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

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

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

\(\Rightarrow x=\dfrac{2}{5}\)

\(b,\dfrac{1}{2}x+\dfrac{1}{2}\left(x-2\right)=\dfrac{3}{4}-2x\)

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

\(\Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{2}+2\right)x=\dfrac{3}{4}+1\)

\(\Rightarrow3x=\dfrac{7}{4}\)

\(\Rightarrow x=\dfrac{7}{4}:3\)

\(\Rightarrow x=\dfrac{7}{12}\)

\(c,\left(x-\dfrac{1}{2}\right)-\dfrac{1}{4}=0\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{4}+\dfrac{2}{4}\)

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

\(d,4^{x-3}+1=17\)

\(\Rightarrow4^{x-3}=17-1\)

\(\Rightarrow4^{x-3}=16\)

\(\Rightarrow4^{x-3}=4^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=2+3\)

\(\Rightarrow x=5\)

#Toru

`3/2 x -1 =1/2x -3/5`

`=> 3/2x -1/2x = -3/5 +1`

`=> 2/2x= -3/5 + 5/5`

`=> x= 2/5`

__

`1/2x +1/2(x-2) = 3/4 -2x`

`=> 1/2x + 1/2x - 2/2 = 3/4 -2x`

`=> 1/2x +1/2x +2x = 3/4 + 1`

`=> 1/2x +1/2x + 4/2x = 3/4 +4/4`

`=> 6/2x = 7/4`

`=> x= 7/4 : 3`

`=>x=7/12`

__

`(x-1/2) -1/4=0`

`=> x-1/2=1/4`

`=> x=1/4 +1/2`

`=> x= 1/4 +2/4`

`=>x=3/4`

__

`4^(x-3) +1=17`

`=> 4^(x-3) =17-1`

`=> 4^(x-3)=16`

`=> 4^(x-3)=4^2`

`=> x-3=2`

`=>x=2+3`

`=>x=5`

chiu

\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)

\(\Leftrightarrow3x^2-6x+3+9x-21=0\)

\(\Leftrightarrow3x^2+3x-18=0\)

\(\Leftrightarrow3\left(x^2+x-6\right)=0\)

\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(S=\left\{2;-3\right\}\)

Phân tích đa thức thành nhân tử à?

1) \(\left(x+y\right)^3-x^3-y^3\)

\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)

\(=3xy\left(x+y\right)\)

2) \(x^3+1-x^2-x\)

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

\(=\left(x+1\right)\left[x^2-x+1-x\right]\)

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

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

( x + y )³ - x³ - y³

= ( x + y )³ - ( x³ + y³ )

= ( x + y )³ - ( x + y )( x² - xy + y² )

= ( x + y )[ ( x + y )² - ( x² - xy + y² ) ]

= ( x + y )( x² + 2xy + y² - x² + xy - y² )

= 3xy( x + y )

x³ + 1 - x² - x

= ( x³ + 1 ) - ( x² + x )

= ( x + 1 )( x² - x + 1 ) - x( x + 1 )

= ( x + 1 )( x² - x + 1 - x )

= ( x + 1 )( x² - 2x + 1 )

= ( x + 1 )( x - 1 )²

    25,1 : 0,25 + 2,7 x 4 - 27,8 x 3,9

= 25,1 x 4 + 2,7 x 4 - 27,8 x 3,9

= 4 x ( 25,1 + 2,7) - 27,8 x 3,9

= 4 x 27,8 - 27,8 x 3,9

= 27,8 x ( 4 - 3,9)

= 27,8 x 0,1

= 2,78 

đề câu c thiếu rồi

A,x thuộc{-3,-2,-1,0,1,2,3,4}

B,x thuộc{-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6}

câu c mk chịu

ĐK:..........

Bình phương 2 vế ta được

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

\(\Leftrightarrow2\sqrt{\left(1-2x\right)\left(1-x\right)}=4x+2\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)\left(1-x\right)}=2x+1\)

\(\Leftrightarrow\left(1-2x\right)\left(1-x\right)=4x^2+4x+1\)

\(\Leftrightarrow1-3x+2x^2=4x^2+4x+1\)

\(\Leftrightarrow2x^2+7x=0\)

\(\Leftrightarrow x\left(2x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}\)

Vậy.........................................

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

\(=\dfrac{2x^2+4x+2}{3\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge0\)

Do đó: \(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\)(1)

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

\(=\dfrac{-2x^2+4x-2}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)

Do đó: \(\dfrac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ (1)và (2) suy ra ĐPCM

\(\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\) có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

dạ cho e hỏi sao k xét cái căn x cộng 2 ạ. e cảm ơn