\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

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

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

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

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+\dfrac{1}{x}=1\)

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

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm

Để \(\frac{2x\left(3x-5\right)}{x^2+1}< 0\)

ta thấy x²+1 luôn dương với mọi x 

nên 2x(3x-5) <0

TH1: \(\orbr{\begin{cases}2x< 0\\3x-5>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0\\3x>5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x< 0\\x>\frac{5}{3}\end{cases}\left(ktm\right)}}\)

TH2: \(\orbr{\begin{cases}2x>0\\3x-5< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0\\3x< 5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x>0\\x< \frac{5}{3}\end{cases}\left(tm\right)}}\)

vậy \(0< x< \frac{5}{3}\)

 THẤY ĐÚNG CHO MK 1 NẾU KO HIỂU THÌ ib NHA

\(\frac{2x\left(3x-5\right)}{x^2+1}< 0\)

\(\Rightarrow2x\left(3x-5\right)< 0\)  ( vì \(x^2+1>0\))

\(\Rightarrow\hept{\begin{cases}2x< 0\\3x-5>0\end{cases}}\)  hoặc \(\hept{\begin{cases}2x>0\\3x-5< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< 0\\x>\frac{5}{3}\end{cases}}\)  hoặc \(\hept{\begin{cases}x>0\\x< \frac{5}{3}\end{cases}}\)

\(\Rightarrow0< x< \frac{5}{3}\)

\(\Leftrightarrow2x^3-3x^2+6x+2x^2-3x+6=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x+6=0\left(vn\right)\end{matrix}\right.\)

Ai chẳng biết chuyển vế đổi dấu :v

a) \(x-7=4x+10\)

\(x-4x=10+7\)

\(-3x=17\)

\(x=\dfrac{17}{-3}\)

Vậy \(x=\dfrac{17}{-3}\)

b) \(2x+5=-3x+7\)

\(2x+3x=7-5\)

\(5x=2\)

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

Vậy \(x=\dfrac{2}{5}\)

c) \(x-\left(3x+7\right)=6x-1\)

\(x-3x-7=6x-1\)

\(-2x-7=6x+1\)

\(-7-1=6x+2x\)

\(-8=8x\)

\(x=\dfrac{-8}{8}=-1\)

Vậy \(x=-1\)

d) \(x+\left(5x-1\right)=15\)

\(x+5x-1=15\)

\(6x=15+1\)

\(6x=16\)

\(x=\dfrac{16}{6}=\dfrac{8}{3}\)

Vậy \(x=\dfrac{8}{3}\)

1 , x - 7 = 4x + 10

x - 4x = 10 + 7

- 3x = 17

x = 17 : ( - 3 )

x = \(\dfrac{-17}{3}\)

2 , 2x + 5 = -3x + 7

2x + 3x = 7 -5

5x = 2

x = 2 : 5

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

3 , x - ( 3x + 7 ) = 6x - 1

x - 3x - 7 = 6x - 1

x - 3x -6x = -1 +7

-8x = 6

x = 6 : ( -8 )

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

4 , x + ( 5x -1 ) = 15

x + 5x - 1 = 15

x + 5x = 15 + 1

6x = 16

x = 16 : 6

x = \(\dfrac{8}{3}\)

5 , / x + 1 / = / 2x - 5 /

TH 1 : x + 1 = 2x - 5

x - 2x = -5 -1

- x = -4

= > x = 4

TH 2 : -x -1 = -2x + 5

-x + 2x = 5 + 1

x = 6

6 , / 3x + 8 / - / x -10 / = 0

3x + 8 - x + 10 = 0

3x - x = 0 - 10 - 8

2 x = -18

x = -18 : 2

x = - 9

\(3x^2+2y^2=7xy\)

\(\Leftrightarrow3x^2-7xy+2y^2=0\)

\(\Leftrightarrow3x^2-6xy-xy+2y^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-y=0\\x-2y=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x=y\\x=2y\end{matrix}\right.\)

+) TH1 : \(y=3x\)

\(\Leftrightarrow A=\dfrac{3x+y}{7y-x}+\dfrac{6x-9y}{2x+y}\)

\(=\dfrac{3x+3x}{7.3x-x}+\dfrac{6x-9.3x}{2x+3x}\)

\(=\dfrac{9x}{20x}+\dfrac{-21x}{5x}\)

\(=-\dfrac{15}{4}\)

+) TH2 : \(x=2y\)

\(\Leftrightarrow A=\dfrac{3x+y}{7y-x}+\dfrac{6x-9y}{2x+y}\)

\(=\dfrac{3.2y+y}{7y-2y}+\dfrac{6.2y-9y}{2.2y+y}\)

\(=\dfrac{7y}{5y}+\dfrac{3y}{5y}\)

\(=2\)

Vậy...

Câu 2: 

\(\left(A\cup B\right)\cap C=A\cap C=[1;+\infty)\cap\left(0;4\right)=[1;4)\)

Tập này có 3 phần tử nguyên