Cho thỏa mãn . Giá trị là
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Có:
\(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:a2+b2+5=2a+4b
⇔ (a2-2a+1)+(b2-4b+4)=0
⇔ (a-1)2+(b-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)
Thay vào P ta có:
\(P=\left|2.1-3.2\right|+1+5=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :18=2.32; 135=32 .5.7
UCLN(18,315)=32=9
B(9)={0;9;18;27;....}
mà 5 < x ≤11
⇒x= 9 (tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\lim\dfrac{1+a+...+a^n}{1+b+...+b^n}=\lim\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}=\lim\dfrac{\left(1-a^n\right)\left(1-b\right)}{\left(1-b^n\right)\left(1-a\right)}=\dfrac{1-b}{1-a}\)
\(\Rightarrow\dfrac{1-b}{1-a}=\dfrac{2}{3}\Leftrightarrow3-3b=2-2a\)
\(\Leftrightarrow2a-3b=-1\)
\(2a:2=4\)
\(2a=4.2\)
\(2a=8\)
\(a=8:2\)
\(=>a=4\)
\(2:2=1\)