Ta có:

\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\)

\(\Rightarrow2A=2.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\right)=2.\frac{2015}{2017}\)

\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{4030}{2017}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{4030}{2017}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}=\frac{4030}{2017}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{4030}{2017}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{4030}{2017}\)

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Đề đúng rồi. co giao minh cung vua giang roi

Tên của bạn làm sai

e)

=> (x-2) . (x+7) = ( x-1 ) . ( x +4)

=> x² +7x - 2x -14 = x² - x + 4x - 4

x² + 5x - 14 = x² + 3x - 4

=> 5x - 14  = 3x - 4

=> 5x  - 3x = 14-4

=> 2x         = 10 => x = 10 : 2 => x = 5

c)

=>( x-1) . 7 = ( x + 5 ) . 6

=> 7x - 7 = 6x + 30

=> 7x - 6x=  30 + 7

=> x         = 37

a,x=\(\frac{5}{2}\)

b,x=\(\frac{13}{176}\)

c,x=37

d, x=\(\frac{12}{5}\)

e, x=5

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

=> x \(\in\) {-1;0;1;2;3;4;5;6}

\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)

\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)

\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)

\(\Leftrightarrow-1\le x< 7\)

Mà x nguyên

=>x={-1;0;1;2;3;4;5;6}

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

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)

\(\Leftrightarrow x=-1\)

\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\\ \Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\\ \Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Gọi \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\)là A

Vì A ≠0

⇒\(x+1=0\Rightarrow x=-1\)

Vậy \(x=-1\)

Câu a hình như sai đề mk sửa nha

a)\(A=\left(2x+\frac{1}{3}\right)^4-1\)

         Vì \(\left(2x+\frac{1}{3}\right)^4\ge0\)

      Suy ra:\(\left(2x+\frac{1}{3}\right)^4-1\ge-1\)

                   Dấu = xảy ra khi \(2x+\frac{1}{3}=0\)

                                               \(2x=-\frac{1}{3}\)

                                                \(x=-\frac{1}{6}\)

Vậy Min A=-1 khi \(x=-\frac{1}{6}\)

b)\(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)

    \(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)

           Vì \(-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le0\)

                     Suy ra:\(3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\le3\)

Dấu = xảy ra khi \(\frac{4}{9}x-\frac{2}{15}=0\)

                            \(\frac{4}{9}x=\frac{2}{15}\)

                            \(x=\frac{3}{10}\)

     Vậy Max B=3 khi \(x=\frac{3}{10}\)

\(1a,\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{12x^2+12x+3}{15}-\frac{5x^2-10x+5}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(x=-\frac{1}{12}\)

Vậy ................

\(b,\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

\(\Leftrightarrow\frac{5\left(7x-1\right)}{30}+\frac{30.2x}{30}=\frac{6\left(16-x\right)}{30}\)

\(\Leftrightarrow35x-5+60x=96-6x\)

\(\Leftrightarrow101x=101\)

\(\Leftrightarrow x=1\)

Vậy ....................

Bài 1:

c) \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{8.\left(x-2\right)^2}{8.3}-\frac{3.\left(2x-3\right).\left(2x+3\right)}{3.8}+\frac{4.\left(x-4\right)^2}{4.6}=0\)

\(\Leftrightarrow\frac{8.\left(x^2-4x+4\right)}{24}-\frac{3.\left(4x^2-9\right)}{24}+\frac{4.\left(x^2-8x+16\right)}{24}=0\)

\(\Rightarrow8.\left(x^2-4x+4\right)-3.\left(4x^2-9\right)+4.\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-\left(12x^2-27\right)+4x^2-32x+64=0\)

\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

\(\Leftrightarrow123-64x=0\)

\(\Leftrightarrow64x=123-0\)

\(\Leftrightarrow64x=123\)

\(\Leftrightarrow x=123:64\)

\(\Leftrightarrow x=\frac{123}{64}.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{123}{64}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x+3=0\)

\(\Leftrightarrow x=12\)

Vậy phương trình có nghiệm là x = 12

\(\Leftrightarrow\dfrac{6}{5}< \dfrac{2x-3}{2}< \dfrac{12}{5}\)

=>12<5(2x-3)<24

\(\Leftrightarrow5\left(2x-3\right)\in\left\{15;20\right\}\)

\(\Leftrightarrow2x-3=3\)

hay x=3

bn đăng từng câu 1 thôi nhe