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\(x^2-\frac{7}{6}x+\frac{1}{3}=0\)

=> \(\left[x^2-2\cdot x\cdot\frac{7}{12}+\left(\frac{7}{12}\right)^2\right]-\frac{1}{144}=0\)

=> \(\left(x-\frac{7}{12}\right)^2-\frac{1}{144}=0\)

=> \(\left(x-\frac{7}{12}\right)^2-\left(\frac{1}{12}\right)^2=0\)

=> \(\left(x-\frac{7}{12}+\frac{1}{12}\right)\left(x-\frac{7}{12}-\frac{1}{12}\right)=0\)

=> \(\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)

=> x = 1/2 hoặc x = 2/3

Bài làm :

\(\frac{125^6.3^{61}.8^{10}}{4^{15}.25^9.9^{30}}\)

\(=\frac{\left(5^3\right)^6.3^{61}.\left(2^3\right)^{10}}{\left(2^2\right)^{15}.\left(5^2\right)^9.\left(3^2\right)^{30}}\)

\(=\frac{5^{18}.3^{61}.2^{30}}{2^{30}.5^{18}.3^{60}}\)

\(=3\)

Học tốt nhé

        Bài làm :

Ta có :

\(\frac{125^6.3^{61}.8^{10}}{4^{15}.25^9.9^{30}}\)

\(=\frac{\left(5^3\right)^6.3^{61}.\left(2^3\right)^{10}}{\left(2^2\right)^{15}.\left(5^2\right)^9.\left(3^2\right)^{30}}\)

\(=\frac{5^{18}.3^{61}.2^{30}}{2^{30}.5^{18}.3^{60}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

               Bài làm :

Ta có hình vẽ :

Ta có :

\(\hept{\begin{cases}\widehat{AOD}+\widehat{BOC}=100^o\\\widehat{AOD}=\widehat{BOC}\left(\text{2 góc đối đỉnh}\right)\end{cases}\Rightarrow\widehat{AOD}=\widehat{BOC}=\frac{100}{2}=50^O}\)

\(\Rightarrow\widehat{BOD}=\widehat{COA}=180-50=130^O\)

Vì \(\widehat{AOD}=\widehat{BOC}\)(2 góc đối đỉnh) mà \(\widehat{AOD}+\widehat{BOC}=100^0\Rightarrow\widehat{AOD}=\widehat{BOC}=\frac{100^0}{2}=50^0\)

Tương tự: \(\widehat{AOC}=\widehat{BOD}\)(2 góc đối đỉnh) mà \(\widehat{AOC}+\widehat{AOD}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{BOD}=\widehat{AOC}=180^0-50^0=130^0\)

\(\left(\frac{1}{4}-\frac{2}{5}\right):\frac{23}{25}+\left(\frac{3}{4}-\frac{3}{5}\right):\frac{23}{25}\)

\(=\left(\frac{1}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right).\frac{25}{23}\)

\(=-\frac{25}{23}\)

20/23

\(\frac{3}{5}.\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right).\frac{3}{5}\)

\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right)\text{]}\)

\(=\frac{3}{5}.\text{[}\frac{5}{3}-\frac{2}{7}-\frac{7}{3}+\frac{3}{7}\text{]}\)

\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{7}{3}\right)-\left(\frac{2}{7}-\frac{3}{7}\right)\text{]}\)

\(=\frac{3}{5}.\text{[}\frac{-2}{3}-\frac{-1}{7}\text{]}\)

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

\(=\frac{3}{5}.\left(\frac{-14}{21}+\frac{3}{21}\right)\)

\(=\frac{3}{5}.\frac{-11}{21}\)

\(=\frac{3.\left(-11\right)}{5.21}\)

\(=\frac{-11}{5.7}=\frac{-11}{35}\)

Chúc bạn học tốt

            Bài làm :

\(1\text{)}8x=7,8x+25\Leftrightarrow8x-7,8x=25\Leftrightarrow0,2x=25\Leftrightarrow x=25\div0,2=125\)

\(2\text{)}\left|x+7\right|-3=2\Leftrightarrow\left|x+7\right|=5\Leftrightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)

\(3\text{)}\left(2x-5\right)\div3=7\Leftrightarrow2x-5=21\Leftrightarrow2x=26\Leftrightarrow x=13\)

a,Ta có :     8x=7,8x+25

   <=> 8x-7,8x=25

               0,2x=25

                    x=25:0,2

                    x=25:1/5

                    x=25.5

             =>   x=125

b,  Ta có :  |x+7|-3=2

          <=>    |x+7|=2+3

                    |x+7|=5

TH1: x+7=5                                                                             TH2:   x+7=-5                      

             x=5-7                                                                                        x=(-5)-7

             x=-2                                                                                          x=-12

c,  Ta có: [2x-5]:3=7

        <=>      2x-5=7.3

                    2x-5=21

                        2x=21+5

                        2x=26

                          x=26:2

                     => x=13

Ta có: \(\frac{2000}{-2001}=-\frac{2000}{2001}=-\left(\frac{2001-1}{2001}\right)=-\left(\frac{2001}{2001}-\frac{1}{2001}\right)=-\left(1-\frac{1}{2001}\right)=-1+\frac{1}{2001}\)

       \(-\frac{2003}{2002}=-\left(\frac{2002+1}{2002}\right)=-\left(\frac{2002}{2002}+\frac{1}{2002}\right)=-\left(1+\frac{1}{2002}\right)=-1-\frac{1}{2002}\)

Vì \(\frac{1}{2001}>-\frac{1}{2002}\) nên \(-1+\frac{1}{2001}>-1-\frac{1}{2002}\)

hay \(\frac{2000}{-2001}>-\frac{2003}{2002}\)

Ta có : \(\left|2x+1\right|\ge0\forall x\)

\(\Rightarrow\left|2x+1\right|=2x+1\)

\(\Rightarrow2x+1=3-4x\)

\(2x+4x=3-1\)

\(6x=2\)

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

Vậy \(x=\frac{1}{3}\)

| 2x + 1 | = 3 - 4x (*)

+) Với x < -1/2 

(*) <=> -( 2x + 1 ) = 3 - 4x

     <=> -2x - 1 = 3 - 4x

     <=> -2x + 4x = 3 + 1

     <=> 2x = 4

     <=> x = 2 ( không thỏa mãn )

+) Với x ≥ -1/2

(*) <=> 2x + 1 = 3 - 4x

     <=> 2x + 4x = 3 - 1

     <=> 6x = 2

     <=> x = 2/6 = 1/3

Vậy x = 1/3