Áp dụng bđt AM-GM cho 2 số không âm ta có:

\(x^4+y^4\ge2\sqrt{x^4y^4}=2x^2y^2\)

\(TT:y^4+z^4\ge2y^2z^2\)

\(z^4+x^4\ge2z^2x^2\)

Cộng vế với vế ta được:

\(2\left(x^4+y^4+z^4\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\) (1)

Tương tự:\(x^2y^2+y^2z^2+z^2x^2=\left(xy\right)^2+\left(yz\right)^2+\left(xz\right)^2\ge xy.yz+yz.xz+xz.xy=xy^2z+xyz^2+x^2yz=xyz\left(x+y+z\right)\left(2\right)\)

Từ (1) và (2) ta có đpcm

Dấu "=" xảy ra khi x = y = z

3) Sửa ab+bc+ca/3 thành ab+bc+ca/2; Thêm đk: a;b;c > 0

Đặt \(A=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)

\(A=\dfrac{\dfrac{1}{a^2}}{a\left(b+c\right)}+\dfrac{\dfrac{1}{b^2}}{b\left(c+a\right)}+\dfrac{\dfrac{1}{c^2}}{c\left(a+b\right)}\)

Áp dụng bđt Cauchy-Schwarz dạng Engel ta có:

\(A\ge\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\)

\(A\ge\dfrac{\dfrac{\left(bc+ac+ab\right)^2}{abc^2}}{2\left(ab+bc+ca\right)}=\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2}\)

Dấu "=" xảy ra khi a = b = c = 1

\(2a^3+b^3=2017\)\(\Rightarrow2a^3\le2017\)

\(\Rightarrow a^3\le1008\)

\(\Rightarrow a\le10\). Mà \(a\in N\) nên \(a\in\left\{0;1;2;3;4;5;6;7;8;9;10\right\}\)

Thử lần lượt với các giá trị của a ta được: a = 7; b = 11 thỏa mãn

Ta có:

\(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{ac}=\dfrac{2}{a}\)

\(\dfrac{1}{ab}+\dfrac{1}{b^2}+\dfrac{1}{bc}=\dfrac{2}{b}\)

\(\dfrac{1}{ac}+\dfrac{1}{bc}+\dfrac{1}{c^2}=\dfrac{2}{c}\)

Cộng vế với vế ta được:

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\)\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{c+a+b}{abc}=2.2\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)

Ta có: \(a+b=\dfrac{-1+\sqrt{2}}{2}+\dfrac{-1-\sqrt{2}}{2}=-1\)

\(ab=\dfrac{-1+\sqrt{2}}{2}.\dfrac{-1-\sqrt{2}}{2}=\dfrac{-1}{4}\)

\(\left(a+b\right)^2=a^2+b^2+2ab=1\)

\(\Rightarrow a^2+b^2+2.\dfrac{-1}{4}=1\)\(\Rightarrow a^2+b^2=\dfrac{3}{2}\) (1)

\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-1\)

\(\Rightarrow a^3+b^3+3.\dfrac{-1}{4}.\left(-1\right)=-1\)\(\Rightarrow a^3+b^3=\dfrac{-7}{4}\) (2)

Từ (1) và (2) suy ra \(\left(a^2+b^2\right)\left(a^3+b^3\right)=\dfrac{3}{2}.\dfrac{-7}{4}=\dfrac{-21}{8}\)

\(\Rightarrow a^5+a^3b^2+a^2b^3+b^5=\dfrac{-21}{8}\)

\(\Rightarrow a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{-21}{8}\)

\(\Rightarrow a^5+b^5+\dfrac{1}{16}.\left(-1\right)=\dfrac{-21}{8}\)\(\Rightarrow a^5+b^5=\dfrac{-41}{16}\) (3)

Từ (1) và (3) suy ra \(\left(a^2+b^2\right)\left(a^5+b^5\right)=\dfrac{3}{2}.\dfrac{-41}{16}=\dfrac{-123}{32}\)

\(\Rightarrow a^7+a^5b^2+a^2b^5+b^7=\dfrac{-123}{32}\)

\(\Rightarrow a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{-123}{32}\)

\(\Rightarrow a^7+b^7+\dfrac{1}{16}.\dfrac{-7}{4}=\dfrac{-123}{32}\)\(\Rightarrow a^7+b^7=\dfrac{-239}{64}\)

Vậy \(a^7+b^7=\dfrac{-239}{64}\)

\(A=\sqrt{10+6\sqrt{2}-2\sqrt{6}-2\sqrt{3}}-\sqrt{7-2\sqrt{6}}\)

\(A=\sqrt{1^2+\left(\sqrt{3}\right)^2+\left(\sqrt{6}\right)^2+2\sqrt{3.6}-2\sqrt{1.6}-2\sqrt{1.3}}-\sqrt{1-2\sqrt{6}+6}\)

\(A=\sqrt{\left(\sqrt{6}+\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}\)

\(A=\sqrt{6}+\sqrt{3}-1-\left(\sqrt{6}-1\right)=\sqrt{3}\)

Giả sử Q(x) có nghiệm nguyên là a

Ta có: Q(a) = a³ - a + 13 = 0

a³ chia hết cho a; a chia hết cho a; 0 chia hết cho a nên 13 chia hết cho a

a nguyên nên \(a\in\left\{1;-1;13;-13\right\}\)

Cứ vậy thay vào thôi

Tên: LTKH

Lớp: 7A

Link của nik: Góc học tập của soyeon_Tiểubàng giải | Học trực tuyến

\(\dfrac{2002}{\sqrt{2003}}+\dfrac{2003}{\sqrt{2002}}\)

\(=\dfrac{2002+1}{\sqrt{2003}}+\dfrac{2013-1}{\sqrt{2002}}+\dfrac{1}{\sqrt{2002}}-\dfrac{1}{\sqrt{2003}}\)

\(=\sqrt{2003}+\sqrt{2002}+\dfrac{1}{\sqrt{2002}}-\dfrac{1}{\sqrt{2003}}\)

\(>\sqrt{2003}+\sqrt{2002}+\dfrac{1}{\sqrt{2003}}-\dfrac{1}{\sqrt{2003}}=\sqrt{2003}+\sqrt{2002}\left(đpcm\right)\)

VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)

\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)

\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)

\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)

\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)