a: \(x\cdot\dfrac{2}{5}+\dfrac{1}{2}\cdot x=9\)

=>\(x\left(\dfrac{2}{5}+\dfrac{1}{2}\right)=9\)

=>\(x\cdot\dfrac{9}{10}=9\)

=>\(x=9:\dfrac{9}{10}=10\)

b: \(\dfrac{1}{9}:x+\dfrac{3}{9}:x=\dfrac{5}{7}\)

=>\(\left(\dfrac{1}{9}+\dfrac{3}{9}\right):x=\dfrac{5}{7}\)

=>\(\dfrac{4}{9}:x=\dfrac{5}{7}\)

=>\(x=\dfrac{4}{9}:\dfrac{5}{7}=\dfrac{4}{9}\cdot\dfrac{7}{5}=\dfrac{28}{45}\)

b. (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 4x + 16 - (x² - 1) = 16

<=> x² + 4x + 16 - x² + 1 - 16 = 0

<=> x² - x² + 4x = 16 - 16 - 1

<=> 4x = -1

<=> x = \(\dfrac{-1}{4}\)

\(a,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\\ \Leftrightarrow x=\dfrac{23}{24}\\ b,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

2(x⁴+3)-(9)=17

⇒2x⁴+6+9=17

⇒2x⁴+15=17

⇒ 2x⁴=2

⇒ x⁴=1

⇒ x=\(\pm1\)

b) 5x².x+1-3.4²=-47

⇒5x³+1-48=-47

⇒5x³-47=-47

⇒5x³=0

⇒x³=0

⇒x=0

a) \(2\left(x^4+3\right)-\left(-9\right)=17\)

              \(2x^4+6+9=17\)

                           \(2x^4=2\)

                             \(x^4=1\)

                       ⇒     \(x=1\)

a) 2*(x⁴+3)-(-9)=17

=>2x⁴+2*3+9=17

=>2x⁴+6=17-9=8

=>2x⁴=8-6=2

=>x⁴=2/2=1

=>x=1

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15

<=> 40x = 40 <=> x = 1

Cái này mới chuẩn

b) (2x+1)(1-2x)+(1-2x)²=18 <=> 1 - 4x² + 4x² - 4x + 1 = 18

<=> -4x = 16 <=> x = -4

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a, \(x\) : \(\dfrac{13}{3}\) = -2,5

    \(x\)         = -2,5 . \(\dfrac{13}{3}\)

    \(x\)         = \(\dfrac{65}{6}\)

b,\(\dfrac{3}{5}\)\(x\)         = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)

   \(\dfrac{3}{5}x\)         = \(\dfrac{-3}{20}\)

      \(x\)          = \(\dfrac{-3}{20}\) :  \(\dfrac{3}{5}\)

      \(x\)          = \(\dfrac{-1}{4}\)

c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)

              \(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)

               \(\dfrac{12}{13}x=2\)

                    \(x=2:\dfrac{12}{13}\)

                    \(x=\dfrac{13}{6}\)

a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)

\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)

b)TH1:  \(x\ge3\)

\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)

TH2: \(-5\le x< 3\)

\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)

TH3: \(x< -5\)

\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)

\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)

\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)

\(2.|3x-1|=\dfrac{4}{5}\)

\(|3x-1|=\dfrac{4}{5}:2\)

\(|3x-1|=\dfrac{2}{5}\)

\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)

\(3x-1=\dfrac{2}{5}\)

\(3x=\dfrac{2}{5}+1\)

\(3x=\dfrac{7}{5}\)

\(x=\dfrac{7}{5}:3\)

\(x=\dfrac{7}{15}\)

\(3x-1=-\dfrac{2}{5}\)

\(3x=-\dfrac{2}{5}+1\)

\(3x=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:3\)

\(x=\dfrac{1}{5}\)

\(a,\Rightarrow\left|3x-1\right|=\dfrac{4}{5}:2=\dfrac{2}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\left(x\ge\dfrac{1}{3}\right)\\1-3x=\dfrac{2}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{5}\left(x\ge\dfrac{1}{3}\right)\\3x=\dfrac{3}{5}\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\left(tm\right)\\x=\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)

\(b,\Rightarrow\left[{}\begin{matrix}-x-5+3-x=9\left(x< -5\right)\\x+5+3-x=9\left(-5\le x< 3\right)\\x+5+x-3=9\left(x\ge3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\left(tm\right)\\0x=1\left(ktm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)