<=>  x^3 - 5x^2 +3x -4 =0

Bài này sai đề rồi 

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

a) \(x^2-10x+4y^2-4y+26=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)+\left(4y^2-4y+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2=0\)

Mà \(\Leftrightarrow\left(x-5\right)^2+\left(2y-1\right)^2\ge0\)

Dấu "="\(\Leftrightarrow\hept{\begin{cases}x-5=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{1}{2}\end{cases}}\)

\(\left(5-x\right).\left(3x-\frac{1}{4}\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}5-x>0\\3x-\frac{1}{4}>0\end{cases}}\) hoặc \(\hept{\begin{cases}5-x< 0\\3x-\frac{1}{4}< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x< 5\\x>\frac{1}{12}\end{cases}}\)  hoặc  \(\hept{\begin{cases}x>5\\x< \frac{1}{12}\end{cases}}\) (vô lí)

Vậy \(\frac{1}{12}< x< 5\)

b, \(\left(x^2+2015\right).\left(x-2016\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x^2==-2015\\x=2016\end{cases}}\)( \(x^2=-2015\)loại do \(x^2\ge0\))

Vậy x= 2016

a, \(xy+3x-7y=21\)

\(\Leftrightarrow x.\left(y+3\right)-7y-21=0\)

\(\Leftrightarrow x.\left(y+3\right)-7.\left(y+3\right)=0\)

\(\Leftrightarrow\left(y+3\right).\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-7\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)

a, xy + 3x - 7y = 21

=> x(y + 3) - 7y - 21 = 21 - 21

=> x(y + 3) - (7y + 21) = 0

=> x(y + 3) - 7(y + 3) = 0

=> (x - 7)(y + 3) = 0

=> \(\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)

Vậy x = {7;-3}

b, (x² + 2015)(x - 2016) = 0

\(\Rightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=2015\left(loại\right)\\x=2016\end{cases}}}\)

Vậy x = 2016

a, \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\Leftrightarrow x^2+6x+9-x^2+4=11\Leftrightarrow6x+2=0\Leftrightarrow x=-\frac{1}{3}\)

b, \(x^2-6x-7=0\Leftrightarrow\left(x+1\right)\left(x-7\right)=0\Leftrightarrow x=-1;x=7\)

c, \(\left(2x+1\right)^2-\left(3x-2\right)^2=0\Leftrightarrow\left(-x+3\right)\left(5x-1\right)=0\Leftrightarrow x=\frac{1}{5};x=3\)

x=3 nha

câu c bạn chờ mình làm lại nhé

a, ( x+ 3 )² -(x + 2 ) ( x - 2) = 11

X² + 6x + 9 - ( x² -4 ) = 11

x² + 6x + 9 - x² + 4 = 11

6x + 13 = 11

6x = -2

x= -1/3

b, x² -6x -7 =0

x² -6x = 7

x( x-6) =7

x-6 = 7/x

x - 7/x = 6

x( 1- 7)=6

-6x = 6 => x=-1

c, ( 2x + 1 ) ² - (3x -2 ) ² =0 ( câu này chờ mình làm lại đã )

4x² + 4x + 1 - ( 9x² - 12x + 4 ) =0

4x² + 4x + 1 - 9x² + 12x - 4 =0

-5x² + 16x -3 =0

-5x² + 16x =3

x( -5x + 16 ) = 3

-5x+ 16 = 3/x 

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

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

\(\Leftrightarrow x\left(3x-1\right)+2\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+2\right)=0\)

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

Vậy ...

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

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

\(\Leftrightarrow3x\left(x+2\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}\)

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

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

\(\left(3x^2+6x\right)-\left(x+2\right)=0\)

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

\(\left(3x-1\right)\left(x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

Vậy ....

Nếu x < -2

=> |x + 2| = -(x + 2) = -x - 2

=> |x - 5| = -(x - 5) = -x + 5

Khi đó |x + 2| + |x - 5| = 3x (1)

<=> -x - 2 - x + 5 = 3x

=> 3 = 5x

=> x = 0,6 (loại)

Nếu \(-2\le x\le5\)

=> |x + 2| = x + 2 

=> |x - 5| = -(x - 5) = -x + 5

Khi đó (1) <=> x + 2 - x + 5 = 3x

=> 3x = 7

=> x = 7/3 (tm)

Nếu x > 5

=> |x + 2| = x + 2

=> |x - 5| = x - 5

Khi đó (1) <=> x + 2 + x - 5 = 3x

=> 2x - 3 = 3x

=> x = -3 (loại) 

Vậy x = 7/3

| x + 2 | ≥ 0 <=> x + 2 ≥ 0 => x ≥ -2

| x - 5 | ≥ 0 <=> x - 5 ≥ 0 => x ≥ 5

Vậy để giải phương trình trên ta xét ba trường hợp

1/ x < -2

Pt trở thành : 

-( x + 2 ) - ( x - 5 ) = 3x

<=> -x - 2 - x + 5 = 3x

<=> -2x + 3 = 3x

<=> -2x - 3x = -3

<=> -5x = -3

<=> x = 3/5 ( không tmđk )

2/ -2 < x < 5

Pt trở thành

( x + 2 ) - ( x - 5 ) = 3x

<=> x + 2 - x + 5 = 3x

<=> 7 = 3x

<=> x = 7/3 ( tmđk )

3/ x ≥ 5

Pt trở thành :

x + 2 + x - 5 = 3x

<=> 2x - 3 = 3x

<=> 2x - 3x = 3

<=> -x = 3

<=> x = -3 ( không tmđk )

Vậy phương trình có nghiệm duy nhất là x = 7/3