Khẳng định nào sau đây sai?

1.

a) \(\frac{4}{5} - \frac{6}{4}x = 2\frac{1}{2}\)

\(\frac{4}{5} - \frac{6}{4}x = \frac{5}{2}\)

\[\frac{6}{4}x = \frac{4}{5} - \frac{5}{2}\]

\[\frac{6}{4}x = \frac{{ - 17}}{{10}}\]

\[x = \frac{{ - 17}}{{10}}:\frac{6}{4}\]

\(x = \frac{{ - 17}}{{15}}\)

Vậy \(x = \frac{{ - 17}}{{15}}\).

b) \(\left( {x + 1} \right)\left( {2\sqrt x - 6} \right) = 0\)

Trường hợp 1: \(x + 1 = 0\)

                     \(x = 0 - 1\)

                     \(x = - 1\)

Trường hợp 2: \(2\sqrt x - 6 = 0\)

\(2\sqrt x = 6\)

\(\sqrt x = 3\)

\(x = {3^2}\)

\(x = 9\)

Vậy \(x \in \left\{ { - 1;\,\,9} \right\}\).

2.

a) \(\frac{3}{8} - \frac{{25}}{{16}} + \frac{5}{8} - \frac{7}{{16}} + \left| { - \frac{5}{4}} \right|\) \( = \left( {\frac{3}{8} + \frac{5}{8}} \right) - \left( {\frac{7}{{16}} + \frac{{25}}{{16}}} \right) + \left| { - \frac{5}{4}} \right|\)         

\( = \frac{8}{8} - \frac{{32}}{{16}} + \frac{5}{4} = 1 - 2 + \frac{5}{4}\)\( = \frac{4}{4} - \frac{8}{4} + \frac{5}{4} = \frac{1}{4}\).        

b) \(\sqrt {\frac{{49}}{{64}}} .{\left( { - 2} \right)^3} - \sqrt {\frac{{36}}{{81}}} :{\left( {\frac{{ - 1}}{3}} \right)^2}\)\( = \frac{7}{8}.\left( { - 8} \right) - \frac{6}{9}:\frac{1}{9}\)

\( =  - 7 - \frac{6}{9}.\frac{9}{1}\)\( = - 7 - 6 = - 13\).

\(A = 1 + 2 + {2^2} + {2^3} + ... + {2^{100}} = {2^0} + {2^1} + {2^2} + {2^3} + ... + {2^{100}}\)

Ta có: \(2A = 2.\left( {{2^0} + {2^1} + {2^2} + {2^3} + ... + {2^{100}}} \right)\)

\(2A = {2.2^0} + {2.2^1} + {2.2^2} + {2.2^3} + ... + {2.2^{100}}\)

\(2A = {2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{101}}\).

Ta có: \(2A - A = \left( {{2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{101}}} \right) - \left( {{2^0} + {2^1} + {2^2} + {2^3} + ... + {2^{100}}} \right)\)

\(A = {2^1} + {2^2} + {2^3} + {2^4} + ... + {2^{101}} - {2^0} - {2^1} - {2^2} - {2^3} - ... - {2^{100}}\)

\(A = - {2^0} + \left( {{2^1} - {2^1}} \right) + \left( {{2^2} - {2^2}} \right) + \left( {{2^3} - {2^3}} \right) + ... + \left( {{2^{100}} - {2^{100}}} \right) + {2^{101}}\)

\(A = {2^{100}} - {2^0} = {2^{100}} - 1\).

Vậy \(A = {2^{100}} - 1\).