1, Cho \(a^2+b^2+c^2=a+2b+3c=14\)
Tính :\(A=abc\)
2,Cho \(a^3-a^2+a-5=0\)và \(b^3-2b^2+2b+4=0\)
Tính \(B=\left(a+b\right)^{2019}\)
1. Cho \(a,b,c>0\) và \(ab+bc+ca=abc\). Chứng minh rằng:
\(\dfrac{1}{a+3b+2c}+\dfrac{1}{b+3c+2a}+\dfrac{1}{c+3a+2b}\le\dfrac{1}{6}\)
2. Cho \(a,b\ge0\) và \(a+b=2\) Tìm Max
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+20ab\)
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
2,
\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)
\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(ab=x\Rightarrow0\le x\le1\)
\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)
\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)
Bài 1.Cho \(x+y+z=0\)
Tính \(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Bài 2. Cho \(a+b+c=1;a^2+b^2+c^2=1;\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
CMR: \(xy+yz+zx=0\)
Bài 3. Cho \(3x-y=2z\)
\(2x+y=7z\)
Tính \(S=\frac{x^2-2xy}{x^2+y^2}\)với \(x,y\ne0\)
Bài 4. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(E=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 5. Cho \(abc\ne0\)thỏa mãn: \(2ab+6bc+2ac=0\)
Tính \(A=\frac{\left(a+2b\right)\left(2b+3c\right)\left(3c+a\right)}{6abc}\)
Bài 6. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(Y=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-c^2a^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}\)
Bài 7. Cho \(\hept{\begin{cases}10a^2-3b^2+5ab=0\\9a^2-b^2\ne0\end{cases}}\)
Tính \(B=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
2. a + b + c = 1 \(\Rightarrow\)( a + b + c )2 = 1 \(\Rightarrow\)a2 + b2 + c2 + 2 ( ab + bc + ac ) = 1 \(\Rightarrow\)ab + bc + ac = 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)
\(\Rightarrow\)x = a ( x + y + z ) ; y = b ( x + y + z ) ; z = c ( x + y + z )
Ta có : xy + yz + xz = ab ( x + y + z )2 + bc ( x + y + z )2 + ac ( x + y + z )2 = ( x + y + z )2 ( ab + bc + ac ) = 0
3. sửa đề : 3x - y = 3z
Ta có : \(\hept{\begin{cases}3x-y=3z\\2x+y=7z\end{cases}\Rightarrow\hept{\begin{cases}\left(3x-y\right)+\left(2x+y\right)=3z+7z\\2x+y=7z\end{cases}\Rightarrow}\hept{\begin{cases}5x=10z\\y=7z-2x\end{cases}\Rightarrow}\hept{\begin{cases}x=2z\\y=3z\end{cases}}}\)
\(\Rightarrow\)\(S=\frac{x^2-2xy}{x^2+y^2}=\frac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8z^2}{13z^2}=\frac{-8}{13}\)
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(C=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(c=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC
Cho abc=36,\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) .Tính
Q=\(\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\cdot\frac{b^2\left(c^2+a^2\right)-c^2a^2}{a^2b^2c^2}\cdot\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Cho abc = 36 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\); \(B=\frac{b^2\left(c^2+a^2\right)-c^2a^2}{a^2b^2c^2}\); \(C=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Tính A; B; C
\(Cho\hept{\begin{cases}a^3-a^2+a-5=0\\b^3-2b^2+2b+4=0\end{cases}}\) . Tính \(\left(a+b\right)^{2017}\)
1. CMR: Nếu a,b,c là độ dài 3 cạnh tam giác thì:
\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\)
2. PTĐT thành nhân tử
a) \(a^6+a^4+a^2b^2+b^4+b^6\)
b) \(a^3+3ab+b^3-1\)
c) \(a^2b^2\left(b-a\right)+b^2c^2\left(c-b\right)-c^2a^2\left(c-a\right)\)
d) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
1.
\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)
Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)
Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)
Từ đó ta được đpcm
2.
\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)
\(c,=a^2b^2\left(b-a\right)+b^2c^2\left(c-a+a-b\right)-c^2a^2\left(c-a\right)\\ =-a^2b^2\left(a-b\right)+b^2c^2\left(a-b\right)+b^2c^2\left(c-a\right)-c^2a^2\left(c-a\right)\\ =\left(a-b\right)\left(b^2c^2-a^2b^2\right)+\left(c-a\right)\left(b^2c^2-c^2a^2\right)\\ =b^2\left(a-b\right)\left(c-a\right)\left(c+a\right)+c^2\left(c-a\right)\left(b-a\right)\left(b+a\right)\\ =\left(a-b\right)\left(c-a\right)\left[b^2\left(c+a\right)-c^2\left(b+a\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b^2c+ab^2-bc^2-ac^2\right)\\ =\left(a-b\right)\left(c-a\right)\left[bc\left(b-c\right)+a\left(b-c\right)\left(b+c\right)\right]\\ =\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(bc+ab+ac\right)\)
Cho các số thực a,b,c thỏa mãn a>1 , b>\(\dfrac{1}{2}\) , \(c>\dfrac{1}{3}\) và \(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\). Tìm GTLN của bt \(P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)