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Tìm x nguyên thỏa mãn$x^2\left(x^2-1\right)\left(x^2-5\right)\left(x^2-10\right)<0$x2(x2−1)(x2−5)(x2−10)<0và $\left|x\right|<5$|x|<5Bài này của lớp 6 nhưng lập bảng xét dấu
xin lỗi em mới học lớp 5
nên ko làm đựơc
nếu ai cũng vậy thì k cho nhé
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
\(a^2+b^2+c^2=\left(a+b+c\right)^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)
Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
Thay vào A ta được :
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(A=0\)
Ta có: \(a+b+c=\frac{1}{2}\) \(\Rightarrow\left\{\begin{matrix}a=\frac{1}{2}-b-c\\b=\frac{1}{2}-a-c\\c=\frac{1}{2}-a-b\end{matrix}\right.\) hay \(\left\{\begin{matrix}a+b=\frac{1}{2}-c\\b+c=\frac{1}{2}\\a+c=\frac{1}{2}-b\end{matrix}\right.\)
\(P=\frac{\left(2ab+c\right)\left(2bc+a\right)\left(2ac+b\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)
\(=\frac{\left(2ab+\frac{1}{2}-a-b\right)\left(2bc+\frac{1}{2}-b-c\right)\left(2ca+\frac{1}{2}-a-c\right)}{\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)}\)
\(=\frac{2\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right).2\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right).2\left(ca+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)}{\left(ac+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right)\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right)}\)
\(=2.2.2=8\)
Vậy với \(a+b+c=\frac{1}{2}\) và \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\) thì \(P=8\)
8