1) 152 + 37 + 28 + 238 + 163

= (152 + 138) + (163 + 37) + 28 

= 290 + 200 + 28

= 390 + 28

= 418

2) 26 x 54 + 54 x 73 

= 54 x (26 + 73)

= 54 x 99

= 54 x (100 - 1) 

= 54 x 100 - 54

= 5400 - 54

= 5346 

3) 23 x 75 + 25 x 23 +180

= 23 x (75 + 25) + 180

= 23 x 100 + 180

= 2300 + 180

= 2480 

4) 12 x 79 + 79 x 172 - 79 x 84

= 79 x (12 + 172 - 84)

= 79 x (184 - 84)

= 79 x 100

= 7900 

5) 28 x 47 + 72 x 29 + 28 x 43 + 72 x 61

= 28 x (47 + 43) + 72 x (29 + 61) 

= 28 x 90+ 72 x 90

= 90 x (28 + 72)

= 90 x 100

= 9000 

6) 24 x 40 + 10 x 8 x 4 + 34 x 80

= 12 x 2 x 40 + 10 x 8 x 4 + 34 x 80

= 12 x 80 + 80 x 4 + 34 x 80

= 80 x (12 + 4 + 34)

= 80 x 50

= 4000

7) (2 + 4 + 6 + ... + 2024) x (125125 x 127 - 127127 x 125) 

= (2 + 4 + 6 + ... + 2024) x (125 x 1001 x 127 - 127 x 1001 x 125)

= (2 + 4 + 6 + ... + 2024) x 0

= 0 

8) (1 + 3 + 5 + ... + 2021) x (13431 x 131 - 14541 x 121)

= (1 + 3 + 5 + ... + 2021) x (111 x 121 x 131 - 111 x 131 x 121) 

= (1 + 3 + 5 + ... + 2021) x 0

= 0 

diện tích mảnh đất là: 37 x 10 = 370 (m²)
mảnh đất còn lại sau khi xây nhà là: 370 - 100 = 270 (m²)
diện tích đất làm vườn là: 270 x 2/3 = 180 (m²)
diện tích đất làm sân là: 270 - 180 = 90 (m²)

Diện tích mảnh đất là:

37 x 10 = 370 `(m^2)`  

Diện tích không dùng để xây nhà là:

370 - 100 = 270 `(m^2)` 

Diện tích dùng để làm vườn là: 

\(\dfrac{2}{3}\times270=180\left(m^2\right)\)

Diện tích dùng làm sân là:

\(270-180=90\left(m^2\right)\)

ĐS: ...  

a) 

\(A=1000\times1020=\left(1010-10\right)\times\left(1010+10\right)\\ =1010\times1010-10\times1010+10\times1010-10\times10\\ =1010\times10101-10\times10< 1010\times1010=B\\ \Rightarrow A< B\) 

b) 

\(B=1200\times1600=\left(1400-200\right)\left(1400+200\right)\\ =1400\times1400-200\times1400+200\times1400-200\times200\\ =1400\times1400-200\times200< 1400\times1400=A\\ \Rightarrow B>A\) 

Xét ΔABE có: \(\widehat{BAE}+\widehat{ABE}+\widehat{AEB}=180^o\)

\(\Rightarrow90^o+x+x=180^o\Rightarrow2x=180^o-90^o=90^o\)

\(\Rightarrow x=\dfrac{90^o}{2}=45^o\) 

Xét ΔABC có: \(\widehat{ABC}+\widehat{BAC}+\widehat{CAB}=180^o\)

\(\Rightarrow\left(x+y\right)+90^o+30^o=180^o\)

\(\Rightarrow\left(x+y\right)+120^o=180^o\)

\(\Rightarrow45^o+y=180^o-120^o\)

\(\Rightarrow45^o+y=60^o\)

\(\Rightarrow y=60^o-45^o=15^o\)

∆ABE vuông tại A (gt)

⇒ ∠ABE + ∠AEB = 90⁰

⇒ x + x = 90⁰

⇒ x = 90⁰ : 2

= 45⁰

∆ABC vuông tại A (gt)

⇒ ∠ABC + ∠ACB = 90⁰

⇒ ∠ABC = 90⁰ - ∠ACB

= 90⁰ - 30⁰

= 60⁰

⇒ y = ∠ABC - x

= 60⁰ - 45⁰

= 15⁰

Diện tích mảnh đất HCN:

