專題小卷（10）解一元一次方程的技巧類型一分?jǐn)?shù)系數(shù)方程轉(zhuǎn)化為整數(shù)系數(shù)方程1.(2024廣東梅州期末,18,★★☆)解方程:

+

=1.去分母,得3(x-1)+2(2x+1)=6,去括號(hào),得3x-3+4x+2=6,移項(xiàng)、合并同類項(xiàng),得7x=7,系數(shù)化為1,得x=1.2.(★★☆)解方程:

(y+2)=1-

(2y-1).去分母,得2(y+2)=6-(2y-1),去括號(hào),得2y+4=6-2y+1,移項(xiàng),得2y+2y=6+1-4,合并同類項(xiàng),得4y=3,系數(shù)化為1,得y=

.3.(2024甘肅武威涼州期末,20(2),★★☆)解方程:

=

-

.去分母,得4(5y+1)=3(9y+1)-8(1-y),去括號(hào),得20y+4=27y+3-8+8y,移項(xiàng)、合并同類項(xiàng),得-15y=-9,系數(shù)化為1,得y=

.4.(★★☆)解方程:x+

=3-

.去分母,得36x+3(x-20)=108-4(7-x),去括號(hào),得36x+3x-60=108-28+4x,移項(xiàng),得36x+3x-4x=108-28+60,合并同類項(xiàng),得35x=140,系數(shù)化為1,得x=4.5.(2024內(nèi)蒙古呼和浩特期末,19(4),★★☆)解方程:

-

=1-

.去分母,得10(3x+2)-5(2x-1)=20-4(2x+1),去括號(hào),得30x+20-10x+5=20-8x-4,移項(xiàng),得30x-10x+8x=20-20-4-5,合并同類項(xiàng),得28x=-9,系數(shù)化為1,得x=-

.類型二靈活去掉多重括號(hào)解方程6.(★★☆)解方程:4x-2[x-5(x-1)-4]=1.去小括號(hào),得4x-2(x-5x+5-4)=1,去括號(hào),得4x-2x+10x-10+8=1,移項(xiàng),得4x-2x+10x=10-8+1,合并同類項(xiàng),得12x=3,系數(shù)化為1,得x=

.7.(★★★)解方程:

=1.方程兩邊同乘7,得

+6=7,移項(xiàng)、合并同類項(xiàng),得

=1,方程兩邊同乘5,得

+4=5,移項(xiàng)、合并同類項(xiàng),得

=1,方程兩邊同乘3,得x+2=3,所以x=1.8.(★★★)解方程:

+x=4.去中括號(hào),得(3x+15)-6+x=4,去括號(hào),得3x+15-6+x=4,移項(xiàng),得3x+x=4-15+6,合并同類項(xiàng),得4x=-5,系數(shù)化為1,得x=-

.9.(★★★)在解一元一次方程時(shí),巧妙利用整體法,可以達(dá)到簡(jiǎn)化計(jì)算的效果.例如,在解方程3{2x-

1-[3(2x-1)+3]}=5時(shí),把2x-1看成一個(gè)整體.令a=2x-1,得3[a-(3a+3)]=5,去括號(hào),得3a-9a-9=5,移項(xiàng)、合并同類項(xiàng),得-6a=14,系數(shù)化為1,得a=-

,故2x-1=-

,解得x=-

.根據(jù)以上材料,請(qǐng)用同樣的方法解方程:4

=1.令a=2x+4,得4

=1,去括號(hào),得4a-2a-20=1,移項(xiàng),得4a-2a=1+20,合并同類項(xiàng),得2a=21,系數(shù)化為1,得a=

,故2x+4=

,解得x=

.類型三靈活求解分子或分母中含有小數(shù)的方程10.(★★☆)解方程:

=

.方程整理得

=

.去分母,得3(3x+5)=2(2x-1),去括號(hào),得9x+15=4x-2,移項(xiàng),得9x-4x=-2-15,合并同類項(xiàng),得5x=-17,系數(shù)化為1,得x=-

.11.(★★☆)解方程:

-2=

.去分母,得1.2x+9-1.2=0.9-2x,移項(xiàng)、合并同類項(xiàng),得3.2x=-6.9,解得x=-

.12.(★★☆)解方程:

-

=5.方程整理得

-

=5,分子、分母約分,得4x+4-(2x-4)=5,去括號(hào)、移項(xiàng),得4x-2x=5-4-4,合并同類項(xiàng),得2x=-3,系數(shù)化為1,得x=-

.13.(★★☆)解方程:2x-

=

.原方程可變?yōu)?x-

=10+2x,即-

=10,去分母,得-10+2x=40,解得x=25.14.(2024四川綿陽安州期末,20