=(25x4) x 895
= 100 x 895
= 89500

chịu

\(\left(-\dfrac{7}{23}\right).\left(\dfrac{13}{28}\right)+\left(-\dfrac{7}{23}\right).\left(\dfrac{25}{28}\right)\)

\(=\left(-\dfrac{7}{23}\right).\left(\dfrac{13}{28}+\dfrac{25}{28}\right)\)

\(=\left(-\dfrac{7}{23}\right).\left(\dfrac{38}{28}\right)\)

\(=\left(-\dfrac{7}{23}\right).\left(\dfrac{19}{14}\right)\)

\(=-\dfrac{19}{46}\)

\(\left(-\dfrac{7}{23}\right)\cdot\left(\dfrac{13}{28}\right)+\left(-\dfrac{7}{13}\right)\cdot\left(\dfrac{25}{28}\right)\)

\(=\left(-\dfrac{7}{23}\right)\cdot\left(\dfrac{13}{28}+\dfrac{25}{28}\right)\)

\(=\left(-\dfrac{7}{13}\right)\cdot\dfrac{38}{28}\)

\(=\left(-\dfrac{7}{13}\right)\cdot\dfrac{19}{14}\)

\(=-\dfrac{19}{26}\)

23.

Ta sẽ tìm điểm \(I\left(a;b;c\right)\) sao cho \(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}\) (1)

\(\left\{{}\begin{matrix}\overrightarrow{IA}=\left(-2-a;2-b;6-c\right)\\\overrightarrow{IB}=\left(-3-a;1-b;8-c\right)\\\overrightarrow{IC}=\left(-1-a;-b;7-c\right)\\\overrightarrow{ID}=\left(1-a;2-b;3-c\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\left(-5-4a;5-4b;24-4c\right)\)

(1) thỏa mãn khi: \(\left\{{}\begin{matrix}-5-4a=0\\5-4b=0\\24-4c=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{5}{4}\\b=\dfrac{5}{4}\\c=6\end{matrix}\right.\)

\(\Rightarrow I\left(-\dfrac{5}{4};\dfrac{5}{4};6\right)\)

Khi đó:

\(T=MA^2+MB^2+MC^2+MD^2=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IB}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IC}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{ID}\right)^2\)

\(=4MI^2+IA^2+IB^2+IC^2+ID^2+2\overrightarrow{MI}\left(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}\right)\)

\(=4MI^2+IA^2+IB^2+IC^2+ID^2\) (do \(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{0}\))

\(IA^2+IB^2+IC^2+ID^2\) cố định nên \(T_{min}\) khi \(MI_{min}\)

\(\Leftrightarrow M\) trùng I

\(\Rightarrow M\left(-\dfrac{5}{4};\dfrac{5}{4};6\right)\Rightarrow x+y+z=-\dfrac{5}{4}+\dfrac{5}{4}+6=6\)

24.

\(a+b=4\Rightarrow b=4-a\)

ABCD là hình chữ nhật \(\Rightarrow\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Rightarrow C\left(a;a;0\right)\)

Tương tự ta có: \(C'\left(a;a;b\right)\)

M là trung điểm CC' \(\Rightarrow M\left(a;a;\dfrac{b}{2}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{A'B}=\left(a;0;-b\right)=\left(a;0;a-4\right)\\\overrightarrow{A'D}=\left(0;a;-b\right)=\left(0;a;a-4\right)\\\overrightarrow{A'M}=\left(a;a;-\dfrac{b}{2}\right)=\left(a;a;\dfrac{a-4}{2}\right)\end{matrix}\right.\)

Theo công thức tích có hướng:

\(\left[\overrightarrow{A'B};\overrightarrow{A'D}\right]=\left(-a^2+4a;-a^2+4a;a^2\right)\)

\(\Rightarrow V=\dfrac{1}{6}\left|\left[\overrightarrow{A'B};\overrightarrow{A'D}\right].\overrightarrow{A'M}\right|=\dfrac{1}{6}\left|a\left(-a^2+4a\right)+a\left(-a^2+4a\right)+\dfrac{a^2\left(a-4\right)}{2}\right|\)

\(=\dfrac{1}{4}\left|a^3-4a^2\right|=\dfrac{1}{4}\left(4a^2-a^3\right)\)

Xét hàm \(f\left(a\right)=\dfrac{1}{4}\left(4a^2-a^3\right)\) trên \(\left(0;4\right)\)

\(f'\left(a\right)=\dfrac{1}{4}\left(8a-3a^2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\left(loại\right)\\a=\dfrac{8}{3}\end{matrix}\right.\)

\(\Rightarrow f\left(a\right)_{max}=f\left(\dfrac{8}{3}\right)=\dfrac{64}{27}\)

Hình vẽ bài 24:

a) 3,54 x 73 + 0,23 x25 +3,54 x 27 +3,54 x 27 +0,17 x 25
   = 258,42 + 5,75 + 95,58 + 95,58 + 4,25
   = 459,58
b) 563 x 23 +23 x 36 +23
   = 12949 + 828 + 23
   = 13800

a)Bị lỗi đề bài nhá,nếu là tính nhanh thì vậy,mình xin sửa và làm:

       \(3,54\times73+0,23\times25+3,54\times27+0,17\times25\)

\(=3,54\times\left(73+27\right)+25\times\left(0,17+0,23\right)\)

\(=3,54\times100+25\times0,5\)

\(=354+12,5\)

\(=366,5\)

b)

       \(563\times23+23\times36+23\)

\(=563\times23+23\times37\)

\(=23\times\left(563+37\right)\)

\(=23\times600\)

\(=13800\)

Cái phần lỗi đề bài đấy mai mình giải không sửa đề.

Không biết nó lỗi thật hay sự cố nhưng mà vẫn : CHÚC BẠN HỌC TỐT

Bài 2:

a. $=58(75+50-25)=58.100=5800$
b. $=27.8+4.9-5=216+36-5=247$

Bài 3:

a. 

$(x-38):16=12$

$x-38=16\times 12=192$
$x=192+38 = 230$
b.

$10+2x=4^5:4^3=4^2=16$

$2x=16-10=6$

$x=6:2=3$
c.

$440+2(125-x)=546$

$2(125-x)=546-440=106$

$125-x=106:2=53$

$x=125-53=72$

d.

$(x-15):5+20=22$
$(x-15):5=22-20=2$

$x-15=2\times 5=10$

$x=10+15=25$

e.

$2x-138=2^3.3^2=8.9=72$

$2x=72+138 = 210$

$x=210:2=105$

Bài 3:

f.

$2.3^x-5=7^2$

$2.3^x=7^2+5=54$
$3^x=54:2=27=3^3$

$\Rightarrow x=3$

g.

$7x-3^3=2^7:2^4$

$7x-27=2^3=8$

$7x=27+8=35$

$x=35:7=5$
h.

$427-(x+100)=227$

$x+100=427-227 = 200$

$x=200-100=100$

i.

$2x-138=2^3.3^2=8.9=72$
$2x=72+138 = 210$

$x=210:2=105$

k.

$(6x-18):3+25.2=78$

$(2x-6)+50=78$

$2x-6=78-50=28$

$2x=28+6=34$

$x=34:2=17$

mình sẽ tích cho người nhanh nhất

6352+6372=12724

6352 + 6372 = 12724

=9/13-13/23-x-7/69=-4/23-7/69-x=-19/69-x

2434+3325+6453+6352==18564

\(2434+3325+6453+6352=18564\)

2434+3325+6453+6352=18564

ai k mình mình k lại