làm hộ mik
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![](https://rs.olm.vn/images/avt/0.png?1311)
#include <bits/stdc++.h>
using namespace std;
double a,b,c,p,s;
int main()
{
cin>>a>>b>>c;
p=(a+b+c)/2;
s=sqrt(p*(p-a)*(p-b)*(p-c));
cout<<fixed<<setprecision(2)<<p;
return 0;
}
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
uses crt;
var a,b,c,max,min:longint;
begin
clrscr;
readln(a,b,c);
max=a;
if max<b then max:=b;
if max<c then max:=c;
min:=a;
if min>c then min:=c;
if min>b then min:=b;
writeln(max,' ',min);
readln;
end.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a)\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}.\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3.\\ \Rightarrow x+\dfrac{2}{3}=\dfrac{5}{4}.\\ \Leftrightarrow x=\dfrac{7}{12}.\)
\(b)\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}.\\\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right) ^3.\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{2}{7}.\\ \Leftrightarrow x=\dfrac{11}{14}.\)
Bài 2:
\(a)\left(x-\dfrac{1}{3}\right)^2=\dfrac{25}{9}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{5}{3}\right)^2.\\\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{-5}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{5}{3}.\\x-\dfrac{1}{3}=\dfrac{-5}{3}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-4}{3}.\end{matrix}\right.\)
\(b)\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{7}{4}\right)^2.\\\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{-7}{4}\right)^2.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}.\\x-\dfrac{3}{4}=\dfrac{-7}{4}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}.\\x=-1.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
Nếu $m=3$ thì pt trở thành:
$x^2+4x-5=0$
$\Leftrightarrow (x-1)(x+5)=0$
$\Leftrightarrow x=1$ hoặc $x=-5$
b.
Để pt có 2 nghiệm pb $x_1,x_2$ thì:
$\Delta'=4+m^2-4>0\Leftrightarrow m^2>0\Leftrightarrow m\neq 0$
PT có 2 nghiệm $(-2+m, -2-m)$
Khi đó:
\(x_2=x_1^3+4x_2^2\Leftrightarrow \left[\begin{matrix} -2+m=(-2-m)^3+4(-2+m)^2\\ -2-m=(-2+m)^3+4(-2-m)^2\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} -m^3+2m^2-29m+10=0\\ m^3-2m^2+29m+10=0\end{matrix}\right.\)
Nghiệm khá xấu, cảm giác đề cứ sai sai bạn ạ.
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Thay a=-2 vào pt, ta được:
\(-2x^2-2\cdot\left(-2-1\right)x-2+1=0\)
\(\Leftrightarrow-2x^2+6x-1=0\)
\(\Leftrightarrow2x^2-6x+1=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot2\cdot1=36-8=28>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{7}}{2}=3-\sqrt{7}\\x_2=3+\sqrt{7}\end{matrix}\right.\)
b: Để phương trình có hai nghiệm phân biệt thì
\(\left\{{}\begin{matrix}\left(-2a+2\right)^2-4a\left(a+1\right)>0\\a< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a^2-8a+4-4a^2-4a>0\\a< >0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-12a>-4\\a< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a< >0\\a< \dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x+5}}=\dfrac{2\sqrt{x}+10}{\sqrt{x}+5}+\dfrac{7}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\)
Để \(A\) ∈ \(Z\) thì \(\dfrac{7}{\sqrt{x}+5}\) phải ∈ \(Z\)
=> \(\sqrt{x}+5\) ∈ \(Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
# Với \(\sqrt{x}+5=-7=>\sqrt{x}=-12\)(Loại)
#Với \(\sqrt{x}+5=-1=>\sqrt{x}=-6\)(Loại)
#Với \(\sqrt{x}+5=1=>\sqrt{x}=-4\left(Loại\right)\)
#Với \(\sqrt{x}+5=7=>\sqrt{x}=2< =>x=4\left(Nhận\right)\)
Vậy \(x=4\) thì \(A\)∈\(Z\)
\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}3\)
\(Ta\) \(Có\) : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}=\sqrt[3]{\dfrac{a^6}{ab.ab\left(a^2-ab+b^2\right)}}=\dfrac{a^2}{\sqrt[3]{ab.ab.\left(a^2-ab+b^2\right)}}\)
\(Áp\) \(dụng\) \(bđt\) \(AM-GM\)
\(\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}\text{≤}\) \(\dfrac{ab+ab+a^2-ab+b^2}{3}\)
\(=>\dfrac{a^2}{\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}}\) \(\text{≥}\) \(\dfrac{3a^2}{a^2+ab+b^2}\) \(Hay\) \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}\text{≥}\dfrac{3a^2}{a^2+ab+b^2}\)
Tương tự ta cũng có :
\(\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\text{≥}\dfrac{3b^2}{b^2+bc+c^2}\)
\(\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+a^2\right)}}\text{≥}\dfrac{3c^2}{a^2+ac+c^2}\)
\(=>\text{}\text{}\)\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\)
Cần c/m \(\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\) ≥ \(1\)
Ta có : \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
\(< =>3a^2\text{≥}a^2+ab+b^2\) \(< =>2a^2-b\left(a+b\right)\text{≥}0\) (1)
Lại có : \(a^2\text{≥}-b\left(a+b\right)\) (2)
Từ (1) và (2) => \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
Tương tự ta cũng có :
\(\dfrac{b^2}{b^2+bc+c^2}\text{≥}\dfrac{1}{3}\)
\(\dfrac{c^2}{a^2+ac+c^2}\text{≥}\dfrac{1}{3}\)
Do đó \(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\text{≥}1\)
Suy ra : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\)
Đẳng thức xảy ra <=> \(a=b=c=1\)
\(A=\frac{1+2+2^2+2^3+...+2^{100}}{1+2+2^2+2^3+...+2^{99}}\)
\(A=\frac{1+2\left(1+2+2^2+...+2^{99}\right)}{1+1\left(1+2+2^2+...+2^{99}\right)}\)
\(A=\frac{1+2.1}{1+1.1}=\frac{1+2}{1}=\frac{3}{1}=3_{\left(1\right)}\)
Làm tương tự với B ta được :
\(B=\frac{1+3.1}{1+1.1}=\frac{1+3}{1}=\frac{4}{1}=4_{\left(2\right)}\)
Từ ( 1 ) và ( 2 )
\(\Rightarrow B>A\left(4>3\right)\)