3.

Do M là trung điểm BC \(\Rightarrow\overrightarrow{CM}=\dfrac{1}{2}\overrightarrow{CB}\)

N là trung điểm AC \(\Rightarrow\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AC}\)

K là trung điểm AB \(\Rightarrow\overrightarrow{BK}=\dfrac{1}{2}\overrightarrow{BA}\)

Do đó:

\(\overrightarrow{AN}+\overrightarrow{CM}-\overrightarrow{KB}=\overrightarrow{AN}+\overrightarrow{CM}+\overrightarrow{BK}=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}\)

\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BA}=\overrightarrow{0}\)

4.

\(\overrightarrow{BC}=\left(6;-2\right)\)

Gọi \(A'\left(x;y\right)\Rightarrow\overrightarrow{BA'}=\left(x+3;y-1\right)\)

Do A' thuộc BC \(\Rightarrow\overrightarrow{BA'}\) và \(\overrightarrow{BC}\) cùng phương

\(\Rightarrow\dfrac{x+3}{6}=\dfrac{y-1}{-2}\Rightarrow x=-3y\)

\(\Rightarrow A'\left(-3y;y\right)\Rightarrow\overrightarrow{AA'}=\left(-3y-2;y-4\right)\)

Mà AA' vuông góc BC \(\Rightarrow\overrightarrow{AA'}.\overrightarrow{BC}=0\)

\(\Rightarrow6\left(-3y-2\right)-2\left(y-4\right)=0\Rightarrow y=-\dfrac{1}{5}\)

\(\Rightarrow A'\left(\dfrac{3}{5};-\dfrac{1}{5}\right)\)

250:25+25.10

=10+250

=260

250:25+25x10
=10+250
=260

Trả lời :
250 : 25 + 25 x 10

= 10 + 250

= 260

~^Hok tốt^~

=4(x²+2x+1)=4(x+1)²

4x² + 8x + 4

= 4 ( x² + 2x + 1 )

= 4 ( x + 1 )²

=4(x+1)²

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

Bài nào em nhỉ?

\(d\left(G;\left(ABCD\right)\right)=\dfrac{1}{3}d\left(S;\left(ABCD\right)\right)=\dfrac{1}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{6}\)

\(S_{\Delta ACD}=\dfrac{1}{2}S_{ABCD}=\dfrac{a^2}{2}\)

\(\Rightarrow V=\dfrac{1}{3}.\dfrac{a^2}{2}.\dfrac{a\sqrt{3}}{6}=\dfrac{a^3\sqrt{3}}{36}\)

23.17-23.14

=23.(17-14)

=23.3 = 69

23.(17-14)=23.3=69

\(\frac{7}{2}\)-\(|\frac{2x-3}{4}\)\(|\)=\(\frac{-7}{4}\)

\(|\frac{2x-3}{4}|\)=\(\frac{7}{2}\)+\(\frac{7}{4}\)

\(|\frac{2x-3}{4}|\)=\(\frac{21}{4}\)

=>\(|2x-3|\)=\(21\)

=>2x-3=21                  hay                                              2x-3=-21

2x=21+3                                                                           2x=-21+3

2x=24                                                                                 2x=-18

x=12                                                                                         x=-9

\(=\left[\dfrac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\cdot\left(\sqrt{7}-\sqrt{5}\right)\\ =\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=7-5=2\)