Giá trị của biểu thức: A=\(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}-\dfrac{b^2}{\left(b-a\right)\left(c-b\right)}-\dfrac{c^2}{\left(c-a\right)\left(b-c\right)}\)
Cho ba số thực dương a,b,c thỏa mãn a+b+c ≤ 2 . Tìm giá trị nhỏ nhất của biểu thức : P = \(\dfrac{b\left(a^2+1\right)^2}{a^2\left(b^2+1\right)}+\dfrac{c\left(b^2+1\right)^2}{b^2\left(c^2+1\right)}+\dfrac{a\left(c^2+1\right)^2}{c^2\left(a^2+1\right)}\)
Giúp mình với mình
\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)
Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)
Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)
\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)
\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)
\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)
\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)
\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)
\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)
\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)
\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)
Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)
\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)
Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)
\(=\dfrac{a^2b^2c^2+\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)+1}{abc}\)
\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)
\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)
\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)
Cho a,b,c là cái số thực dương thỏa mãn a + b + c = 1 . Tìm giá trị nhỏ nhất của biểu thức : Q = \(\dfrac{\left(1-c\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}+\dfrac{\left(1-a\right)^2}{\sqrt{2\left(c+a\right)^2+ca}}\) + \(\dfrac{\left(1-b\right)^2}{\sqrt{2\left(a+b\right)^2+ab}}\)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Cho a, b, c >0 và dãy tỉ số \(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính giá trị của biểu thức P=\(\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2b+2c+2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)
\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)
\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)
cho biết \(\dfrac{a}{2}-b=c\dfrac{2}{3}\)và a,b,c khác 0. Tính giá trị biểu thức Q=2018-\(\left(\dfrac{c}{a}-\dfrac{1}{3}\right)^5.\left(\dfrac{a}{2}-2\right)^5.\left(\dfrac{3}{2}+\dfrac{b}{c}\right)^5\)
Tính giá trị của biểu thức với a,b,c khác nhau:
\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\)
\(M=\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a-b\right)\left(b-c\right)}.\dfrac{\left(b+c\right)\left(c+a\right)}{\left(b-c\right)\left(c-a\right)}.\dfrac{\left(c+a\right)\left(a+b\right)}{\left(c-a\right)\left(a-b\right)}\) (loạn mắt quá !!! )
\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)\left(b+c\right)\left(c+a\right)\left(a-b\right)\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Thôi để bữa sau a làm (đau mắt quá :)))
cho 3 số a,b,c thỏa mãn
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
tính giá trị biểu thức A=\(\dfrac{a^2+b^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{b^2+c^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{c^2+a^2}{\left(c+b\right)\left(a+b\right)}\)
\(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}=a+b+c\)
\(\Leftrightarrow\dfrac{abc}{a^2}+\dfrac{abc}{b^2}+\dfrac{abc}{c^2}=a+b+c\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{a+b+c}{abc}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)
\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Thay vào A r tính thôi
Cho 3 số hữu tỉ dương a;b;c thỏa mãn: \(\dfrac{a+b-2c}{c}=\dfrac{b+c-2a}{a}=\dfrac{c+a-2b}{b}\)
Tính giá trị biểu thức: P = \(\left(1+\dfrac{a}{b}\right)\left(2+\dfrac{b^2}{c^2}\right)\left(3+\dfrac{c^3}{a^3}\right)\)
Cho a,b,c là các số nguyên khác nhau đôi một. CMR biểu thức sau có giá trị là 1 số nguyên: \(P=\dfrac{a^3}{\left(a-b\right).\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right).\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right).\left(c-b\right)}\)
Bài này mình làm một lần ở trường rồi nhưng không có điện thoại chụp được:((
Ta có: \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)+b^3\left(a-c\right)-c^3\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{a^3\left(c-b\right)+b^3a-b^3c-c^3a+c^3b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c^3-b^3\right)+bc\left(c^2-b^2\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c-b\right)\left(a^2+bc+b^2\right)+bc\left(c-b\right)\left(c+b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left(a^3-ac^2-abc-ab^2+bc^2+b^2c\right)}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}=\dfrac{\left(c-b\right)\left[a\left(a^2-b^2\right)-c^2\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left[a\left(a-b\right)\left(a+b\right)-c\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left(a^2+ab-c-bc\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)
\(\dfrac{\left(c-b\right)\left(a-b\right)\left[a^2-c^2+ab-bc\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left(a-c\right)\left(a+b+c\right)}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)\(=a+b+c\)
Vì a, b, c là các số nguyên
=> a+b+c là các số nguyên
=> Đpcm.
Đấy mình làm chi tiết tiền tiệt lắm luôn, không hiểu thì mình chịu rồi, trời lạnh mà đánh máy nhiều thế này buốt tay lắm luôn:vv