gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8

=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)

=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)

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

=>\(\frac{1}{2x}=\frac{1}{4}\)

=> \(2x=4\)

=> \(x=2\)

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z+3}{-4}=\frac{2x-2+3y-6-z+3}{9}=\frac{56-5}{9}\)\(=\frac{17}{3}\)

\(\Rightarrow x=\frac{37}{3},y=19,z=\frac{77}{3}\)

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\); \(2x+3y-z=56\)

\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4};2x+3y-z=56\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{56-2-6+3}{9}=\frac{51}{9}=\frac{17}{3}\)

\(\Leftrightarrow x=\frac{37}{3};y=19;z=\frac{77}{3}\)

Vậy \(x=\frac{37}{3};y=19;z=\frac{77}{3}\)

a/ ĐK x-1 khác 0 ; x^2+x khác 0 ; x^3-x khác 0 ; 1-x^2 khác 0 

=> x khác {1;0;-1} 

b/ \(B=\frac{1}{x-1}-\frac{x^3-x}{x^2+x}.\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)

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

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

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

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

a) \(3\left(x-4\right)+5=2\left(x+1\right)-8\)

\(\Leftrightarrow3x-12+5=2x+2-8\)

\(\Leftrightarrow x=1\)

Vậy : \(S=\left\{1\right\}\)

b) \(5\left(x+1\right)^2+2x=5x^2-3\)

\(\Leftrightarrow5x^2+10x+5+2x=5x^2-3\)

\(\Leftrightarrow12x=-8\)

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

Vậy : \(S=\left\{-\frac{2}{3}\right\}\)

c) \(\frac{4\left(x+2\right)}{15}=\frac{13x-9}{40}\)

\(\Leftrightarrow32\left(x+2\right)=3\left(13x-9\right)\)

\(\Leftrightarrow32x-39x=-27-64\)

\(\Leftrightarrow-7x=-91\)

\(\Leftrightarrow x=13\)

Vậy : \(S=\left\{13\right\}\)

2/3 x ( 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/70.73 - 2x = 1

2/3 x ( 1 - 1/4 + 1/4 - 1/7 + ... + 1/70 - 1/73 ) - 2x = 1

2/3 x ( 1- 1/73 ) - 2x = 1

2/3 x 72/73 - 2x = 1

48/73 - 2x = 1

2x = 48/73 - 1

2x = -25/73

x = -25/73 : 2 

x = -25/146 

Vậy x = -25/146

Tk nha !!

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

\(\Rightarrow\left(2x-1\right)\cdot\left(2x-1\right)=2\cdot8\)

\(\Rightarrow\left(2x-1\right)^2=16\)

\(\Rightarrow\left(2x-1\right)^2=\left(\pm4\right)^2\)

\(\Rightarrow2x-1=4\) hoặc \(2x-1=-4\)

\(2x=4+1\) hoặc \(2x=-4+1\)

\(2x=5\) hoặc \(2x=-3\)

\(x=5:2\) hoặc \(x=-3:2\)

\(x=2,5\) hoặc \(x=-1,5\)

\(\frac{2x-1}{2}=\frac{8}{2x-1}.\)

\(\Rightarrow\left(2x-1\right).\left(2x-1\right)=2.8\)

\(\Rightarrow\left(2x-1\right)^2=16\)

\(\Rightarrow\left(2x-1\right)^2=\left(\pm4\right)^2\)

\(\Rightarrow2x-1=\pm4.\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5:2\\x=\left(-3\right):2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{2};-\frac{3}{2}\right\}.\)

Chúc bạn học tốt!

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

\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=2.8\)

\(\Leftrightarrow\left(2x-1\right)^2=16\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(\pm4\right)^2\)

\(\Leftrightarrow2x-1=\pm4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=4+1\\2x=-4+1\end{matrix}\right.\)

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

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

Vậy \(x=\frac{5}{2}\)hoặc \(x=\frac{-3}{2}\)

Chúc bạn hok tốt!!! Nguyễn Thanh Trúc. Chúc bạn mai thi tốt nha!!! Nguyễn Thanh Trúc