a,b khac 0 sao cho a + \(\frac{1}{b}=1\) va \(a^2+\frac{1}{b^2}=3\) Tinh \(N=\frac{a^4b^4+a^2b^2+1}{b^4}\)
Những câu hỏi liên quan
cho a khac 0 b khac 0 va a+b=1 chung minh rang \(\frac{b}{a^3-1}-\frac{a}{b^3-1}=\frac{2\left(a-b\right)}{a^2b^2+3}\)
chi a,,b,c thoa man (a+2b)(2b+3c)(3c+a)khac 0 va
\(\frac{a^2}{a+2b}+\frac{4b^2}{2b+3c}+\frac{9c^2}{3c+a}=\frac{a^2}{2b+3c}+\frac{4b^2}{a+3c}+\frac{9c^2}{a+2b}\)
cmr;\(\frac{a}{6}=\frac{b}{3}=\frac{c}{2}\)
Cho a,b,c>0 thỏa mãn a+2b+3c=1
CMR: \(\frac{2ab}{a^2+4b^2}+\frac{6bc}{4b^2+9c^2}+\frac{3ac}{9c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}\right)\ge\frac{15}{4}\)
cho a,b,c khac 0 va \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\).tinh P= \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
cho a,b,c khac 0 va a+b+c=0 . tinh Q=\(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
a + b + c = 0 => c = -a - b ; b= -a - c ; a = - b - c
Thay vào Q ta có :
\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)
\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)
Đúng 0
Bình luận (0)
1. Cho a3+2b2-4b+3=0; a2+a2b2-2b=0
Tính S=a2017+b2016
2. Tính \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+...+\frac{2011}{2011^4+2011^2+1}\)
Giúp với ạ. Cảm ơn nhiều ạ!!!
Câu 1) Ta có\(a^3+2b^2-4b+3=0\Leftrightarrow a^3=-2.\left(b-1\right)^2-1\)\(\le-1\Rightarrow a^3\le-1\Rightarrow a\le-1\Rightarrow a^2\ge1\)
\(\Rightarrow\hept{\begin{cases}a^2\ge1\\a^2b^2\ge b^2\end{cases}}\)\(\Rightarrow a^2+a^2b^2-2b\ge1+b^2-2b\)\(\Leftrightarrow\left(b-1\right)^2\le0\)
Mà \(\left(b-1\right)^2\ge0\)với mọi b nên \(\left(b-1\right)^2=0\)\(\Rightarrow b=1\)
Thay b=1 vào 2 pt ban đầu được \(\hept{\begin{cases}a^3+2-4+3=0\\a^2+a^2-2=0\end{cases}}\)<=> a=1(tm)
Vậy (a,b)=(1;1)
Câu 2 bạn xem ở đây nhé http://olm.vn/hoi-dap/question/716469.html
Đúng 0
Bình luận (0)
cho a,b khac 0 thoa man a+b=1. chung minh \(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)
Cho biểu thức sau:
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
a, Rút gọn biểu thức P
b, Tính giá trị của P khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)
\(=\frac{8ab}{a^4b^4-16}\)
b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
=> (a2 + 4).9 = a2(b2 + 9)
=> 9a2 + 36 = a2b2 + 9a2
=> a2b2 = 36
=> (ab)2 = 36
=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)
Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)
Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)
cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) va a,b,c khac 0. Rut gon bieu thuc N=\(\frac{1}{a^2+2ab}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\)