2x-3=x+1/2

a,2x-3=x+1/2                       b,4x-(x+1/2)=2x+(1/2-5)                           c,2/3-1/3(x-2/3)-1/2(2x+1)=5

2x-x =1/2+3                           4x-x-1/2=2x+1/2-5                             d,(x+1/2).(x-3/4)=0

x=7/2                                4x-x-2x  =1/2-5+1/2                                 \(\orbr{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

                                            x=-4

e,(2x-1)(3x+1/5)=0

\(\orbr{\begin{cases}2x-1=0\\3x+\frac{1}{5}=0\end{cases}}\orbr{\begin{cases}2x=1\\3x=\frac{1}{5}\end{cases}}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{15}\end{cases}}\)

f, 4x²-2x=0

Các câu mk chưa làm thì bạn cứ chờ để mk suy nghĩ.

Bài 1.

1) ( 2x + 1 )³ - ( 2x + 1 )( 4x² - 2x + 1 ) - 3( 2x - 1 ) = 15

<=> 8x³ + 12x² + 6x + 1 - [ ( 2x )³ - 1³ ] - 6x + 3 = 15

<=> 8x³ + 12x² + 4 - 8x³ + 1 = 15

<=> 12x² + 15 = 15

<=> 12x² = 0

<=> x = 0

2) x( x - 4 )( x + 4 ) - ( x - 5 )( x² + 5x + 25 ) = 13

<=> x( x² - 16 ) - ( x³ - 5³ ) = 13

<=> x³ - 16x - x³ + 125 = 13

<=> 125 - 16x = 13

<=> 16x = 112

<=> x = 7

Bài 2.

A = ( x + 5 )( x² - 5x + 25 ) - ( 2x + 1 )³ - 28x³ + 3x( -11x + 5 )

= x³ + 5³ - ( 8x³ + 12x² + 6x + 1 ) - 28x³ - 33x² + 15x

= -27x³ + 125 - 8x³ - 12x² - 6x - 1 - 33x² + 15x

= -33x³ - 45x² + 9x + 124 ( có phụ thuộc vào biến )

B = ( 3x + 2 )³ - 18x( 3x + 2 ) + ( x - 1 )³ - 28x³ + 3x( x - 1 )

= 27x³ + 54x² + 36x + 8 - 54x² - 36x + x³ - 3x² + 3x - 1 - 28x³ + 3x² - 3x

= 7 ( đpcm )

C = ( 4x - 1 )( 16x² + 4x + 1 ) - ( 4x + 1 )³ + 12( 4x + 1 )³ + 12( 4x + 1 ) - 15

= ( 4x )³ - 1³ - [ ( 4x + 1 )³ - 12( 4x + 1 )³ - 12( 4x + 1 ) ] - 15

= 64x³ - 1 - ( 4x + 1 )[ ( 4x + 1 )² - 12( 4x + 1 )² - 12 ] - 15

= 64x³ - 16 - ( 4x + 1 )[ 16x² + 8x + 1 - 12( 16x² + 8x + 1 ) - 12 ]

= 64x³ - 16 - ( 4x + 1 )( 16x² + 8x - 11 - 192x² - 96x - 12 )

= 64x³ - 16 - ( 4x + 1 )( -176x² - 88x - 23 )

= 64x³ - 16 - ( -704x³ - 528x² - 180x - 23 )

= 64x³ - 16 + 704x³ + 528x² + 180x + 23 

= 768x³ + 528x² + 180x + 7 ( có phụ thuộc vào biến )

a)\(x^2+3x+2\)

\(=x^2+x+2x+2\)

\(=x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+2\right)\left(x+1\right)\)

k)\(4x^2+4x+1\)

\(=\left(2x\right)^2+2.2x+1^2\)

\(=\left(2x+1\right)^2\)

j) \(-5x^2+4x+1\)

\(=-5x^2+5x-x+1\)

\(=-5x\left(x-1\right)-\left(x-1\right)\)

\(=\left(-5x-1\right)\left(x-1\right)\)

\(\left(4x-1\right)^3=\frac{-1}{27}\)

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

\(\Rightarrow4x-1=\frac{-1}{3}\)

\(\Rightarrow4x=\frac{-1}{3}+1\)

\(\Rightarrow4x=\frac{2}{3}\)

\(\Rightarrow x=\frac{2}{3}:4\)

\(\Rightarrow x=\frac{1}{6}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)

\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)

\(\Leftrightarrow12x-9=29x-145\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x+136=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(S=\left\{8\right\}\)

\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)

\(\Rightarrow2x-1=2\left(5-3x\right)\)

\(\Leftrightarrow2x-1=10-6x\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x-11=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)

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

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)

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

\(\Rightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-5-3x+2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)

\(\Rightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\)

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

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

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)

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

\(\Leftrightarrow4x-5-2x+2+2x=0\)

\(\Leftrightarrow4x=3\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

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

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

\(\Leftrightarrow4x-x-\frac{1}{2}=2x-\frac{1}{2}x+5\)

\(\Leftrightarrow4x-x-2x+\frac{1}{2}x=5-\frac{1}{2}\)

\(\Leftrightarrow\frac{3}{2}x=\frac{9}{2}\Leftrightarrow x=3\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)