规定一种新运算符号*的运算过程为a*b=1\3a-1\4b求解方程2*(2*x)=1*x和5*(-5)
a*b=1/3 a-1/4 b2*(2*x)=1*x
解
2*(2/3-x/4)=1/3-x/4
2/3-1/4(2/3-x/4)=1/3-x/4
1/2+x/16=1/3-x/4
5/16 x=-1/6
解得x= -8/15
5*(-5)=1/3×5-1/4×(-5)=5/3+5/4=35/12
绛旓細瑙勫畾鏂拌繍绠楃鍙*鐨勮繍绠楄繃绋涓篴*b=1/3a-1/4銆 (1)姹5*(-5)鐨勫; (2)瑙f柟绋: 瑙勫畾鏂拌繍绠楃鍙*鐨勮繍绠楄繃绋嬩负a*b=1/3a-1/4銆(1)姹5*(-5)鐨勫;(2)瑙f柟绋:2*(2*x)=1*銆... 瑙勫畾鏂拌繍绠楃鍙*鐨勮繍绠楄繃绋嬩负a*b=1/3a-1/4銆 (1)姹5*(-5)鐨勫; (2)瑙f柟绋:2*(2*x)=1*銆
绛旓細锛1锛2鈥(-3)=锛3*2-1锛/銆2*锛-3锛+1銆=-1 锛2锛1鈥伙紙2鈥粁锛=4 2鈥粁=5/(2x+1)1鈥伙紙2鈥粁锛=2/[2*5/(2x+1)+1]=4瑙e緱x=-21/2
绛旓細瑙勫畾涓绉嶆柊鐨勮繍绠楃鍙鈥*",瀹氫箟a*b=a-b/ab锛岃鏍规嵁绗﹀彿鈥*鈥濈殑鎰忎箟瑙g瓟銆傦紙1锛4*2 =(4-2)/4x2 =1/4 锛2锛夛紙2*3)*(-2)=(2-3)/2x3*锛-2锛=-1/6*(-2)=(-1/6+2)/(1/6x2)=11/2
绛旓細鏅,鎶5,-5寰閲岄潰甯,5*锛-5锛=1/(3x5)-1/[4x(-5)]=1/15+1/20=7/60
绛旓細1/锛3脳2锛-1/锛4脳x锛+1/锛3脳1锛-1/锛4脳x锛=0 甯告暟鍔犱竴璧锋湭鐭ユ暟鍦ㄤ簡鍙︿竴杈 1/2=1/2x x=1
绛旓細锛1锛5*锛-5锛=13脳5-14脳锛-5锛=53+54=3512锛涳紙2锛夆埖2*x=23-x4锛屸埓2*锛23-x4锛=23-14锛23-x4锛=23-16+x16=12+x161*x=13-x4锛屸埓12+x16=13-x4锛屽幓鍒嗘瘝寰楋紝24+3x=16-12x锛岀Щ椤瑰緱锛15x=-8锛岀郴鏁板寲涓1寰楋紝x=-815锛
绛旓細鍙樺舰 1/锛3*2锛-1/锛4x锛+1/锛3*1锛-1/锛4x锛=0 1/6+1/3-1/锛2x锛=0 1/2=1/锛2x锛夊緱x=1
绛旓細2*(2*X)=1*X;2*(2/3-X/4)=1/3-X/4;2/3-(1/4)x(2/3-X/4)=1/3-X/4;2/3-(1/6-X/16)=1/3-X/4;X= -8/15.
绛旓細2*x=2/3-x/4 鎵浠2脳1/3-(2/3-x/4)/4=1脳1/3+x/4 2/3-1/6+x/16=1/3+x/4 x/4-x/16=2/3-1/6-1/3 3x/16=1/6 x=1/6梅3/16 x=8/9
绛旓細濡傛灉a*b=7/[12(a-b)]1瑙o細=7/12(5+5)=7/120 2瑙:=8x[7/12(2-x)]+7/12(3-x)=[56x^2+161x+14]/[12(2-x)(3-x)]濡傛灉a*b=[7(a-b)]/12 1瑙o細=7(5+5)/12=70/12=35/6 2瑙:=8x[7(2-x)/12]+7(3-x)/12 ...