a) \(5-\left[x+1\right]=29\)

\(x+1=5-29\)

\(x+1=-24\)

\(x=-24-1\)

\(x=-25\)

Vậy \(x=-25\) là giá trị cần tìm

b) \(\left(-1\right)+3+\left(-5\right)+7+\left(-9\right)+........+x=600\)

\(\Rightarrow\left[\left(-1\right)+3\right]+\left[\left(-5\right)+7\right]+............+\left[-\left(x-2\right)+x\right]=600\)

\(\Rightarrow2+2+2+.......+2=600\)

\(\Rightarrow2\left(1+1+......+1\right)=600\)

\(\Rightarrow1+1+........+1=300\)

Số dấu ngoặc [] là :

\(\dfrac{x-3}{4+1}\)

\(\Rightarrow\dfrac{x-3}{4+1}=300\)

\(\Rightarrow\dfrac{x-3}{4}=299\)

\(\Rightarrow x-3=299.4\)

\(\Rightarrow x=299.4+3\)

\(\Rightarrow x=1199\)

\(5-\left|x+1\right|=29\)

\(\left|x+1\right|=5-29\)

\(\left|x+1\right|=-24\Leftrightarrow x\in\left\{\varnothing\right\}\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

hay x=-2

bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

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

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

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

Lời giải:

a) 

$1+3+5+...+(2x-1)=750$

$\frac{(2x-1+1)x}{2}=750$

$x^2=750$

$x=\sqrt{750}$ (vô lý?!!)

b) 

$1+5+9+13+...+x=501501$

$\frac{1}{2}(\frac{x-1}{4}+1)(x+1)=501501$

$(x+1)(x+3)=4012008$$x(x+4)=4012005=2001.2005$

$\Rightarrow x=2001$

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

a) (2x+3)-(5x-17)=9

2x+3-5x+17=9

-3x+20=9

-3x=9-20=-11

=>x=11/3

b) \(\left(\frac{1}{5}+\frac{4}{5}x\right)-\left(\frac{2}{5}x+\frac{1}{9}\right)=\frac{3}{7}\)

=> \(\frac{1}{5}+\frac{4}{5}x-\frac{2}{5}x-\frac{1}{9}=\frac{3}{7}\)

=> \(\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{4}{5}x-\frac{2}{5}x\right)=\frac{3}{7}\)

\(\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\)

\(\frac{2}{5}x=\frac{3}{7}-\frac{4}{45}=\frac{107}{315}\)

x=107/315:2/5=107/126

dễ tí nữa tôi lấy nk khác của tôi làm thì và kb nhé

a) = 2x-5x+3+17=9\(\Rightarrow\)-3x+20=9\(\Rightarrow\)-3x=-11\(\Rightarrow\)x=11/3

b) = \(\frac{1}{5}-\frac{1}{9}+\frac{4}{5}x-\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{4}{45}+\frac{2}{5}x=\frac{3}{7}\Rightarrow\frac{2}{5}x=\frac{107}{315}\Rightarrow x=\frac{107}{126}\)