  \(28\times18=504\left(m^2\right)\)

Diện tích phần đất còn lại chiếm số phần so với diện tích mảnh đất HCN là:

 \(1-\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{11}{20}\)

Diện tích phần đất còn lại là:

 \(504\times\dfrac{11}{20}=277,2\left(m^2\right)\)

Diện tích của HCN là:

28 x 18 = 504 (cm2)

Diện tích trồng hoa là:

504 : 5 = 100,8 (cm2)

diện tích trồng cây ăn quả là:

504: 4= 126 (cm2)

diện tích phần đất còn lại là:

504 - 126 -100, 8 = 277,2 (cm2)

ĐS: 277,2 cm2 

a) 

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,3\right)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0,01\\ \Rightarrow\left(\dfrac{1}{12}+\dfrac{19}{6}-\dfrac{303}{10}\right)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right)\cdot100\\ \Rightarrow-\dfrac{541}{20}x-8=\dfrac{51}{50}\cdot100\\ \Rightarrow-\dfrac{541}{20}x-8=102\\ \Rightarrow-\dfrac{541}{20}x=110\\ \Rightarrow x=110:-\dfrac{541}{20}\\\Rightarrow x=-\dfrac{2200}{541}\approx4,067\)

b) 

\(\left(\dfrac{1}{12}+2\dfrac{1}{12}-10,75\right)x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+0,225\right):0,1\\ \Rightarrow\left(\dfrac{1}{12}+\dfrac{25}{12}-\dfrac{43}{4}\right)x-7=\left(\dfrac{2}{5}+\dfrac{3}{8}+\dfrac{9}{40}\right)\cdot10\\ \Rightarrow-\dfrac{103}{12}x=1\cdot10\\ \Rightarrow x=10:-\dfrac{103}{12}\\ \Rightarrow x=-\dfrac{120}{103}\\ \Rightarrow x\approx-1,165\) 

c) 

\(\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\left(x-3\right)=\dfrac{21}{13}\\\Rightarrow \dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow\dfrac{84}{305}\left(x-3\right)=\dfrac{21}{13}\\ \Rightarrow x-3=\dfrac{21}{13}:\dfrac{84}{305}\\ \Rightarrow x-3=\dfrac{305}{52}\\ \Rightarrow x=\dfrac{305}{52}+3\\ \Rightarrow x=\dfrac{461}{52}\\ \Rightarrow x\approx8,865\)

x.10 - x + 3.x - 9x

= (10 - 1 + 3 - 9)x

= 3x

3x

Bài 10:

\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{\left(2^2\right)^6\cdot3^6+\left(2^3\right)^4\cdot3^5}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\\ =\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}\\ =\dfrac{3-1}{3\cdot\left(3+1\right)}\\ =\dfrac{2}{3\cdot4}\\ =\dfrac{1}{6}\)

\(B=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\\ =\dfrac{2\cdot3\cdot5\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot\left(3^2\right)^7-2^2\cdot3\cdot\left(2^3\right)^5\cdot7^5\cdot}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{17}\cdot3^{14}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{15}\cdot3^{31}-2^{17}\cdot7^5\cdot3}\\ =\dfrac{5\cdot\left(3^{30}\cdot2^{15}-2^{17}\cdot7^5\right)}{3\cdot\left(2^{15}\cdot3^{30}-2^{17}\cdot7^5\right)}\\ =\dfrac{5}{3}\)

Bài 8:

\(\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y+z=\dfrac{1}{3}\\x+z=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{3}-z+\dfrac{1}{6}-z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-2z=\dfrac{1}{2}\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2z=0\\y=\dfrac{1}{3}-z\\x=\dfrac{1}{6}-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=0\\y=\dfrac{1}{3}-0=\dfrac{1}{3}\\x=\dfrac{1}{6}-0=\dfrac{1}{6}\end{matrix}\right.\)

Bài 5:

\(A=\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)

\(=\dfrac{15-4+1}{10}:\dfrac{9-4+1}{12}\)

\(=\dfrac{12}{10}\cdot\dfrac{12}{6}=\dfrac{6}{5}\cdot2=\dfrac{12}{5}\)