+) \(P=\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{x^2+xy+y^2}\)

\(\ge\text{Σ}\frac{x^2}{y^2+\frac{y^2+z^2}{2}+z^2}=\frac{2}{3}\text{Σ}\frac{x^2}{y^2+z^2}\)

+) Đặt \(a=x^2;b=y^2;c=z^2\)

Ta có: \(A=\text{Σ}\frac{x^2}{y^2+z^2}=\text{Σ}\frac{a}{b+c}=\text{Σ}\frac{a^2}{ab+ac}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3}{2}\)(BDT Nesbitt)

Vậy \(P=\frac{2}{3}A\ge1\)

Dấu = xảy ra khi x = y = z

bệnh tật

Trả lời : 

Muỗi anophen truyền bệnh sốt rét cho còn người

Học tốt!!!

+) \(P=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2}\)

\(\le\frac{1-x^2+\frac{3}{4}}{\sqrt{3}}+\frac{1-y^2+\frac{3}{4}}{\sqrt{3}}+\frac{1-z^2+\frac{3}{4}}{\sqrt{3}}\)

\(=\frac{\frac{21}{4}-x^2-y^2-z^2}{\sqrt{3}}\)

+) \(1=xy+yz+xz+2xyz\le\frac{\left(x+y+z\right)^2}{3}+\frac{2\left(x+y+z\right)^3}{27}\)

Đặt \(a=x+y+z\), ta được \(2a^3+9a^2-27\ge0\Leftrightarrow\left(2a-3\right)\left(a+3\right)^2\ge0\Rightarrow a\ge\frac{3}{2}\)

+) \(A=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{\frac{9}{4}}{3}=\frac{3}{4}\)

+) \(P\ge\frac{\frac{21}{4}-A}{\sqrt{3}}=\frac{\frac{21}{4}-\frac{3}{4}}{\sqrt{3}}=\frac{9}{2\sqrt{3}}=\frac{3\sqrt{3}}{2}\)

Dấu = xảy ra khi x = y = z = 1/2

a) 60-3(x-2)=51

         3(x-2)=60-51

         3(x-2)=9

            x-2 = 9:3

           x-2 = 3

          x     = 3+2

         x     5

b)   4x-20=2^5:2^2

      4x-20=8

      4x     =8+20

      4x     =28

        x     =28:4

       x     = 7

các bạn giúp mình với mình đang cần đáp án gấp

1) a.Ta có \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)

Vì \(3\inℤ\Rightarrow\frac{21}{n-4}\inℤ\Rightarrow21⋮n-4\Rightarrow n-4\inƯ\left(21\right)\)

=> \(n-4\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\)

=> \(n\in\left\{5;3;8;1;11;-3;25;-17\right\}\)

b) Ta có B = \(\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)

Vì \(3\inℤ\Rightarrow\frac{8}{2n-1}\inℤ\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)(1)

lại có với mọi n nguyên => 2n \(⋮\)2 => 2n - 1 không chia hết cho 2 (2)

Kết hợp (1) ; (2) => \(2n-1\in\left\{1;-1\right\}\Rightarrow n\in\left\{1;0\right\}\)

2) Ta có : \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

=> \(\frac{20+xy}{4x}=\frac{1}{8}\)

=> 4x = 8(20 + xy)

=> x = 2(20 + xy)

=> x = 40 + 2xy

=> x - 2xy = 40

=> x(1 - 2y) = 40

Nhận thấy : với mọi y nguyên => 1 - 2y là số không chia hết cho 2 (1)

mà x(1 - 2y) = 40

=> 1 - 2y \(\inƯ\left(40\right)\)(2)

Kết hợp (1) (2) => \(1-2y\in\left\{1;5;-1;-5\right\}\)

Nếu 1 - 2y = 1 => x = 40

=> y = 0 ; x = 40

Nếu 1 - 2y = 5 => x = 8

=> y = -2 ; x = 8 

Nếu 1 - 2y = -1 => x = -40

=> y = 1 ; y = - 40

Nếu 1 - 2y = -5 => x = -8

=> y = 3 ; x =-8

Vậy các cặp (x;y) thỏa mãn là : (40 ; 0) ; (8; - 2) ; (-40 ; 1) ; (-8 ; 3)

4) \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{-4}{3}}=\frac{-\frac{5}{60}}{\frac{2}{5}}=-\frac{5}{60}:\frac{2}{5}=-\frac{5}{24}\)

b) \(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)

\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

c) \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

1. He shouldn't play with the ......LIGHTER............ because he may get a burn.

2. They shouldn't jump into the river because thay may ...down.................... .

3. She should eat a lot of ........FRESH............. vegetable .

4. You should wash your .....HANDS......... before and after meals.

5.Why shouldn't she go ....barefoot................. ?

Điền từ thích hợp trong khung vào chỗ trống:

1. He shouldn't play with the .......lighter........... because he may get a burn.

2. They shouldn't jump into the river because thay may .......down.................. .

3. She should eat a lot of .........fresh............ vegetable .

4. You should wash your ....hands.......... before and after meals.

5.Why shouldn't she go .........barefoot............. ?

\(B=2^{2005}-2^{2004}-2^{2003}-...-2-1\)

\(B=2^{2005}-\left(2^{2004}+2^{2003}+...+2+1\right)\)

Đặt \(A=1+2+...+2^{2004}\) \(\Rightarrow2A=2+2^2+...+2^{2005}\)

\(2A-A=\left(2+2^2+...+2^{2005}\right)-\left(1+2+...+2^{2004}\right)\)

\(A=2^{2005}-1\). Thay A vào B, ta có : 

\(B=2^{2005}-\left(2^{2005}-1\right)=2^{2005}-2^{2005}+1=1\)

Ta có : B = 2²⁰⁰⁵ - 2²⁰⁰⁴ - 2²⁰⁰³ - ... - 2 - 1

= B = 2²⁰⁰⁵ - (2²⁰⁰⁴ + 2²⁰⁰³ + ... + 2 + 1)

Đặt A = 2²⁰⁰⁴ + 2²⁰⁰³ + ... + 2 + 1

=> 2A = 2²⁰⁰⁵ + 2²⁰⁰⁴ + .... + 2² + 2

Lấy 2A trừ A theo vế ta có 

2A - A = (2²⁰⁰⁵ + 2²⁰⁰⁴ + .... + 2² + 2) - (2²⁰⁰⁴ + 2²⁰⁰³ + ... + 2 + 1)

       A = 2²⁰⁰⁵ - 1

Khi đó B = 2²⁰⁰⁵ - (2²⁰⁰⁵ - 1) = 1

f=61/36

\(f+\frac{5}{9}=\frac{9}{4}\)

\(f=\frac{9}{4}-\frac{5}{9}\)

\(f=\frac{61}{36}\)

vậy \(\frac{61}{36}\)