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1. 化简下列各式 :

1. 化简下列各式 :. 1. 6. (2) 2(lg 2 ) 2 +lg 2 · lg5+ (lg 2 ) 2 - lg2+1 ;. (3) lg5(lg8+lg1000)+(lg2 ) 2 +lg +lg0.06. (2) 原式 =lg 2 (2lg 2 +lg5)+ (lg 2 - 1) 2. =lg 2 +1 - lg 2. =lg 2 (lg2+lg5)+(1 - lg 2 ). (1) (lg5) 2 +lg2 · lg50;. 解 : (1) 原式 =(lg5) 2 +lg2(lg2+2lg5).

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1. 化简下列各式 :

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  1. 1.化简下列各式: 1 6 (2) 2(lg 2)2+lg 2 ·lg5+ (lg 2)2-lg2+1 ; (3) lg5(lg8+lg1000)+(lg2 )2+lg +lg0.06. (2)原式=lg 2(2lg 2 +lg5)+ (lg 2-1)2 =lg 2+1-lg 2 =lg 2(lg2+lg5)+(1-lg 2) (1) (lg5)2+lg2·lg50; 解: (1)原式=(lg5)2+lg2(lg2+2lg5) =(lg5)2+(lg2)2+2lg2lg5 =(lg5+lg2)2 =1. =1. (3)原式=lg5(3lg2+3)+3lg22-lg6+lg6-2 =3lg5lg2+3lg5+3lg22-2 =3lg2(lg5+lg2)+3lg5-2 =3(lg2+lg5)-2 =1.

  2. 解:原式即为: loga[(x2+4)·(y2+1)]=loga[5(2xy-1)]. x ∴(x2+4)(y2+1)=5(2xy-1). 整理得x2y2+x2+4y2-10xy+9=0. 配方得(xy-3)2+(x-2y)2=0. x= 6, x=- 6, ∴ 或 xy-3=0, 6 6 y=, y=- , ∴ 2 2 x-2y=0, ∴ log8 =log8 =- . ∴ = . 1 1 1 2 3 2 y y x x 2.loga(x2+4)+loga(y2+1)=loga5+loga(2xy-1), 求 log8的值. y

  3. 8 3 x 3.已知lgx+lgy=2lg(x-2y), 求log 的值. y 解:注意到 lgab·lgba=1, 又已知 lgab+lgba=, 100 = -4= . 9 x x ∴ =4. ∴ log =log 4=4. y y 10 64 ∴lgab-lgba=- . 3 9 解: 由已知x>0, y>0, x-2y>0, ∴x>2y>0. ∵lgx+lgy=2lg(x-2y), ∴lg(xy)=lg(x-2y)2. ∴xy=(x-2y)2. ∴x2-5xy+4y2=0. ∴(x-y)(x-4y)=0. ∴x=y(舍去)或x=4y. 4.已知a>b>1, 且3lgab+3lgba=10, 求lgab-lgba的值. ∴(lgab-lgba)2=(lgab+lgba)2-4lgab·lgba ∵a>b>1, ∴lgab-lgba<0.

  4. 课堂互动讲练 5

  5. 课堂互动讲练

  6. 6(1)化简: (2)化简: (3)已知loga2=m,loga3=n,求a2m+n的值.

  7. 解 (1)原式= (2) (3)方法一 ∵loga2=m,∴am=2. ∵loga3=n,∴an=3. 故a2m+n=(am)2·an=4×3=12. 方法二 ∵loga2=m,loga3=n,

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