Thu gọn: (a+2b-3c-d)*(a+2b+3c+d)
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\(\left(a+2b-3c-d\right)\left(a+2b+3c+d\right)\)
\(=\left[\left(a+2b\right)-\left(3c-d\right)\right]\left[\left(a+2b\right)+\left(3c-d\right)\right]\)
\(=\left(a+2b\right)^2-\left(3c-d\right)^2\)
\(=a^2+4ab+4b^2-9c^2+6cd-d^2\)
a)(a+2b-3c-d)(a+2b+3c+d)=[(a+2b)-(3c+d)][(a+2b)+(3c-d)]
=(a+2b)2-(3c-d)2=a2+4ab+4b2-9c2+6cd-d2
câu b tương tự
a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)
b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)
c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)
d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)
e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
\(\Rightarrow\dfrac{2a+3c}{2a-3c}=\dfrac{2b+3d}{2b-3d}\)
\(\Rightarrow dpcm\)
Ta có: \(\left(a+2b-3c-d\right)\left(a+2b+3c+d\right)\)
\(=\left[\left(a+2b\right)-\left(3c+d\right)\right]\cdot\left[\left(a+2b\right)+\left(3c+d\right)\right]\)
\(=\left(a+2b\right)^2-\left(3c+d\right)^2\)
\(=a^2+4ab+4b^2-9c^2-6cd-d^2\)
( a + 2b - 3c - d )( a + 2b + 3c + d )
= [ ( a + 2b ) - ( 3c + d ) ][ ( a + 2b ) + ( 3c + d ) ]
= ( a + 2b )2 - ( 3c + d )2
= a2 + 4ab + 4b2 - ( 9c2 + 6cd + d2 )
= a2 + 4ab + 4b2 - 9c2 - 6cd - d2