Ta có : 2a=3b  => \(a=\frac{3}{2}b\)

           5b=7c  => \(c=\frac{5}{7}b\)

Theo đề bài : 3a+5c-7b=30

=> \(3\left(\frac{3}{2}b\right)+5\cdot\left(\frac{5}{7}b\right)-7b=30\)

<=> \(\frac{9}{2}b+\frac{25}{7}b-7b=30\)

<=> \(\frac{15}{14}b=30\)

<=> b=28

=> a=\(28\cdot\frac{3}{2}=42\); c=\(28\cdot\frac{5}{7}=20\)

Vậy a=42;b=28;c=20

Ta có: 2*x+1/2*x+1/4*x=99

=>      (2+1/2+1/4)*x       =99

=>           11/4*x               =99

=>                 x                   =99:11/4

=>                 x                   =36

Vậy x=36

a) Ta có: \(\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\)\(\frac{2\cdot\left(-1\right)\cdot13\cdot3\cdot3\cdot2\cdot5}{3\cdot\left(-1\right)2\cdot2\cdot\left(-1\right)\cdot5\cdot2\cdot13}\)=\(-3.\)

b)Ta có: \(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}\)=\(\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

      = \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a^2bc+abc^2+ab^2c}{a^2b^2c^2}\)(1)

Mà a+c+b=0(2)

Từ (1)(2)=>\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)      =  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{abc\left(a+b+c\right)}{a^2b^2c^2}\)= \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

=>  \(\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)(đpcm)

Ta có: a+b=1(1)

=> (a+b)³=1

<=> \(a^3+3a^2b+3ab^2+b^3=1\)

<=> \(a^3+b^3+3ab\left(a+b\right)=1\)(2)

Từ (1)(2)=> \(a^3+b^3+3ab=1\)

<=> \(a^3+b^3=1-3ab\)(đpcm)

Is

Theo đề bài : \(\left|\frac{2x+1}{3}\right|=\frac{x+5}{2}\)

TH1: \(\frac{2x+1}{3}=\frac{x+5}{2}\)

=> \(4x+2=3x+15\)

<=> x=13

TH2: \(-\frac{2x+1}{3}=\frac{x+5}{2}\)

=> \(-4x-2=3x+15\)

<=>7x=-17

<=> \(x=-\frac{17}{7}\)

Vậy x\(\in\left\{-\frac{17}{7};13\right\}\)