Ta có : m và n là các số nguyên dương

Và \(A=\frac{2+4+6+...+2m}{m}=\frac{2.\left(1+2+....+m\right)}{m}=\frac{2.\left(m-1\right).m}{m}=2.\left(m-1\right)\)

B = \(\frac{2+4+6+...+2n}{n}=\frac{2.\left(1+2+3+...+n\right)}{n}=\frac{2.\left(n-1\right).n}{n}=2.\left(n-1\right)\)

Mà A < B 

Nên 2 . ( m - 1 ) < 2 . ( n - 1 )

Do đó m - 1 < n - 1 

Và m < n

Vậy m < n

1/

\(\frac{2n+1}{n-3}+\frac{3n-5}{n-3}-\frac{4n-5}{n-3}=\frac{2n+1+\left(3n-5\right)-\left(4n-5\right)}{n-3}=\frac{2n+1+3n-5-4n+5}{n-3}=\frac{n+1}{n-3}=\frac{n-3+4}{n-3}=\frac{n-3}{n-3}+\frac{4}{n-3}=1+\frac{4}{n-3}\)

Để S là số nguyên <=> n - 3 thuộc Ư(4) = {1;-1;2;-2;4;-4}

Vậy...

câu 2 dễ ẹt

2/

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

...........

\(\frac{1}{99^2}< \frac{1}{98.99}=\frac{1}{98}-\frac{1}{99}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}=1-\frac{1}{99}< 1\)

=> ĐPCM

3/

\(\frac{m}{9}-\frac{3}{n}=\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{m}{9}-\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{2m}{18}-\frac{1}{18}\)

=> \(\frac{3}{n}=\frac{2m-1}{18}\)

=> n(2m - 1) = 3.18 = 54

=> n và 2m - 1 thuộc Ư(54) = {1;-1;2;-2;3;-3;6;-6;9;-9;18;-18;27;-27;54;-54}

Mà 2m - 1 là số lẻ => 2m - 1 thuộc {1;-1;3;-3;9;-9;27;-27}

                                  n thuộc {2;-2;6;-6;18;-18;54;-54}

Ta có bảng:

Vậy các cặp (m;n) là (1;54) ; (0;-54) ; (2;18) ; (-1;-18) ; (5;6) ; (-4;-6) ; (14;2) ; (-13;-2)

bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)

Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)

\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)

\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)

\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)

\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)

\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)

Vì m+n+p=0=>m+n=-p

\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)

\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)

\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)

\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)

\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)

\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)

\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)

\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)

\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)

\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)

\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)

Từ (1),(2),(3) suy ra :

\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)

\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)

*Tới đây để tính được m³+n³+p³,ta cần CM được bài toán phụ sau:

Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)

Từ m+n+p=0=>m+n=-p

Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)

\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)

Vậy ta đã CM được bài toán phụ

*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)

Vậy A=9

bài 2)

a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:

\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)

\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)

\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)

suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)

Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)

\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)

...........................

\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)

\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)

\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)

Vậy A=2036/37

b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 1⁴=1 mà

Nhận thấy các thừa số của B có dạng tổng quát:

\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)

\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)

\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)

Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)

Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)

Vậy B=1/221

1. Tổng các hệ số của đa thức là: 1²⁰⁰⁴.2²⁰⁰⁵=2²⁰⁰⁵

2.Cần chứng minh x⁴+x³+x²+x+1=0 vô nghiệm.

Nhận thấy x = 1 không là nghiệm của phương trình .

Nhân cả hai vế của pt cho (x−1)≠0 được : 

(x−1)(x⁴+x³+x²+x+1)=0⇔x⁵−1=0⇔x=1(vô lí)

Vậy pt trên vô nghiệm.

1. Tổng các hệ số của đa thức là: 

1²⁰¹⁴ . 2²⁰¹⁵ = 2²⁰¹⁵

2 . Cần chứng minh. 

\(x4 + x3 + x2 + x + 1 = 0\)

Vô nghiệm. 

Ta nhận thấy \(x + 1 \) không là nghiệm của phương trình. 

Nhân cả hai vế của phương trình cho:

\(( x - 1 ) \) \(\ne\) \(0\) được :

\(( x-1). (x4+x3+x2+x+1)=0\)

\(\Leftrightarrow\)\(5x-1=0\) \(\Leftrightarrow\) \(x = 1\)

Vô lí. 

Vậy phương trình trên vô nghiệm. 

Cảm ơn nhiều ạ

\(\frac{x+7}{3}+\frac{x+5}{4}=\frac{x+3}{5}+\frac{x+1}{6}\)

\(\Rightarrow\frac{x+7}{3}+2+\frac{x+5}{4}+2=\frac{x+3}{5}+2+\frac{x+1}{6}+2\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}=\frac{x+13}{5}+\frac{x+13}{6}\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}-\frac{x+13}{5}-\frac{x+13}{6}=0\)

\(\Rightarrow\left(x+13\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì \(\left(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}\right)\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>0\)

\(\Rightarrow x+13=0\Leftrightarrow x=-13\)

\(\frac{x+m}{n+p}+\frac{x+n}{p+m}+\frac{x+p}{n+m}+3=0\)

\(\Rightarrow\frac{x+m}{n+p}+1+\frac{x+n}{p+m}+1+\frac{x+p}{n+m}+1=0\)

\(\Rightarrow\frac{x+m+n+p}{n+p}+\frac{x+m+n+p}{p+m}+\frac{x+m+n+p}{n+m}=0\)

\(\Rightarrow\left(x+m+n+p\right)\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)=0\)

Vì m,n,p là số dương nên \(\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)>0\)

\(\Rightarrow x+m+n+p=0\Rightarrow x=-\left(m+n+p\right)\)

\(\frac{5x+\frac{3x-4}{5}}{15}=\frac{\frac{3-x}{15}+7x}{5}+1-x\)

\(\Rightarrow\frac{\frac{25x+3x-4}{5}}{15}=\frac{\frac{3-x+105x}{15}}{5}+1-x\)

\(\Rightarrow\frac{\frac{28x-4}{5}}{15}=\frac{\frac{3+104x}{15}}{5}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x}{75}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x+75-75x}{75}\)

\(\Rightarrow\frac{28x-4}{75}=\frac{78+29x}{75}\)

\(\Rightarrow28x-4=78+29x\)

\(\Rightarrow x=-82\)

có cần full ko :3

có chứ anh

\(a,\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}\)

\(\Leftrightarrow\frac{a+2b-3c}{2+6-12}=\frac{-20}{-4}=5\)

Vậy: \(a=5.2=10\)

\(b=5.6:2=15\)

\(c=5.12:3=20\)

sao con gai ma de anh con trai vay cau trac nghich lam nhi

b, Ta có :

\(\frac{m}{2}=\frac{n}{3}=\frac{p}{4}=\frac{m^2}{4}=\frac{n^2}{9}=\frac{p^2}{16}\)

\(\Rightarrow\frac{m^2}{4}=\frac{n^2}{9}=\frac{2p^2}{32}=\frac{m^2-n^2+2p^2}{4-9+32}=\frac{108}{27}=4\)

Do đó :

\(\frac{m^2}{4}=4\Leftrightarrow m^2=16\Leftrightarrow m=\pm4\)

\(\frac{n^2}{9}=4\Leftrightarrow n^2=36\Leftrightarrow n=\pm6\)

\(\frac{2p^2}{32}=4\Leftrightarrow2p^2=128\Leftrightarrow p^2=64\Leftrightarrow p=\pm8\)

Vậy : ...