tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^4+2x^3-3x^2-8x-4=0\)

\(\Leftrightarrow x^4-4x^2+2x^3-8x+x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2-4\right)+2x\left(x^2-4\right)+\left(x^2-4\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-2;1\right\}\)

b) \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)-72=0\)

Đặt \(t=x^2-4\), ta có :

\(t\left(t-6\right)-72=0\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-12=0\\t+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-16=0\left(tm\right)\\x^2+2=0\left(ktm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm4\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;-4\right\}\)

c) \(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)

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

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

\(\Leftrightarrow\)\(x+1=0\)

hoặc \(2x+1=0\)

hoặc \(x+2=0\)

\(\Leftrightarrow\)\(x=-1\)

hoặc \(x=-\frac{1}{2}\)

hoặc \(x=-2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-2;-\frac{1}{2}\right\}\)

a, \(x^4+2x^3-3x^2-8x-4=0\)

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

TH1 : \(x+1=0\Leftrightarrow x=-1\)

TH2 : \(x^3+x^2-4x-4=0\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)

=> \(x=-1;x=\pm2\)

b, \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40=72\)

\(\Leftrightarrow x^4-14x^2-32=0\) Đặt \(x^2=t\left(t\ge0\right)\)

Ta có pt mới : \(t^2-14t-32=0\) Tự xử 

a,  7\(x\).(2\(x\) + 10) = 0

        \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\){-5; 0}

b, - 9\(x\) : (2\(x\) - 10) = 0

      - 9\(x\) = 0

           \(x\) = 0

c, (4 - \(x\)).(\(x\) + 3) = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

    \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-2023; 2024}

a: =>14x+20+5=6x-9-9x

=>14x+25=-3x-9

=>17x=-34

=>x=-2

b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)

=>-28x-30=-50

=>-28x=-20

=>x=20/28=5/7

c: =>2x+x^3-x=x^3+1

=>x=1

d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22

=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22

=>-5x^2+2x-1+2x^2+x+22=0

=>-3x^2+3x+21=0

=>x^2-x-7=0

=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)

\(14x\left(x+3\right)-7x^2\left(3+x\right)=\left(x+3\right)\left(14x-7x^2\right)=7x\left(2-x\right)\left(x+3\right)\\ \left(2x-5\right)\left(3+4x\right)-4x+10=\left(2x-5\right)\left(3+4x-2\right)=\left(2x-5\right)\left(4x+1\right)\)

a/ x + 2x + 3x = 36

=> 6x = 36

=> x = 6

Vậy ..

b/ 7x - 2x - x = 24

=> 4x = 24

=> x = 6

Vậy ...

c/ (x : 5) - 10 = 5

=> x : 5 = 15

=> x = 75

Vậy ...

d/ 7. (x+3) = 49

=> x + 3 = 7

=> x = 4

Vậy ....

e/ (x - 5) . 4 = 32

=> x - 5 = 8

=> x = 13

Vậy. ..

g/ 7x - 3 = 25

=> 7x = 28

=> x = 4

Vậy ...

h/ 17 - (x :3) = 10

=> x : 3 = 17 - 10

=>x : 3 = 7

=> x = 21

Vậy ..

a) x + 2x + 3x = 36 

=> 6x = 36

=> x = 6

b)7x - 2x - x = 24

=> 4x = 24 

=> x = 6

c) (x : 5 ) - 10 = 5

=> x : 5 = 15

=> x = 3

d) 7 . ( x + 3 ) = 49

=> x + 3 = 7

=> x = 4

e) ( x - 5 ) .  4 =32 

=> x - 5 = 8

=> x = 3

g) 7x - 3 = 25 

=> 7x = 28

=> x = 4

h) 17 - ( x : 3 ) = 10

=> x : 3 = 7

=> x = 21

a) x + 2x + 3x = 36

=> 6x = 36 =>  x = 36 : 6 = 6.

b) 7x - 2x -x = 24

=> 4x = 24 => x = 24 : 4 = 6.

c)(x : 5) - 10 = 5

=> x : 5 = 5 + 10 = 15

=> x  = 15 x 5 = 75.

d) 7. (x + 3) = 49

=> x + 3 = 49 : 7 = 7

=> x = 7 - 3 = 4.

e)(x - 5) . 4 = 32

=> x - 5 = 32 : 4 = 8

=> x = 8 + 5 = 13.

g) 7x - 3 = 25

=> 7x = 25 + 3 = 28 => x = 28 : 7 = 4

h) 17 - ( x : 3 ) = 10

=> x : 3 = 17 - 10 = 7

=> x = 7.3 = 21

a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)

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

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

b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)

\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)

\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)