Cho a^2 \(\ne\)b^2 và M =\(\frac{a+b}{a-b}+\frac{a-b}{a+b}.TínhN=\frac{a^4+b^4}{a^4-b^4}+\frac{a^4-b^4}{a^4+b^4}theoM\)
Cho \(a^2\ne b^2và\:x=\frac{a+b}{a-b}+\frac{a-b}{a+b}\)
Tính giá trị của biểu thức:
\(A=\frac{a^4+b^4}{a^4-b^4}+\frac{a^4-b^{.4}}{a^4+b^4}\) theo \(x\)
áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
Cho 2 số thực a,b thỏa mãn a^2 khác b^2.
Đặt A=\(\frac{a+b}{a-b}+\frac{a-b}{a+b}\) . Tính B=\(\frac{a^4+b^4}{a^4-b^4}+\frac{a^4-b^4}{a^4+b^4}\) theo A.
Rút gọn :
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)
Cho 2 số thực a,b thỏa mãn : a^2 khác b^2
Đặt A=\(\frac{a+b}{a-b}+\frac{a-b}{a+b}\) Tính : B=\(\frac{a^4+b^4}{a^4-b^4}+\frac{a^4-b^4}{a^4+b^4}\) theo A.
Chân thành cảm ơn!
\(A=\frac{4ab}{a^2-b^2}=\frac{4.\frac{a}{b}}{\left(\frac{a}{b}\right)^2-1}\Leftrightarrow A\left(\frac{a}{b}\right)^2-4\frac{a}{b}-A=0\Leftrightarrow At^2-4t+\frac{4}{A}=A+\frac{4}{A}\)
\(t=\frac{2}{A^2}+-\sqrt{\frac{A^2+4}{A^3}}\)
\(B=\frac{4a^4b^4}{a^8-b^8}=\frac{4t^4}{t^8-1}=..\)
Nhật Minh xem lại đi cậu!
A=\(\frac{2\left(a^2+b^2\right)}{a^2-b^2}\) chứ!
1. Cho a,b,c > 0 thõa mãn abc = 1. CM: \(\frac{a}{a+b^4+c^4}+\frac{b}{b+c^4+a^4}+\frac{c}{c+a^4+b^4}\le1\)
2. CHo 1 < = a,b,c < = 3. thõa mãn a + b + c = 3. CM: \(a^2+b^2+c^2\le14\)
1.
Ta có: \(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\)
\(\Rightarrow VT\le\frac{a}{a+bc\left(b^2+c^2\right)}+\frac{b}{b+ca\left(c^2+a^2\right)}+\frac{c}{c+ab\left(a^2+b^2\right)}\)
\(\Rightarrow VT\le\frac{a^2}{a^2+abc\left(b^2+c^2\right)}+\frac{b^2}{b^2+abc\left(a^2+c^2\right)}+\frac{c^2}{c^2+abc\left(a^2+b^2\right)}\)
\(\Rightarrow VT\le\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
dạ mọi người giúp em bài Toán 8 này với ạ! Dạ em cảm ơn ạ
Cho a,b,c > 0 thỏa mãn abc = 1. Chứng minh
a) \(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{a^2+c^2}{a+c}\ge\:3\)
b) \(\frac{1}{a+b^4+c^4}+\frac{1}{b+a^4+c^4}+\frac{1}{c+b^4+a^4}\le\:1\)
a/
\(VT\ge\frac{\frac{1}{2}\left(a+b\right)^2}{a+b}+\frac{\frac{1}{2}\left(b+c\right)^2}{b+c}+\frac{\frac{1}{2}\left(c+a\right)^2}{c+a}=a+b+c\ge3\sqrt[3]{abc}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ Ta có: \(x^4+y^4\ge\frac{1}{2}\left(x^2+y^2\right)\left(y^2+y^2\right)\ge xy\left(x^2+y^2\right)\)
\(\Rightarrow VT\le\frac{1}{a+bc\left(b^2+c^2\right)}+\frac{1}{b+ca\left(a^2+c^2\right)}+\frac{1}{c+ab\left(a^2+b^2\right)}\)
\(VT\le\frac{1}{a+\frac{1}{a}\left(b^2+c^2\right)}+\frac{1}{b+\frac{1}{b}\left(a^2+c^2\right)}+\frac{1}{c+\frac{1}{c}\left(a^2+b^2\right)}\)
\(VT\le\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}=\frac{a+b+c}{a^2+b^2+c^2}\)
\(VT\le\frac{a+b+c}{\frac{1}{3}\left(a+b+c\right)^2}=\frac{3}{a+b+c}\le\frac{3}{3\sqrt[3]{abc}}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Cho a, b, c > 0 và a + b + c = 3. Tìm GTNN của:
a, \(A=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\)
b, \(B=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\)
c, \(C=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\)
d, \(D=a^4+b^4+c^4+a^2+b^2+c^2+2020\)
Bạn tham khảo:
Câu hỏi của Nobody - Toán lớp 8 | Học trực tuyến
a) \(x^4+y^4\ge xy\left(x^2+y^2\right)\)với mọi x,y b) cho a,b,c>0 thoả mãn abc=1 tìm GTLN : A = \(\frac{a}{b^4+c^4+a}+\frac{b}{c^4+a^4+b}+\frac{c}{a^4+b^4+c}\)
a) \(x^4+y^4\ge xy\left(x^2+y^2\right)\)
\(\Leftrightarrow x^4+y^4-x^3y-xy^3\ge0\)
\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(x^3-y^3\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow x=y\)
b) Áp dụng câu a) ta có :
\(b^4+c^4+a\ge bc\left(b^2+c^2\right)+a\)
Mặt khác : \(abc=1\Leftrightarrow bc=\frac{1}{a}\)
\(\Rightarrow b^4+c^4+a\ge\frac{b^2+c^2}{a}+a=\frac{a^2+b^2+c^2}{a}\)
\(\Rightarrow\frac{a}{b^4+c^4+a}\le\frac{a}{\frac{a^2+b^2+c^2}{a}}=\frac{a^2}{a^2+b^2+c^2}\)
Chứng minh tương tự :
\(\frac{b}{c^4+a^4+b}\le\frac{b^2}{a^2+b^2+c^2};\frac{c}{a^4+b^4+c}\le\frac{c^2}{a^2+b^2+c^2}\)
Cộng theo vế của 3 bất đẳng thức
\(\Rightarrow A\le\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)