x=24 nên x+1=25

Sửa đề: \(f\left(x\right)=x^{50}-25x^{49}+25x^{48}-...+25x^2-25x+18\)

\(=x^{50}-x^{49}\left(x+1\right)+x^{48}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+18\)

\(=x^{50}-x^{50}-x^{49}+x^{49}+...+x^3+x^2-x^2-x+18\)

=-x+18=-24+18=-6

Đoạn cuối là \(+25x^2+25x+18\) hay \(+25x^2-25x+18\) em?

\(x^2+4y^2-2xy+2x-14y+9=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+2\left(x-y\right)+3y^2-12y+12-3=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x-y\right)+1+3\left(y-2\right)^2-4=0\)

\(\Leftrightarrow\left(x-y+1\right)^2+3\left(y-2\right)^2=4\) (1)

Do \(\left(x-y+1\right)^2\ge0;\forall x;y\)

\(\Rightarrow3\left(y-2\right)^2\le4\)

\(\Rightarrow\left(y-2\right)^2\le\dfrac{4}{3}\Rightarrow\left[{}\begin{matrix}\left(y-2\right)^2=0\\\left(y-2\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=2\\y=3\\y=1\end{matrix}\right.\)

Thế vào (1): 

Với \(y=1\) \(\Rightarrow x^2=1\Rightarrow x=\pm1\)

Với \(y=2\Rightarrow\left(x-1\right)^2=4\Rightarrow x=\left\{3;-1\right\}\)

Với \(y=3\Rightarrow\left(x-2\right)^2=1\Rightarrow x=\left\{3;1\right\}\)

Vậy \(\left(x;y\right)=\left(-1;1\right);\left(1;1\right);\left(-1;2\right);\left(3;2\right);\left(1;3\right);\left(3;3\right)\)

\(8⋮3n+1\)

=>\(3n+1\in\left\{1;2;4;8\right\}\)

=>\(3n\in\left\{0;1;3;7\right\}\)

=>\(n\in\left\{0;\dfrac{1}{3};1;\dfrac{7}{3}\right\}\)

mà n là số tự nhiên

nên \(n\in\left\{0;1\right\}\)

FFbh

\(\overline{93ab}⋮2;\overline{93ab}⋮5\)

=>b=0

=>Số cần tìm có dạng là \(\overline{93a0}\)

\(\overline{93a0}⋮3\)

=>\(9+3+a+0⋮3\)

=>\(a+12⋮3\)

=>\(a\in\left\{0;3;6;9\right\}\)

\(E=2015\cdot2017=\left(2016-1\right)\left(2016+1\right)\)

\(=2016^2-1=F-1\)

=>E<F

\(C=123\cdot137137=123\cdot137\cdot1001\)

\(D=137\cdot123123=137\cdot123\cdot1001\)

Do đó: C=D

\(C=123.137137=123.137.1001\)

\(D=137.123123=137.123.1001\)

\(\Rightarrow C=D\)

b.

Em kiểm tra lại đề, \(F=1016.2016\) hay \(F=2016.2016?\)

286:(38-2x)=13

=>38-2x=286:13=22

=>2x=38-22=16

=>x=16:2=8

\(286:\left(38-2x\right)=13\)

\(38-2x=286:13\)

\(38-2x=22\)

\(2x=38-22\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

Ta có: 

A = 123 x 123

= (121 + 2) x 123

= 121 x 123 + 2 x 123 

= 121 x (124 - 1) + 246

= 121 x 124 - 121 + 246  

= 121 x 124 + 125 > 121 x 124

=> A > B 

\(B=121\cdot124=\left(123-2\right)\left(123+1\right)\)

\(=123^2+123-2\cdot123-2\)

\(=123^2-123-2=A-125\)
=>B<A

\(187:\left(2x-1\right)=11\\ =>2x-1=\dfrac{187}{11}\\ =>2x-1=17\\ =>2x=17+1\\ =>2x=18\\ =>x=\dfrac{18}{2}\\ =>x=9\)

\(187:\left(2x-1\right)=11\)

\(2x-1=187:11\)

\(2x-1=17\)

\(2x=1+17\)

\(2x=18\)

\(x=18:2\)

\(x=9\)

\(0< a< 2\Rightarrow a\left(a-2\right)< 0\Rightarrow a^2< 2a\)

Tương tự: \(\left\{{}\begin{matrix}b\left(b-2\right)< 0\\c\left(c-2\right)< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b^2< 2b\\c^2< 2c\end{matrix}\right.\)

\(\Rightarrow a^2+b^2+c^2< 2\left(a+b+c\right)\)

\(\Rightarrow a^2+b^2+c^2< 2.3=6\)

a: Xét ΔAEB vuông tại E và ΔAFC vuông tại F có

\(\widehat{EAB}\) chung

Do đó: ΔAEB~ΔAFC

=>\(\dfrac{AE}{AF}=\dfrac{AB}{AC}\)

=>\(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

=>\(AE\cdot AC=AF\cdot AB\)

b: Xét ΔADB vuông tại D có DM là đường cao

nên \(AM\cdot AB=AD^2\left(1\right)\)

Xét ΔADC vuông tại D có DN là đường cao

nên \(AN\cdot AC=AD^2\left(2\right)\)

Từ (1),(2) suy ra \(AM\cdot AB=AN\cdot AC\)

=>\(\dfrac{AB}{AC}=\dfrac{AN}{AM}\)

=>\(\dfrac{AN}{AM}=\dfrac{AE}{AF}\)

=>\(\dfrac{AE}{AN}=\dfrac{AF}{AM}\)

=>\(AN\cdot AF=AM\cdot AE\)

c: Xét ΔANM có \(\dfrac{AE}{AN}=\dfrac{AF}{AM}\)

nên EF//MN