#include <bits/stdc++.h>

using namespace std;

int main()
{
    string S;
    int x,sum=0;
    cin >> S;
    cin >> x;
    for (int i=0;i<=S.length()-1;i++)
    {
        sum+=((int(S[i])-97)+x)%26;
    }
    cout << sum;
}

Từ pt suy ra \(x\ge0\).

PT \(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}=2x\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+\left|\sqrt{2x-1}+1\right|=2x\). (*)

+) \(\sqrt{2x-1}-1\ge0\Leftrightarrow x\ge1\): Khi đó (*) tương đương \(2\sqrt{2x-1}=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) (thoả mãn)

+) \(\sqrt{2x-1}-1< 0\Leftrightarrow x< 1\): Khi đó (*) tương đương \(2=2x\Leftrightarrow x=1\), vô lí.

Vậy x = 1

\(2x^2-7x+5=\left(2x^2-2x\right)-\left(5x-5\right)=2x\left(x-1\right)-5\left(x-1\right)=\left(2x-5\right)\left(x-1\right)\)

\(3x^2+5x+2=\left(3x^2+3x\right)+\left(2x+2\right)=3x\left(x+1\right)+2\left(x+1\right)=\left(3x+2\right)\left(x+1\right)\)

Ta có: \(\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)=\left(a^2+ab+bc+ca\right)\left(b^2+ab+bc+ca\right)\left(c^2+ab+bc+ca\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

4: Áp dụng bđt AM - GM và Cauchy - Schwarz ta có:

\(a^3\sqrt[3]{\left(\dfrac{bc}{b^2-bc+c^2}\right)^2}+b^3\sqrt[3]{\left(\dfrac{ca}{c^2-ca+a^2}\right)^2}+c^3\sqrt[3]{\left(\dfrac{ab}{a^2-ab+b^2}\right)^2}\)

\(=\dfrac{a^3bc}{\sqrt[3]{bc\left(b^2-bc+c^2\right)^2}}+\dfrac{b^3ca}{\sqrt[3]{ca\left(c^2-ca+a^2\right)^2}}+\dfrac{c^3ab}{\sqrt[3]{ab\left(a^2-ab+b^2\right)^2}}\)

\(\ge\dfrac{3a^2}{2\left(b^2+c^2\right)-bc}+\dfrac{3b^2}{2\left(c^2+a^2\right)-ca}+\dfrac{3c^2}{2\left(a^2+b^2\right)-ab}\)

\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{4\left(a^2b^2+b^2c^2+c^2a^2\right)-a^2bc-b^2ca-c^2ab}\).

Ta chứng minh \(\dfrac{3\left(a^2+b^2+c^2\right)^2}{4\left(a^2b^2+b^2c^2+c^2a^2\right)-a^2bc-b^2ca-c^2ab}\ge3\Leftrightarrow a^4+b^4+c^4+a^2bc+b^2ca+c^2ab\ge2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Leftrightarrow a^2\left(a-b\right)\left(a-c\right)+b^2\left(b-c\right)\left(b-a\right)+c^2\left(c-a\right)\left(c-b\right)\ge0\). (đúng theo Schur bậc 4)

Vậy ta có đpcm.

5:

\(BĐT\Leftrightarrow\dfrac{2\left(x+2y\right)^2}{x^2+xy+y^2}+\dfrac{\left(x+2y\right)^2}{3y^2}\ge9\)

\(\Leftrightarrow\dfrac{2\left(y-x\right)\left(2x+y\right)}{x^2+xy+y^2}+\dfrac{\left(x-y\right)\left(x+5y\right)}{3y^2}\ge0\)

\(\Leftrightarrow\left(x-y\right).\dfrac{\left(x+5y\right)\left(x^2+xy+y^2\right)-6y^2\left(2x+y\right)}{3y^2\left(x^2+xy+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(x^2+7xy+7y^2\right)}{3y^2\left(x^2+xy+y^2\right)}\ge0\). (luôn đúng)

Xét các trường hợp:

+) x = 0: Khi đó \(y^2=2^0+3=4\Rightarrow y=2\).

+) x = 1: Khi đó \(y^2=2^1+3=5\), vô lí

+) x > 1: Khi đó \(2^x⋮4\Rightarrow y^2=2^x+3\equiv3\left(mod4\right)\), vô lí vì số chính phương khi chia cho 4 dư 0 hoặc 1.

Vậy x = 0; y = 2.