Mọi người giúp em câu c bài 3 với câu c,d bài 4 với ạ 
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HL
1
SV
Sinh Viên NEU
CTVVIP
19 tháng 11 2023
III
1 It's colder today than yesterday
2 It takes 4 hours to travel by car and fives hours by train
3 We were busier at work today than everyday
4 Jane's sister cooks worse than her
5 Nobody in this team can play football as well as Tom
IV
1 D
2 A
CC
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2020
\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)
\(=\sqrt{6}-\sqrt{2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=0\)
\(\Rightarrow a=-b\Rightarrow a^5+b^5=0\)
HN
0
27 tháng 1 2022
#include <bits/stdc++.h>
using namespace std;
long long a[50],i,n,t;
int main()
{
cin>>n;
for (i=1; i<=n; i++)
{
cin>>a[i];
if (a[i]<0 || a[i]>10) cin>>a[i];
}
for (i=1; i<=n; i++) cout<<a[i]<<" ";
cout<<endl;
t=0;
for (i=1; i<=n; i++) t=t+a[i];
cout<<t;
return 0;
}
NL
1
7 tháng 1 2022
Bài 1:
a: \(=-10x^3+20x^4-5x\)
b: \(=\dfrac{1}{3}a^2b+7a^5-1\)
c: \(=a^3+8+25-a^3=33\)
d: \(=x^2-16+8-x^3=-x^3+x^2-8\)
e: \(=a^3+1+8-a^3=9\)
f: \(=\dfrac{7-2x+4x-8}{2x+3}=\dfrac{2x-1}{2x+3}\)
g: \(=\dfrac{3}{2\left(x+3\right)}-\dfrac{2}{x\left(x+3\right)}\)
\(=\dfrac{3x-4}{2x\left(x+3\right)}\)
2 tháng 10 2021
a) \(P=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{9}+5}{\sqrt{9}-2}=\dfrac{3+5}{3-2}=8\)
b) \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{5\sqrt{x}-2}{4-x}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c) \(M=\dfrac{Q}{P}=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+5}=\dfrac{\sqrt{x}}{\sqrt{x}+5}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}< 3\sqrt{x}+15\Leftrightarrow\sqrt{x}>-15\left(đúng\forall x\ge0,x\ne4\right)\)
d) \(M=\dfrac{\sqrt{x}}{\sqrt{x}+5}=1-\dfrac{5}{\sqrt{x}+5}\in Z\)
\(\Rightarrow\sqrt{x}+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(x\ge0,x\ne4\)
\(\Rightarrow x\in\left\{0\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 10 2021
\(x^6+x^4-3x^2-4x+6\)
\(=\left(x^6+2x^5+4x^4+6x^3+5x^2\right)-\left(2x^5+4x^4+8x^3+12x^2+10x\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)
\(=x^2\left(x^4+2x^3+4x^2+6x+5\right)-2x\left(x^4+2x^3+4x^2+6x+5\right)+\left(x^4+2x^3+4x^2+6x+5\right)+1\)
\(=\left(x^4+2x^3+4x^2+6x+5\right)\left(x^2-2x+1\right)+1\)
\(=\left[\left(x^4+2x^3+x^2\right)+3\left(x^2+2x+1\right)+2\right]\left(x-1\right)^2+1\)
\(=\left[\left(x^2+x\right)^2+3\left(x+1\right)^2+2\right]\left(x-1\right)^2+1\ge1\)
Dấu "=" xảy ra khi \(x=1\)
Bài 3:
c) Ta có: \(\dfrac{2-x}{5}=\dfrac{x+4}{7}\)
\(\Leftrightarrow14-7x=5x+20\)
\(\Leftrightarrow-7x-5x=20-14\)
\(\Leftrightarrow-12x=6\)
hay \(x=-\dfrac{1}{2}\)