\(1,\Rightarrow\overline{7ab}=36\cdot\overline{ab}\Rightarrow\overline{ab}+700=36\cdot\overline{ab}\\ \Rightarrow35\cdot\overline{ab}=700\Rightarrow\overline{ab}=20\\ 2,\Rightarrow\overline{30abc}=151\cdot\overline{abc}\Rightarrow\overline{abc}+30000=151\cdot\overline{abc}\\ \Rightarrow\overline{abc}\cdot150=30000\Rightarrow\overline{abc}=200\\ 3,\Rightarrow\overline{ab5}=10\cdot\overline{ab}+5=554+\overline{ab}\\ \Rightarrow\overline{ab}\cdot9=549\Rightarrow\overline{ab}=61\\ 4,\Rightarrow\overline{ab78}=\overline{ab}\cdot100+78=1266+\overline{ab}\\ \Rightarrow\overline{ab}\cdot99=1188\Rightarrow\overline{ab}=12\\ 5,\Rightarrow\overline{abc85}=\overline{abc}\cdot100+85=43348+\overline{abc}\\ \Rightarrow\overline{abc}\cdot99=43263\Rightarrow\overline{abc}=437\\ 6,7,\text{Đề thiếu}\)

\(a,\left\{{}\begin{matrix}AB=AC\\BH=HC\\AH\text{ chung}\end{matrix}\right.\Rightarrow\Delta AHB=\Delta AHC\left(c.c.c\right)\\ \Rightarrow\widehat{BAH}=\widehat{CAH}\\ \Rightarrow AH\text{ là p/g }\widehat{BAC}\\ \Delta AHB=\Delta AHC\Rightarrow\widehat{AHB}=\widehat{AHC}\\ \text{Mà }\widehat{AHC}+\widehat{AHB}=180^0\\ \Rightarrow\widehat{AHC}=\widehat{AHB}=90^0\\ \Rightarrow AH\bot BC\)

\(b,\left\{{}\begin{matrix}HK=HA\\BH=HC\\\widehat{AHB}=\widehat{KHC}\left(\text{đối đỉnh}\right)\end{matrix}\right.\Rightarrow\Delta AHB=\Delta KHC\left(c.g.c\right)\\ \Rightarrow\widehat{ABH}=\widehat{HCK}\\ \text{Mà 2 góc này ở vị trí so le trong nên }CK\text{//}AB\\ c,\text{Đề lỗi}\)

\(a,=3x^2-6x\\ b,=4x^2+x^3-x^3+x^2=5x^2\\ c,=x^2-4x+3x-12=x^2-x-12\\ d,=2x^2+2x-5x-5-2x^2+x=-2x-5\\ e,=x^2+12x+36-x^2=12x+36\\ f,=4a^2-8a+4-4=4a^2+8a\\ g,=9b^2-1+1=9b^2\\ h,=x^3+6x^2+12x+8\\ i,=y^3-9y^2+27y-27\)

Trung bình mỗi xe chở \(\left(5\times2,6+4\times3,5\right):\left(4+5\right)=27:9=3\left(\text{tấn gạo}\right)=30\left(\text{tạ gạo}\right)\)

lạc đề rồi

Xét tam giác ABC: \(y=\widehat{C}=180^0-\widehat{A}-\widehat{B}=70^0\)

Theo t/c góc ngoài: \(x=\widehat{C}+\widehat{B}=120^0\)

\(=\left|2-\sqrt{3}\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)

hảo :D