Này thì khó :))Ta có công thức:$C^{n-k}_n=C^k_n$Nên $VT=C^1_x+C^2_x+C^3_x+...+C^{10}_x (*)$Ta xét khai triển: $(1+1)^x=C^0_x+C^1_x+C^2_x+...+C^{10}_x=2^x\Rightarrow (*)=2^x-C^0_x$ĐK: $n\in N^*$Pt $\Leftrightarrow C^1_x+C^2_x+C^3_x+...+C^{10}_x=1023$$\Leftrightarrow 2^x-C^0_x=1023$$\Leftrightarrow 2^x=1024$$\Leftrightarrow 2^x=2^{10}$$\Leftrightarrow x=10$
Pt $\Left
righ
tarrow C^1_x+C^2_x+C^3_x+...+C^{10}_x
=1023$$
\Left
ri
ght
arr
ow C^0_x+C^1_x+C^2_x+...+C^{10}_x=
10
24$$
<
;=>
;C^
0_x+C^
1_x+C^
2_x+...+C^{10}_x=
2^{10
}$
mà $C^0_x
+C^
1_x
+...+C^
x_x=2^
x$$
=>
;x=10$