Bài 2:

#include <bits/stdc++.h>
using namespace std;
double a,b,c,delta,x1,x2;
int main()
{
    //freopen("PTB2.inp","r",stdin);
    //freopen("PTB2.out","w",stdout);
    cin>>a>>b>>c;
    delta=(b*b-4*a*c);
    if (delta<0) cout<<"-1";
    if (delta==0) cout<<fixed<<setprecision(5)<<(-b/(2*a));
    if (delta>0)
    {
        x1=(-b-sqrt(delta))/(2*a);
        x2=(-b+sqrt(delta))/(2*a);
        cout<<fixed<<setprecision(5)<<x1<<" "<<fixed<<setprecision(5)<<x2;
    }
    return 0;
}

Câu 18:

Ta có: \(3\sqrt{8a}+\dfrac{1}{4}\sqrt{\dfrac{32a}{25}}-\dfrac{a}{\sqrt{3}}\cdot\sqrt{\dfrac{3}{2a}}-\sqrt{2a}\)

\(=6\sqrt{2a}-\sqrt{2a}+\dfrac{1}{4}\cdot\dfrac{4\sqrt{2a}}{5}-\dfrac{a}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{2a}}\)

\(=5\sqrt{2a}+\dfrac{1}{5}\sqrt{2a}-\dfrac{1}{2}\sqrt{2a}\)

\(=\dfrac{47}{10}\sqrt{2a}\)

Chọn C

Câu 18
\(=3\sqrt{4}.\sqrt{2a}+\frac{1}{4}\sqrt{\frac{16}{25}}.\sqrt{2a}-\sqrt{\frac{a^2}{3}}.\sqrt{\frac{3}{2a}}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{a}{2}}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\sqrt{\frac{1}{4}}.\sqrt{2a}-\sqrt{2a}\)

\(=6\sqrt{2a}+\frac{1}{5}\sqrt{2a}-\frac{1}{2}\sqrt{2a}-\sqrt{2a}=\frac{47}{10}\sqrt{2a}\)

Đáp án C.

Câu 19:

\(=2\sqrt{a}-\sqrt{(3a)^2}.\sqrt{a}+a\sqrt{a}.\sqrt{16}+\sqrt{\frac{4}{a^4}.36a^5}\)

\(=2\sqrt{a}-3a\sqrt{a}+4a\sqrt{a}+\sqrt{144a}\)

\(=2\sqrt{a}+a\sqrt{a}+\sqrt{144}.\sqrt{a}=2\sqrt{a}+a\sqrt{a}+12\sqrt{a}=14\sqrt{a}+a\sqrt{a}\)

Đáp án A.

Câu 1: 

\(f\left(-x\right)=\left(-x\right)^4-3\cdot\left(-x\right)^2+1\)

\(=x^4-3x^2+1=f\left(x\right)\)

Vậy: f(x) là hàm số chẵn

\(c1:D=R\Rightarrow\forall x\in D\Rightarrow-x\in D\)

\(\Rightarrow f\left(-x\right)=x^4-3x^2+1=f\left(x\right)\Rightarrow hàm\) \(số\) \(chẵn\)

\(c2:cm:\overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{MN}\)

\(M,N\) \(trung\) \(điểm\) \(AB,CD\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}+\overrightarrow{MB}=\overrightarrow{0}\\\overrightarrow{DN}+\overrightarrow{CN}=\overrightarrow{0}\end{matrix}\right.\)

\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}\)

\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN}\)

\(\Rightarrow2\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{MA}+\overrightarrow{BC}+\overrightarrow{AD}+\overrightarrow{CN}+\overrightarrow{DN}\)

\(=\overrightarrow{0}+\overrightarrow{BC}+\overrightarrow{AD}+\overrightarrow{0}\Rightarrow2\overrightarrow{MN}=\overrightarrow{AD}+\overrightarrow{BC}\left(đpcm\right)\)

\(c3:D=R\text{[}-5;\text{+∞.)}\)

\(f\left(x\right)\) \(nghịch\) \(biến\Leftrightarrow a< 0\)

\(\Rightarrow-5\le a< 0\left(a\in Z\right)\Rightarrow a=\left\{-5;-4;-3;-2;-1\right\}\)

\(c4:\) có công thức \(X\cap Y=X+Y-X\cup Y=25+20-36=9\)

1.

\(2sinx+cosx=4\)

\(\Leftrightarrow\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sinx+\dfrac{1}{\sqrt{5}}cosx\right)=4\)

\(\Leftrightarrow sin\left(x+arccos\dfrac{2}{\sqrt{5}}\right)=\dfrac{4}{\sqrt{5}}>1\)

\(\Rightarrow2sinx+4cosx-4\ne0\)

Khi đó: 

\(2P.sinx+P.cosx-4P=sinx-2cosx-3\)

\(\Leftrightarrow\left(2P-1\right)sinx+\left(P+2\right)cosx=4P-3\)

Phương trình có nghiệm khi:

\(\left(2P-1\right)^2+\left(P+2\right)^2\ge\left(4P-3\right)^2\)

\(\Leftrightarrow4P^2-4P+1+P^2+4P+4\ge16P^2+9-24P\)

\(\Leftrightarrow11P^2-24P+4\le0\)

\(\Leftrightarrow\dfrac{2}{11}\le P\le2\)

\(\Rightarrow maxP=2\)

tải quanda về nha

mình chịu

Bạn nên đăng những câu khó nhất hoặc bạn lọc ra những câu tương tự nhau để bản thân có thể vận dụng nhé!