Chuyên đề tìm x

Chuyên đề tìm x: Dạng 1: Tìm x thông thường

Dạng 1: Tìm x thông thường

Bài 1: Tìm x biết:

.a. \[\frac{x}{2} – \left( {\frac{{3x}}{5} – \frac{{13}}{5}} \right) = \left( {\frac{7}{5} + \frac{7}{{10}}x} \right)\]

b. \[\frac{{13}}{{x – 1}} + \frac{5}{{2x – 2}} = \frac{6}{{3x – 3}}\]

Hướng dẫn giải:

a. \[\frac{x}{2} – \left( {\frac{{3x}}{5} – \frac{{13}}{5}} \right) = \left( {\frac{7}{5} + \frac{7}{{10}}x} \right)\]

\[= > \frac{x}{2} – \frac{{3x}}{5} + \frac{{13}}{5} = \frac{7}{5} + \frac{7}{{10}}x = > \frac{x}{2} – \frac{{3x}}{5} – \frac{{7x}}{{10}} = \frac{7}{5} – \frac{{13}}{5}\]

Suy ra: \[\frac{{ – 4}}{5}x = \frac{{ – 6}}{5} = > x = \frac{{ – 6}}{5}:\frac{{ – 4}}{5} = \frac{3}{2}\]

Vậy: \[x = \frac{3}{2}\]

b. \[\frac{{13}}{{x – 1}} + \frac{5}{{2x – 2}} = \frac{6}{{3x – 3}} = > \frac{{78}}{{6x – 6}} + \frac{{15}}{{6x – 6}} = \frac{{12}}{{6x – 6}}\]

\[= > \frac{{93}}{{6x – 6}} = \frac{{12}}{{6x – 6}} = > 93 = 12\] (Vô lý)

Vậy không có giá trị x thỏa mãn

Bài 2: Tìm x biết:

a. \[\frac{{2x – 3}}{3} + \frac{{ – 3}}{2} = \frac{{5 – 3x}}{6} – \frac{1}{3}\]

b. \[\frac{2}{{3x}} – \frac{3}{{12}} = \frac{4}{5} – \left( {\frac{7}{x} – 2} \right)\]

Hướng dẫn giải:

a. \[\frac{{2x – 3}}{3} + \frac{{ – 3}}{2} = \frac{{5 – 3x}}{6} – \frac{1}{3}\]

\[= > \frac{{4x – 6 + \left( { – 9} \right)}}{6} = \frac{{5 – 3x – 2}}{6} = > 4x – 15 = 3 – 3x = > 7x = 18 = > x = \frac{{18}}{7}\]

b. \[\frac{2}{{3x}} – \frac{3}{{12}} = \frac{4}{5} – \left( {\frac{7}{x} – 2} \right)\]

\[= > \frac{2}{{3x}} – \frac{1}{4} = \frac{4}{5} – \frac{7}{x} + 2 = > \frac{2}{{3x}} + \frac{7}{x} = \frac{4}{5} + \frac{1}{4} + 2\]

\[= > \frac{{23}}{{3x}} = \frac{{61}}{{20}}\]

\[ = > 3x = \frac{{460}}{{61}} = > x = \frac{{460}}{{183}}\]

Bài 3: Tìm x biết:

a. \[\frac{1}{{x – 1}} + \frac{{ – 2}}{3}\left( {\frac{3}{4} – \frac{6}{5}} \right) = \frac{5}{{2 – 2x}}\]

b. \[\frac{9}{{17}}x + 15\frac{{13}}{{17}}x – 20\frac{5}{{17}}x = 16\]

Hướng dẫn giải

a. \[\frac{1}{{x – 1}} + \frac{3}{{10}} = \frac{5}{{2 – 2x}} = > \frac{1}{{x – 1}} + \frac{5}{{2\left( {x – 1} \right)}} = \frac{{ – 3}}{{10}}\]

\[= > \frac{7}{{2\left( {x – 1} \right)}} = \frac{{ – 3}}{{10}}\]

\[ = > 2\left( {x – 1} \right) = – \frac{{70}}{3} = > x – 1 = \frac{{ – 35}}{3} = > x = \frac{{ – 32}}{3}\]

b. \[\left( {\frac{9}{{17}} + 15\frac{{13}}{{17}} – 20\frac{5}{{17}}} \right)x = 16 = > – 4.x = 16 = > x = – 4\]

Bài 4: Tìm x biết:

a. \[720:\left[ {41 – \left( {2x – 5} \right)} \right] = {2^3}.5\]

b. \[6\left( {x + 11} \right) – 7\left( {2 – x} \right) = 26\]

Hướng dẫn giải

a. \[720:\left[ {46 – 2x} \right] = 40 = > 46 – 2x = 18 = > 2x = 46:18 = \frac{{23}}{9} = > x = \frac{{23}}{{18}}\]

b. \[6x + 66 – 14 + 7x = 26 = > 13x = – 26 = > x = – 2\]

Bài 5: Tìm x biết:

a, \[ – 4x\left( {x – 5} \right) – 2x\left( {8 – 2x} \right) = – 3\]

b, \[ – 7\left( {x + 9} \right) – 3\left( {5 – x} \right) = 2\]

HD:

a, \[ = > – 4{x^2} + 20x – 16x + 4{x^2} = – 3 = > 4x = – 3 = > x = \frac{{ – 3}}{4}\]

b, \[ – 7x – 63 – 15 + 3x = 2 = > – 4x – 78 = 2 = > – 4x = 80 = > x = – 20\]

Bài 6: Tìm x biết :

\[ – \frac{7}{4}x\left( {\frac{{33}}{{12}} + \frac{{3333}}{{2020}} + \frac{{333333}}{{303030}} + \frac{{33333333}}{{42424242}}} \right) = 22\]

HD:

Ta có : \[ – \frac{7}{4}x\left( {\frac{{33}}{{12}} + \frac{{33}}{{20}} + \frac{{33}}{{30}} + \frac{{33}}{{42}}} \right) = 22\]

=>\[ – \frac{7}{4}x.33\left( {\frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}}} \right) = 22\]

=>\[ – \frac{7}{4}x.33\left( {\frac{1}{3} – \frac{1}{7}} \right) = 22 = > \frac{{ – 7}}{4}x.33.\frac{4}{{21}} = 22 = > x = – 2\]

Bài 7: Tìm x biết:

\[\frac{1}{{2016}}:2015x = \frac{{ – 1}}{{2015}}\]

HD:

\[\frac{1}{{2016.2015}}x = \frac{{ – 1}}{{2015}} = > x = \frac{{ – 1}}{{2015}}:\frac{1}{{2016.2015}} = – 2016\]

Bài 8: Tìm x biết :

\[2\left( {x – 1} \right) – 3\left( {2x + 2} \right) – 4\left( {2x + 3} \right) = 16\]

Bài 9: Tìm x để biểu thức sau nhận giá trị bằng 0:

\[\frac{{x\left( {x + \frac{1}{2}} \right) – \frac{1}{5}x – \frac{1}{{10}}}}{3}.\frac{{x\left( {x – \frac{1}{2}} \right) – \frac{1}{3}x + \frac{1}{6}}}{5}\]

HD :

Quy đồng trên tử ta có : \[x\left( {x + \frac{1}{2}} \right) – \frac{1}{5}x – \frac{1}{{10}} = 0\]

\[= > 10{x^2} – 3x – 1 = 0 = > \left( {2x – 1} \right)\left( {5x + 1} \right) = 0\]

Làm tương tự với tử còn lại