giúp mình với mn 

Ta có

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ,có

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)

=>\(\left\{{}\begin{matrix}a=-2\\b=-3\\c=-4\end{matrix}\right.\)

Nửa chu vi mảnh vườn là:

532 : 2 = 266 (m)

Tổng số phần bằng nhau là:

3 + 4 = 7 (phần)

Chiều dài mảnh vườn là:

266 : 7 x 4 = 152 (m)

Chiều rộng mảnh vườn là:

266 - 152 = 114 (m)

Diện tích mảnh vườn là:

152 x 114 = 17328 (m²)

Đáp số : 17328 m²

mọi người ơi, giúp mik ik

a,  1 a b + 36 = a b 1

100 +  a b + 36 = 10. a b + 1

135 = 9 a b

a b = 135 : 9

a b = 15

Vậy a = 1, b = 5

b,  a b c d + a b c + a b + a = 4321

Ta có  a b c d = 1000 a + 100 b + 10 c + d

a b c = 100 a + 10 b + c

a b = 10 a + b

=>  a b c d + a b c + a b + a = 1111a + 111b + 11c + d

Theo đề ta có 1111a + 111b + 11c + d = 4321 với a,b,c,d ∈ {0,1,2,…,9}, a≠0

+ Nếu a>3 thì VT ≥ 4444 + 111.0 + 11.0 + 0 > VP

+ Nếu a<3 thì VT ≤ 2222 + 111.9 + 11.9 + 9 = 3329 < VP

Vậy a = 3 => VT = 3333 + 111b + 11c + d = 4321
tick mik nhé

\(\left(x^2-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=\left(x^2\right)^2-2.x^2.2y+\left(2y\right)^2+\left(x^2-\left(\dfrac{1}{2}y\right)^2\right)\)

\(=x^4-4x^2y+4y^2+x^2-\dfrac{y^2}{4}\)

\(=x^4-4x^2y+x^2+\dfrac{15y^2}{4}\)

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\\ \Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\\ \Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\\ \Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\\ \Rightarrow\left(x+2012\right).\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\\ \dfrac{1}{2008}>\dfrac{1}{2009}>\dfrac{1}{2010}>\dfrac{1}{2011}\Rightarrow\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)\ne0\\ \Rightarrow x+2012=0\\ \Rightarrow x=-2012\)

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2001}\)

\(\Leftrightarrow\dfrac{x+4}{2008}+\dfrac{x+3}{2009}+2=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}+2\)

\(\Leftrightarrow\) \(\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)

\(\Leftrightarrow\)\(\dfrac{x+4+2008}{2008}+\dfrac{x+3+2009}{2009}=\dfrac{x+2+2010}{2010}+\dfrac{x+1+2011}{2011}\)

\(\Leftrightarrow\) \(\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)

\(\Leftrightarrow\) \(\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)

\(\Leftrightarrow\) \(\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)

Vì: \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}=0\)

Nên: \(x+2012=0\)

\(x=-2012\)