Đề chưa rõ ràng, em nhé.

ab+2023a+2023

\(=ab+a\cdot abc+abc=ab+a^2bc+abc=ab\left(1+ac+c\right)\)

bc+b+2023=bc+b+abc

=b(c+1+ac)

\(2023a^2bc=abc\cdot a^2bc=a^3b^2c^2\)

\(\frac{2023a^2bc}{ab+2023a+2023}\)

\(=\frac{a^3b^2c^2}{ab\left(1+ac+c\right)}=\frac{a^2bc^2}{1+ac+c}\)

\(\frac{ab^2c}{bc+b+2023}=\frac{ab^2c}{b\cdot\left(c+1+ac\right)}=\frac{abc}{c+ac+1}\)

Ta có: \(P=\frac{2023a^2bc}{ab+2023a+2023}+\frac{ab^2c}{bc+b+2023}+\frac{abc^2}{ac+c+1}\)

\(=\frac{a^2bc^2}{1+ac+c}+\frac{abc}{ac+c+1}+\frac{abc^2}{1+ac+c}=\frac{abc\left(ac+1+c\right)}{1+ac+c}\)

=abc

=2023

\(5^{10}\cdot5^{12}\\ =5^{10+12}\\ =5^{22}\)

5^10 . 5^12

=5^10+12

=5^22

a) Số lượng số hạng là: 

(30 - 20) : 1 + 1 = 11 (số hạng)

Tổng của dãy số là: (30 + 20) x 11 : 2 = 275 

b) 2.10 + 2.12 + 2.14 + 2.16 + 2.18 + 2.20 

= 2.(10 + 12 + 14 + 16 + 18 + 20) 

= 2. 90 

= 180 

c) 22 + 24 + 26 + 28 + 30 

= (22 + 28) + (24 + 26) + 30

= 50 + 50 + 30 

= 130 

a: Ta có: \(AE=EB=\dfrac{AB}{2}\)

\(AD=DC=\dfrac{AC}{2}\)

mà AB=AC
nên AE=EB=AD=DC

b: Xét ΔABD và ΔACE có

AB=AC

\(\widehat{BAD}\) chung

AD=AE

Do đó: ΔABD=ΔACE

c: Xét ΔABC có \(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)

nên DE//BC

d: Xét tứ giác BEDC có ED//BC

nên BEDC là hình thang 

Xét hình thang BEDC có BD=EC(ΔABD=ΔACE)

nên BEDC là hình thang cân

\(x\left(2x+\dfrac{-4}{10}\right)\) = 0

\(\left[{}\begin{matrix}x=0\\2x-\dfrac{4}{10}=10\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\2x=\dfrac{4}{10}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{4}{10}:2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {0; \(\dfrac{1}{5}\)}

\(x\left(2x+\dfrac{-4}{10}\right)=0\\ =>x\left(2x+\dfrac{-2}{5}\right)=0\\ =>2x\left(x-\dfrac{1}{5}\right)=0\\ TH1:2x=0\\ =>x=0\\ TH2:x-\dfrac{1}{5}=0\\ =>x=\dfrac{1}{5}\)

\(\dfrac{x}{2}=\dfrac{y}{3}=>\dfrac{x}{8}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{21}\\ =>\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{21}=\dfrac{2x-y+z}{2\cdot8-12+21}=\dfrac{50}{25}=2\\ =>\dfrac{x}{8}=2=>x=2\cdot8=16\\ =>\dfrac{z}{12}=2=>z=2\cdot12=24\\ =>\dfrac{z}{21}=2=>z=2\cdot21=42\)

 tôi thấy nó raats jnjdnjnnv

\(\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{\dfrac{9}{1001}-\dfrac{9}{13}+\dfrac{9}{17}-\dfrac{9}{11}+9}\\ =\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{9+\dfrac{9}{17}-\dfrac{9}{11}+\dfrac{9}{1001}-\dfrac{9}{13}}\\ =\dfrac{3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}}{3\left(3+\dfrac{3}{17}-\dfrac{3}{11}+\dfrac{3}{1001}-\dfrac{3}{13}\right)}\\ =\dfrac{1}{3}\)

\(\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}\\ =\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{2\left(50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}\right)}\\ =\dfrac{1}{2}\) 

Bài 8

a; \(\dfrac{12}{13}\) x \(\dfrac{4}{111}\)

= \(\dfrac{3\times4}{13}\) x \(\dfrac{4}{3\times37}\)

= \(\dfrac{16}{481}\)

b; \(\dfrac{1}{100}\) x \(\dfrac{2}{30}\)

  = \(\dfrac{1}{100}\) x \(\dfrac{2}{2\times15}\)

  =  \(\dfrac{1}{1500}\)

c; \(\dfrac{56}{65}\) x \(\dfrac{45}{54}\)

=  \(\dfrac{56}{65}\) x \(\dfrac{5}{6}\)

= \(\dfrac{28}{39}\)

Còn 3 bài ạ!
Giúp mik nhé!!