(3a + 2)2 + 2(2 + 3a )(1-2b) + (2b - 1)2
Tính:
a, (3a^2-1/2)^3+(a^3+1/4)^2-(a+1)^3
b,(1/3a^2-1/2b).(1/3a^2-1/2b)-(a+1/2b)-(a+1/2b).(a^2-1/2ab)+1/4b^2
\(Q=11a^2b-2a^2b-3a^2b-3a^2\) tại \(a=\dfrac{-1}{3};b=2\dfrac{3}{4}\)
\(Q=6a^2b-3a^2=6\cdot\dfrac{1}{9}\cdot\dfrac{11}{4}-3\cdot\dfrac{1}{9}=\dfrac{3}{2}\)
Rút gọn các biểu thức sau:
a) M = ( 2 a + b ) 2 – ( b – 2 a ) 2 ;
b) N = ( 3 a + 2 ) 2 + 2 ( 2 + 3 a ) ( 1 – 2 b ) + ( 2 b - 1 ) 2 .
a) M = 8ab;
b) N = [ ( 3 a + + 2 ) + ( 1 – 2 b ) ] 2 = ( 3 a – 2 b + 3 ) 2 .
thu gọn các đơn thức sau
a)ab.4/3a^2b^4.7abc
b)a^3b^3.a^2b^2c
c)2/3a^3b.(-1/2ab).a^2b
d)-2 1/3a^3c^21/7ac^2 6abc
e)(-1,5ab^2)1/4bca^2b
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: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)
(3a+2b-1)(a+5)-2b(a-2)=(3a+5)(a+3)+2(7b-10)
Chứng minh hả ? -.-
( 3a + 2b - 1 )( a + 5 ) - 2b( a - 2 ) = ( 3a + 5 )( a + 3 ) + 2( 7b - 10 )
<=> 3a2 + 15a + 2ab + 10b - a - 5 - 2ab + 4b = 3a2 + 14a + 15 + 14b - 10
<=> 3a2 + 14a + 14b - 5 = 3a2 + 14a + 14b - 5
=> đpcm
Cho \(a>0\) , \(b>0\) thỏa mãn: \(\log_{3a+2b+1}\left(9a^2+b^2+1\right)+\log_{6ab+1}\left(3a+2b+1\right)=2\) .
Tính giá trị của biểu thức: \(P=a+2b\)
\(a;b>0\Rightarrow3a+2b+1>1\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\) đồng biến
Mà \(9a^2+b^2\ge2\sqrt{9a^2b^2}=6ab\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)\)
\(\Rightarrow log_{3a+2b+1}\left(9a^2+b^2+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge log_{3a+2b+1}\left(6ab+1\right)+log_{6ab+1}\left(3a+2b+1\right)\ge2\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}log_{6ab+1}\left(3a+2b+1\right)=1\\3a=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6ab+1=3a+2b+1\\b=3a\end{matrix}\right.\)
\(\Rightarrow18a^2+1=3a+6a+1\)
\(\Leftrightarrow18a^2-9a=0\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\)
Cho a, b,c : abc = 1. Chứng minh:
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}+\dfrac{b^2c^2}{2b^2+c^2+3b^2c^2}+\dfrac{c^2a^2}{2c^2+a^2+3a^2c^2}\le\dfrac{1}{2}\)
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
(3a+2b)^2-18a(3a+2b)+81a^2
\(=\left(3a+2b-9a\right)^2=\left(2b-6a\right)^2\)
cho các số a,b,c thỏa mãn 3a-2b/4=2c-4a/3=4b-3c/2 tính giá trị biểu thức A=3a+2b-c/3a-2b+c + 2a^2-b^2+c^2/2a^2+b^2-c^2
làm ơn trả lời hộ mk với ah mai mk phải nộp bài r
1.(a+b+c)(a^2+b^2+c^2-ab-bc-ca)= a^3-b^3+c^3-3abc
2. (3a+2b-1)(a+5)-2b(a-2)=(3a+5)(a+3)+2(7b-10)
chứng minh các đẳng thức
1) a³ + b³ + c³ - 3abc
=(a + b)(a² - ab + b²) + c³ - 3abc
=(a + b)(a² - ab + b²) + c(a² - ab + b²) - 2abc - ca² - cb²
=(a + b + c)(a² - ab + b²) - (abc + b²c + bc² + ac² + abc + c²a) + c³ + ac² + bc²
=(a + b = c)(a² - ab + b²) - (a + b + c)(bc + ca) + c²(a + b + c)
=(a + b + c)(a² + b² + c² - ab - bc - ca)
2) \(\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)=\left(3a+5\right)\left(a-3\right)+2\left(7b-10\right)\left(1\right)\)
\(\Leftrightarrow3a^2+15a+2ab+10b-a-5-2ab+4b=3a^2+14a+15+14b-10\)
\(\Leftrightarrow3a^2+14a+14b-5=3a^2+14a+14b-5\)( đúng)
\(\Rightarrow\left(1\right)\) đúng (đpcm)
1) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\left(đpcm\right)\)