x=30 30.7\ 691+3x=700
2(x-2)-3(4x-1)=9(1-x) 2x-12x+9x=4-3+9-x=10x=-1011x+64-2x=100-9x 11x-2x+9x=-64+10018x=36x=215-(8-5x)=7x+(4-3x) 15-8+5x=7x+4-3x5x-7x+3x=-15+8+4x=-33(x-7)-2[9-4(2-x)]=223x-21-2-8x=223x-8x=22+21+2-5x=45x=-9 3/2[2/3(1/4x-1)-2]-x=21/4x-1/2-1/3-x=21/4x-x=2+1/2+1/33/4x=17/6x=34/9 2(x-2)+2=x+12x-4+2=x+12x-x=4-2+1x=30.4(x-0.2)+1.5=0.7x-0.380.4x-0.08+1.5=0.7x-0.380.4x-0.7x=0.08-1.5-0.38-0.3x=-4.5x=15 30x-10(10-x)=100 30x-100+10x=10030x+10x=100+10040x=200x=54(x+2)=5(x-2) 4x+8=5x-104x-5x=-8-10-x=-18x=18 120-4(x+5)=25 120-4x-20=25-4x=-120+20+25x=18.75 15x+863-65x=54 -50x=-809x=16.18 12.3(x-2)+1=x-(2x-1) 12.3x-24.6+1=x-2x+112.3x-x+2x=24.6-1+113.3x=24.6x=246/133 11x+64-2x=100-9x 11x-2x+9x=-64+10018x=36x=2 14.59+x-25.31=0 x=﹣14.59+25.31x=10.72 x-48.32+78.51=80 x=48.32+80-78.52x=49.8 820-16x=45.5×8 -16x= (x-6)×7=2x 3x+x=180.8+3.2=7.2 12.5-3x=6.5 1.2(x-0.64)=0.54 x+12.5=3.5x 8x-22.8=1.2 1\ 50x+10=60 2\ 60x-30=20 3\ 3^20x+50=1104\ 2x=5x-35\ 90=10+x6\ 90+20x=30 7\ 691+3x=700
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要过程不?
扫描下载二维码已知f(x)=a^x/(a^x+√a)(a>0),求f(1/10)+f(2/10)+……+f(9/10)的值
1/10+9/10=1所以想到求f(x)+f(1-x)的关系f(1-x)=a^(1-x)/[a^(1-x)+√a]上下乘a^x,a^x*a^(1-x)=a所以f(1-x)=a/(a+a^x*√a)上下除以√af(1-x)=√a/(√a+a^x)所以f(x)+f(1-x)=(√a+a^x)/(√a+a^x)=1所以f(1/10)+f(9/10)=1……f(4/10)+f(6/10)=1f(5/10)+f(5/10)=1,所以f(5/10)=1/2所以原式=1+1+1+1+1/2=9/2
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根据经验用逆序相加可以得出结果 f(x)+f(1-x)=1所以原式为4.5
扫描下载二维码这是个机器人猖狂的时代,请输一下验证码,证明咱是正常人~9(x的平方+1)=10怎么算
9(x的平方+1)=109x²+9=109x²-1=0(3x+1)(3x-1)=0x1=-1/3,x2=1/3
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9(x的平方+1)=10x^2+1=10/9x^2=10/9-1x^2=1/9x=±1/3
扫描下载二维码求证:2/x2-1+4/x2-4+6/x2-9+…+20/x2-100=11/(x-1)(x+10)+…+11/(x-10)(x+1)
先证明了左右两边首尾两项的等量关系:2/(x^2-1)+20/(x^2-100)=(22x^-220)/[(x^2-1)(x^2-100)]
=(22x^-220)/[(x+1)(x-1)(x+10)(x-10)]
=22(x^-10)/[(x+1)(x-1)(x+10)(x-10)]11/(x-1)(x+10)+11/(x-10)(x+1)=11(2x^2+20)/[(x+1)(x-1)(x+10)(x-10)]
=22(x^2+10)/[(x+1)(x-1)(x+10)(x-10)]
即2/(x^2-1)+20/(x^2-100)=11/(x-1)(x+10)+11/(x-10)(x+1)同理4/(x^2-4)+18/(x^2-81)=11/(x-2)(x+9)+11/(x-9)(x+2)……所以:2/(x2-1)+4/(x2-4)+6/(x2-9)+…+20/(x2-100)=11/(x-1)(x+10)+…+11/(x-10)(x+1)
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