con chịu bố

a) \(\Leftrightarrow3y^2+2.!x-y+1!-y^2=y^2-y^2\)(-ý^2 hai vế điện giải nhớ ra cho bạn hiểu)

\(\Leftrightarrow y^2\left(3-1\right)+2!x-y+1!=y^2\left(1-1\right)\)đặt y^2 thừa số chung

\(\Leftrightarrow y^2.2+2!x-y+1!=y^2.0\Leftrightarrow2y^2+2!x-y+1!=0\)

\(\hept{\begin{cases}2.y^2=0\\\\2!x-y+1=0\end{cases}}\) tổng hai số không (-) \(\ge0\)chỉ =0 khi cả chi số cùng =0

\(\hept{\begin{cases}y=0\\x-y+1=0\end{cases}}\)\(\hept{\begin{cases}y=0\\\\x=-1\end{cases}}\) ok chưa

A=!3-x!+!2x-1!-8+x=0

phá trị tuyệt đối tham khao neu chua hieu  http://olm.vn/hoi-dap/question/774863.html 

3 1/2

*khi x<1/2

<=> 3-x+-(2x-1)-8+x=0=>-2x-1=3=>x=-2

*khi 1/2<=x<3

<=>3-x+(2x-1)-8+x=0=>2x=-6=>x=-3 (loai)

*khi x>=3

<=>x-3+2x-1-8+x=0=>4x=12=> x=3 nhan

a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)

\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)

\(\Rightarrow2y-1=3+y\)

\(2y-y=3+1\)

\(y=4\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)

\(x-\frac{2}{9}=\frac{4}{9}\)

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

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)

\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)

\(8x^3-1=16x^4-1\)

\(16x^4-8x^3=0\)

\(8x^3\left(2x-1\right)=0\)

Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)

Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)

Vậy x=0 và x=1/2

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

a, Ko viết đề bài :

3 ( 2x + 8 ) - ( 5x + 7 ) = 0

6x + 24 - 5x - 7 = 0

( 6x - 5x ) + ( 24 - 7 ) = 0

x + 17 = 0

x = -17

b, Ko viết đề bài :

( 3x - 1 ) ( y - 2 ) = 5

=> ( 3x - 1 ) và ( y - 2 ) \(\in\)Ư(5 ) = { 1 ; -1 ; 5 ; - 5 }

Ta có bảng :

3x - 1                 1               -1                       5                        -5

y - 2                   5               - 5                      1                         -1

x                      \(\frac{2}{3}\)          0                      2                      \(\frac{-4}{3}\)

y                        7               - 3                      3                        1

Phần b, mk ko chắc

1,(x-1)(y+5)=101

th1:x-1=101

<=>x=102

th2:y+5=101

<=>y=96

2,(x-2)(-y+5)=12

th1:x-2=12

<=>x=14

th2:-y+5=12

<=>-y=7

<=>y=-7

3,(x-5)(2x+6)<0

th1:x-5<0

<=>x<5

th2:2x+6<0

<=>2x<-6

<=>x<-3