cho a+b /a-3 = b+4/b-4
tinh gt cua bieu thuc : D = a^3 +3^3 / b^3 + 4^3
Những câu hỏi liên quan
cho a+b / a- 3=b+4/b-4
tinh gt cua bieu thuc : D = a^3 +3^3 / b^3 +4^3
cho bieu thuc a=-1/3+1/3^2-1/3^3+1/3^4-1/3^5+...+1/3^100 tinh gia tri cua bieu thuc b=4/a/+1/3^100
cho bieu thuc A=3+3^2+3^3+3^4+...+3^100va B=3^301-1. chung minh rang A>B
Ta có: 3A = 3^2 + 3^3 + 3^4 + 3^5 +...+ 3^101
A = 3 + 3^2 + 3^3 + 3^4 +...+ 3^100
=> 3A - A = 3^101 - 3
=> 2A = 3^101 - 3
=> A = \(\frac{3^{101}-3}{2}\)
=> A = \(\frac{3^{101}-1}{2}-\frac{2}{2}=\left(3^{101}-1\right).\frac{1}{2}-1\)
=> A < B
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Tim GTLN hoac( GTNN )cua bieu thuc ;
A=|2x-3/5|+1,(3)
B=1/3-|x-2| (B>0)00
C=-2|1/3x+4|+3/2
D=|x-3|+|x+2/3|
gia tri cua bieu thuc 5\3-3\4 la a 21\12 b 12\12 c 11\12 d 7\12
cho a+b=3 ,a*b=2 tinh gia tri cua bieu thuc 1/a^3-1/b^3
Ta có a + b = 3
=> (a + b)2 = 9
=> a2 + 2ab + b2 = 9
=> a2 + b2 = 5 (ab = 2)
Khi a2 + b2 = 5 => a2 - 2ab + b2 = 1
=> (a - b)2 = 1
=> a - b = \(\pm1\)
Đặt A \(\frac{1}{a^3}-\frac{1}{b^3}=\frac{b^3-a^3}{\left(a.b\right)^3}=\frac{\left(b-a\right)\left(b^2+ab+a^2\right)}{\left(ab\right)^3}=-\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(ab\right)^3}\)
Với a - b = 1 ; ab = 2 ; a2 + b2 = 5 ta có A = \(-\frac{1.\left(5+2\right)}{2^3}=-\frac{7}{8}\)
Với a - b = - 1 ; ab = 2 ; a2 + b2 = 5 ta có A = \(-\frac{\left(-1\right).\left(5+2\right)}{2^3}=\frac{7}{8}\)
Ta có: \(\hept{\begin{cases}a+b=3\\ab=2\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2=9\\ab=2\end{cases}\Leftrightarrow}\hept{\begin{cases}a^2+2ab+b^2=9\\ab=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a^2+b^2=5\\ab=2\end{cases}}\)
Khi đó: \(\frac{1}{a^3}-\frac{1}{b^3}=\frac{b^3-a^3}{a^3b^3}=\frac{\left(b-a\right)\left(a^2+ab+b^2\right)}{8}=\frac{7\left(b-a\right)}{8}\)
Ta có: \(a+b=3\Rightarrow a=3-b\) thay vào: \(\left(3-b\right)b=2\)
\(\Leftrightarrow b^2-3b+2=0\Leftrightarrow\left(b-1\right)\left(b-2\right)=0\Leftrightarrow\orbr{\begin{cases}b=1\Rightarrow a=2\\b=2\Rightarrow a=1\end{cases}}\)
Nếu \(\hept{\begin{cases}a=2\\b=1\end{cases}\Rightarrow}\frac{1}{a^3}-\frac{1}{b^3}=-\frac{7}{8}\)
Nếu \(\hept{\begin{cases}a=1\\b=2\end{cases}}\Rightarrow\frac{1}{a^3}-\frac{1}{b^3}=\frac{7}{8}\)
tim gia tri lon nhat cua bieu thuc :
a) C= 5+ 15/ 4 I 3x+7 I +3
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
a) A= I x+1I + 1,7
b) B= I x-2/3I +3/7
c) C= 5+ -8/ 4x I5x+7I 24
Cho a,b,c thoa man \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
Tinh GT cua bieu thuc A=\(\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)\)
Lời giải:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)
\(\Rightarrow (a+b)(c+a)(c+b)=0\)
Do đó:
\(A=(a^3+b^3)(b^3+c^3)(c^3+a^3)\)
\(=(a+b)(a^2-ab+b^2)(b+c)(b^2-bc+c^2)(c+a)(c^2-ca+a^2)\)
\(=(a+b)(c+a)(c+b)[(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)]=0\)
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Cho phan thuc B=(3\y+3)+(1\y-3)-(18\9-y2)
a)Tim dieu kien cua y de gia tri cua bieu thuc B duoc xac dinh
b)Rut gon bieu thuc B
c)Tinh gia tri cua B de B co gia tri nguyen