\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

__

\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)

(x+2)^2-(x-2)(x+2)=0

=> (x+2)(x+2-x+2)=0

=> (x+2).4=0

=> x+2=0

=> x=-2

mấy câu còn lại tự làm nha

a) (x+2)^2-(x-2)(x+2)=0 

 (x+2).[x+2-x+2]=0

(x+2).4=0

 x+2=0

x=-2

 b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

   4x²-4x+1-4x²+25=18

   26-4x=18

   4x=8

    x=2

 c)( 2x - 1)^2 - 25 = 0

    ( 2x - 1)^2 - 5² = 0

     (2x-1-5)(2x-1+5)=0

    (2x-6)(2x+4)=0

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

 d)9(3x - 2) = x ( 2- 3x)

    9(3x - 2) - x ( 2- 3x) = 0

    9(3x - 2) + x ( 3x-2 ) = 0

    (3x-2)(9+x)=0

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\9+x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)

e)x³ - 0,25x = 0 

  x.(x²-0,25)=0

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-0,25=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\\end{cases}}\)

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)

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

c) \(x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

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

TH1: \(x=0\)

TH2: \(x-3=0\Rightarrow x=3\)

\(x+3=0\Rightarrow x=-3\)

Vậy:..

d) \(\left(5+2x\right)\left(2x-7\right)=4x^2-25\)

\(\Leftrightarrow\left(5+2x\right)\left(2x-7\right)=\left(2x-5\right)\left(2x+5\right)\)

 \(\Leftrightarrow\left(2x+5\right)\left(2x-7-2x+5\right)=0\)

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

\(\Leftrightarrow2x+5=0\)

\(\Leftrightarrow x=-\frac{5}{2}\)

e) \(x^2-11x+30=0\) 

\(\Leftrightarrow x^2-5x-6x+30=0\)

\(\Leftrightarrow x\left(x-5\right)-6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)

a) (x-2)² -(x-3)(x-3)=6

=>x² -4x+4-x²+3=6

=>7-4x=6

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

b)4(x-3)² -(2x-1)(2x+1)=10

=>4(x²-6x+9)-4x²+1=10

=>4x²-24x+36-4x²+1=10

=>37-24x=10 =>24x=27 =>x=\(\frac{9}{8}\)

c)x²-16-3(x+4)=0

=>(x-4)(x+4)-3(x+4)=0

=>(x-7)(x+4)=0

=>\(\orbr{\begin{cases}x-7=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-4\end{cases}}}\)

=>x\(\in\left\{-4;7\right\}\)

d)(x-4)²-(x-2)(x+2)=6

=>x²-8x+16-x²+4=6

=>20-8x=6

=>8x=14 =>x=\(\frac{4}{7}\)

e) 9(x+1)²-(3x-2)(3x+2)=10

=>9(x² +2x+1)-9x²+4=10

=>9x²+18x+9-9x²+4=10

=>18x+13=10

=>18x=-3

=>x=\(\frac{-1}{6}\)

mình chỉ làm bài 1 nha

nhớ chon mk đúng nha

Cảm ơn bạn nha . Ai giúp mình làm bài 2 với TT

2,

B=-2x²+6x-11=-2(x²-3x+\(\frac{4}{9}\)) - \(\frac{91}{9}\)

=-2(x+\(\frac{2}{3}\))²-\(\frac{91}{9}\)

vì (x+\(\frac{2}{3}\))² \(\ge\)0  với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))²\(\le\)0 với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))² -\(\frac{91}{9}\)\(\le-\frac{91}{9}\)với mọi x\(\in\)R

dấu bằng xảy ra khi -2(x+\(\frac{2}{3}\))² =0 hay x+\(\frac{2}{3}\)=0 =>x=-\(\frac{2}{3}\)

vậy GTLN của B=-\(\frac{91}{9}\)khi x=-\(\frac{2}{3}\)

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

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

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

\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn 

Vậy ... 

1./ \(x+y=3\Rightarrow\left(x+y\right)^3=27\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\Rightarrow x^3+y^3+3\cdot2\cdot3=27.\)

\(\Rightarrow x^3+y^3=9\)

2./ \(\left(x+3\right)\left(x^2-3x+3^2\right)-x^3-2x-4=0\)

\(\Leftrightarrow x^3+27-x^3-2x-4=0\Leftrightarrow2x=23\Leftrightarrow x=\frac{23}{2}\)

1/ \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow x^2+4+y^2=9\)

\(\Rightarrow x^2+y^2=5\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.1=3\)

câu đầu bạn dưới làm rồi nên mình k làm lại

(2x+9)²=0

=> 2x+9=0 

=> 2x=-9

=> x=-9/2

(2x-1)³=8

=> 2x-1=2

=> 2x=3

=> x=3/2

(1-3x)²=16

=> 1-3x=4

=> 3x=-3

=> x=-1

(3x+1)+1=-26

=> 3x=-27

=> x=-9

(x+1)+(x+3)+(x+5)+...+(x+2017)=0

(x+x+x+...+x)+(1+3+5+...+2017)=0

=> 1009x+1018081=0

1009x=-1018081

=> x=-1009

\(\left(x+2\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}}\)

Vậy x = -2 hoặc x = 5

lm luôn cho đủ

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

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

vậy x=-3