d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

chứng minh hay tìm bạn

Tìm x bạn

1/ <=> x² - x -(x² - x)/x³ = 0

<=> (x² - x)(1 - 1/x³) = 0 

Phần còn lại bạn làm tiếp nha điều kiện x#0

Mong các bạn và thầy cô giải giùm ạ!

Đặt \(t=\left(x+\frac{1}{x}\right)^2\)\(\Rightarrow\)\(x^2+\frac{1}{x^2}=t-2\)điều kiện t>=0,x # 0

Phương trình trở thành

8t +4(t-2)² - 4(t-2)²t =(x+4)²

8t + 4t² - 16t + 16 -4t³ + 16t² - 16t=(x+4)²

-4t³ + 20t² -24t=x² +8x

-4t(t² -5t +6)=x(x+8)

-4t(t-2)(t-3)=x(x+8)

Mình chỉ giúp dược tới đó

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

\(x-\frac{3}{4}-x.\frac{2}{3}+x:\frac{1}{2}-x:\frac{2}{5}=\frac{11}{4}\)

\(x-x.\frac{2}{3}+x.2-x.\frac{5}{2}=\frac{11}{4}+\frac{3}{4}\)

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

\(x.\frac{-1}{6}=\frac{7}{2}\)

\(x=\frac{7}{2}:-\frac{1}{6}\)

\(x=-21\)

Vậy \(x=-21\)

\(\frac{x-1}{4}+\frac{x-2}{3}+\frac{x-3}{2}+\frac{x-4}{1}=4\)\(\Leftrightarrow\left(\frac{x-1}{4}-1\right)+\left(\frac{x-2}{3}-1\right)+\left(\frac{x-3}{2}-1\right)+\left(\frac{x-4}{1}-1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{4}+\frac{1}{3}+\frac{1}{2}+1\right)=0\Leftrightarrow x=5\)

Đáp án : 2\(\frac{29}{48}\)( 2,6041 )

\(\frac{\left(x+1\right)^2-\frac{x}{2}}{4}=\frac{\left(2x-3\right)^2}{3}-\frac{\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}}{4}\)

\(\Rightarrow3\left[\left(x+1\right)^2-\frac{x}{2}\right]=4\left(2x-3\right)^2-3\left[\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}\right]\)

\(\Rightarrow3\left(x+1\right)^2-\frac{3x}{2}=4\left(2x-3\right)^2-\frac{3\left(x+1\right)}{4}+\frac{3x\left(3-2x\right)}{3}\)

\(\Rightarrow36\left(x+1\right)^2-18x=48\left(2x-3\right)^2-9\left(x+1\right)+12x\left(3-2x\right)\)

=> 36.(x² + 2x + 1) - 18x = 48.(4x² - 12x + 9) - 9(x + 1) + 12x(3 - 2x)

=> 36x² + 72x + 36 - 18x - 192x² + 576x - 432 + 9x + 9 - 36x + 24x² = 0

=> -132x² + 603x - 387 = 0

Có: \(\Delta=603^2-4.\left(-387\right)\left(-132\right)=159273\Rightarrow\sqrt{\Delta}=\sqrt{159273}\)

\(\Rightarrow x=\frac{-603+\sqrt{159273}}{-264}\)          hoặc          \(x=\frac{-603-\sqrt{159273}}{-264}\)

Vậy phương trình có 2 nghiệm : x = \(\left\{\frac{-603+\sqrt{159273}}{-264};\frac{-603-\sqrt{159273}}{-264}\right\}\)

Câu này không có nghiệm nguyên nha bạn.

Cảm ơn bn nhìu

\(\frac{1}{2}x+\frac{4}{5}\cdot\left(2\frac{1}{2}+x\right)=3\frac{3}{4}\)

\(\frac{1}{2}x+\frac{4}{5}\cdot\left(\frac{5}{2}+x\right)=\frac{15}{4}\)

\(\frac{1}{2}x+\frac{4}{5}\cdot\frac{5}{2}+\frac{4}{5}\cdot x=\frac{15}{4}\)

\(\frac{1}{2}x+2+\frac{4}{5}x=\frac{15}{4}\)

\(\frac{1}{2}x+\frac{4}{5}x=\frac{15}{4}-2\)

\(\frac{1}{2}x+\frac{4}{5}x=\frac{7}{4}\)

\(x\left(\frac{1}{2}+\frac{4}{5}\right)=\frac{7}{4}\)

\(x\cdot\frac{13}{10}=\frac{7}{4}\)

\(x=\frac{7}{4}:\frac{13}{10}\)

\(x=\frac{35}{26}\)

\(\frac{1}{2}x+\frac{4}{5}\cdot\left(2\frac{1}{2}+x\right)=3\frac{3}{4}\)

\(\frac{1}{2}x+\frac{4}{5}\cdot\left(\frac{5}{2}+x\right)=\frac{15}{4}\)

\(\frac{1}{2}x+\frac{4}{5}\cdot\frac{5}{2}+\frac{4}{5}\cdot x=\frac{15}{4}\)

\(\frac{1}{2}x+2+\frac{4}{5}x=\frac{15}{4}\)

\(\frac{1}{2}x+\frac{4}{5}x=\frac{15}{4}-2\)

\(\frac{1}{2}x+\frac{4}{5}x=\frac{7}{4}\)

\(x\cdot\left(\frac{1}{2}+\frac{4}{5}\right)=\frac{7}{4}\)

\(x\cdot\frac{13}{10}=\frac{7}{4}\)

\(x=\frac{7}{4}-\frac{13}{10}\)

\(x=\frac{9}{20}\)