Cho a,b,c \(\in\)Q thoã mãn: ab+bc+ca=2018
Chứng minh: \(\sqrt{\left(a^2+2018\right)\left(b^2+2018\right)\left(c^2+2018\right)}\) \(\in\) Q
các bạn tham khảo nhé
a, Cho \(a^{2018}+b^{2018}+c^{2018}=\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Tính \(P=\left(a-b\right)^{2018}+\left(b-c\right)^{2018}+\left(c-a\right)^{2018}\)
b, Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)và \(\frac{2}{ab}-\frac{1}{c^2}=9\)
Tính \(P=\left(a+2b+c\right)^{2018}\)
Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
Mà đẳng thức trên xảy ra dấu =
\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)
Bài kia tí nghĩ nốt, khó v
Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)
Cho ax+by+cz=0 và a+b+c =1/2018 Chứng minh rằng \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}\) =2018
Đặt \(A=\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+cz\left(z-x\right)}\)
Từ ax+by+cz=0
=>(ax+by+cz)2=0
=>a2x2+b2y2+c2z2+2axby+2bycz+2czax=0
=>a2x2+b2y2+c2z2=-2(ax+by+byca+czax)
Xét mẫu thức: \(ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2\)
\(=ab\left(x^2-2xy+y^2\right)+bc\left(y^2-2yz+z^2\right)+ca\left(z^2-2zx+x^2\right)\)
\(=abx^2-2abxy+aby^2+bcy^2-2bcyz+bcz^2+caz^2-2cazx+cax^2\)
\(=\left(abx^2+bcz^2\right)+\left(aby^2+acz^2\right)+\left(acx^2+bcy^2\right)-2\left(abxy+bcyz+cazx\right)\)
\(=\left(aby^2+acz^2\right)+\left(abx^2+bcz^2\right)+\left(acx^2+bcy^2\right)+a^2x^2+b^2y^2+c^2z^2\)
\(=\left(a^2x^2+aby^2+acz^2\right)+\left(abx^2+b^2y^2+bcz^2\right)+\left(acx^2+bcy^2+c^2z^2\right)\)
\(=a\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+c\left(ax^2+by^2+cz^2\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
Do đó: \(A=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{\frac{1}{2018}}=2018\) (dpcm)
Cho ax+by+cz=0 và a+b+c =1/2018 Chứng minh : \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}=2018\)
Cho \(\left(a+b+c\right)\left(ab+bc+ca\right)\)=2018 và \(abc=2018\).Tính P=\(\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+abc\right)\)
Cho : \(\left(a+b+c\right)\left(ab+bc+ca\right)=2018\)và \(abc=2018\)
Tính P = \(\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
Giúp mình rồi mình tik cho
Cho các số thực a, b, c thỏa mãn: \(\left(a+b+c\right)\left(ab+bc+ca\right)=2018\) và \(abc=2018\)
Tính giá trị biểu thức \(A=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
\(\left(a+b+c\right)\left(ab+ac+bc\right)=\left(a+b+c\right)\left(ab+ac+bc+c^2-c^2\right)\)
\(=\left(a+b+c\right)\left(\left(a+c\right)\left(b+c\right)-c^2\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2\left(a+b\right)+c\left(a+c\right)\left(b+c\right)-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2a-c^2b+abc+c^2a+c^2b+c^3-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)+abc=\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018=2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Ta có:
\(A=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
\(A=\left(b^2c+abc\right)\left(c^2a+abc\right)\left(a^2b+abc\right)\)
\(A=bc\left(a+b\right)ac\left(b+c\right)ab\left(a+c\right)\)
\(A=\left(abc\right)^2\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=2018^2.0=0\)
Bài 1: Cho các số thực a, b, c thoả mãn \(\left(a+b+c\right)\left(ab+bc+ca\right)=2018\) và \(abc=2018\). Tính giá trị của biểu thức \(P=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
Cho \(a+b+c=6\) và \(ab+bc+ca=12\). Tính \(M=\left(a-b\right)^{2018}+\left(b-c\right)^{2018}+\left(c-a\right)^{2018}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\Rightarrow6^2=a^2+b^2+c^2+2.12\Rightarrow a^2+b^2+c^2=12\)
Ta có:
\(a^2+b^2+c^2=ab+bc+ca\left(=12\right)\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow M=0}\)
Chúc bạn học tốt.
1.
a) \(A=\frac{\left(\frac{2018}{1}-1\right)\left(\frac{2018}{2}-1\right)...\left(\frac{2018}{1000}-1\right)}{\left(\frac{1000}{1}+1\right)\left(\frac{1000}{2}+1\right)...\left(\frac{1000}{1007}+1\right)}\)
b) Tìm x biết 378% của x kém A 55 đơn vị.
2. Tìm a, b, c sao cho : \(\frac{\overline{ab}.\overline{bc}.\overline{ca}}{\overline{ab}+\overline{bc}+\overline{ca}}=\frac{3321}{11}\)