the boy's father bought him a large toy train__.A.to play with B.to play with itC.which to play with D.at which to play 那个“it”什么时候加什么时候不加?类似的题很多,求解此语法,

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the boy's father bought him a large toy train__.A.to play with B.to play with itC.which to play with D.at which to play 那个“it”什么时候加什么时候不加?类似的题很多,求解此语法,

the boy's father bought him a large toy train__.A.to play with B.to play with itC.which to play with D.at which to play 那个“it”什么时候加什么时候不加?类似的题很多,求解此语法,
the boy's father bought him a large toy train__.A.to play with B.to play with it
C.which to play with D.at which to play
那个“it”什么时候加什么时候不加?类似的题很多,求解此语法,

the boy's father bought him a large toy train__.A.to play with B.to play with itC.which to play with D.at which to play 那个“it”什么时候加什么时候不加?类似的题很多,求解此语法,
这个是非谓语短语作后置定语,修饰a large toy train,
首先,咱先看看a large toy train 非谓语中做什么,
显然,作宾语,
to do 用作主动形式表被动意思,是说火车玩具被玩.
所以选择A.
对于B,如果有了 it ,那么非谓语中啥都不缺了（不考虑逻辑主语阿.）
火车玩具咱就不知道放哪了不是?所以不能选.
对于加不加it ,主要就是看被修饰词在非谓语中作什么成份,
莫要让it 把 修饰词在非谓语中的位置挤掉,
这种找位置的方法在定语从句也管用.
eg.the boy's father bought him a large toy train ——
A .which the boy can play with B .which the boy can play with it
a large toy train 依然在定于从句中作介词with的宾语,不能让it 挤掉它,所以选A

整个句子中没有提到it所指的东西，之前的句子有，则使用
反之一般不用
除了要填入的部分，句子成分已经完整。整个句子的结构是：
sb.1+brought+（sb.2+sth.）双宾语+宾语补足语
A正确。
选C的话，相当于跟一个定语从句，修饰a large toy train，但是从句中少了谓语动词，应为which is used to play with，即...

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整个句子中没有提到it所指的东西，之前的句子有，则使用
反之一般不用
除了要填入的部分，句子成分已经完整。整个句子的结构是：
sb.1+brought+（sb.2+sth.）双宾语+宾语补足语
A正确。
选C的话，相当于跟一个定语从句，修饰a large toy train，但是从句中少了谓语动词，应为which is used to play with，即使这样用了也觉得不太必要（谁都知道，大的玩具火车是用来玩的）。
选B的话，动词不定式作定语不用跟宾语（it）。
选D的话，首先介词错误，玩某物介词应该用play with sth.，其次犯的毛病与A相同，定语从句缺少谓语动词，且此处又不必要用定语从句描述。

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.which to play with D.at which to play

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