câu b c nx đi mừ

câu b c nx đi mà

1) \(4-25x^2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

2) Tính thì phải cho giá trị của x.

\(A=x^3-3x^2+3x-1\)

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

\(=\left[{}\begin{matrix}\left(\dfrac{5}{2}-1\right)^3=\dfrac{27}{8}\\\left(-\dfrac{5}{2}-1\right)^3=-\dfrac{343}{8}\end{matrix}\right.\)

\(a,ĐK:x\ne\pm2\\ A=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\\ ĐK:x\ne-1;x\ne-2\\ B=\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ b,x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \forall x=0\Leftrightarrow A=\dfrac{1}{0-2}=-\dfrac{1}{2}\\ \forall x=-1\Leftrightarrow A=\dfrac{1}{-1-2}=-\dfrac{1}{3}\)

\(x^2+2x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ \Leftrightarrow B=\dfrac{1}{0+2}=\dfrac{1}{2}\)

3x + 2(5 - x) = 0 

=> 3x + 10 - 2x = 0

=> x = -10

(x - 2)(3x + 1) = 0

x - 2 = 0 hoặc 3x + 1 = 0

*) x - 2 = 0

x = 0 + 2

x = 2 (nhận)

*) 3x + 1 = 0

3x = 0 - 1

3x = -1

x = -1/3 (loại)

Vậy x = 2

(x-2)(3x+1)=0
TH1

x-2=0

x=2

TH2

3x+1=0

x=-1/3

`-3x^2+4x+15=0`

`<=>3x^2-4x-15=0`

`<=>3x^2-9x+5x-15=0`

`<=>3x(x-3)+5(x-3)=0`

`<=>(x-3)(3x+5)=0`

`<=>` \(\left[ \begin{array}{l}x-3=0\\3x+5=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-\dfrac53\end{array} \right.\) 

Vậy `S={3,-5/3}`

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

$\textbf{a)}$

Điều kiện: $x\ne0,\ x\ne-1,\ x\ne1.$

Ta có $B=\left(\dfrac{x+1}{2(x-1)}+\dfrac{3x-1}{(x-1)(x+1)}-\dfrac{x+3}{2(x+1)}\right):\dfrac3{x+1}.$

Quy đồng các phân thức trong ngoặc:

$\dfrac{x+1}{2(x-1)}=\dfrac{(x+1)^2}{2(x-1)(x+1)},$

$\dfrac{x+3}{2(x+1)}=\dfrac{(x+3)(x-1)}{2(x-1)(x+1)}.$

Do đó \[\begin{aligned}&\dfrac{(x+1)^2+2(3x-1)-(x+3)(x-1)}{2(x-1)(x+1)}\\&=\dfrac{x^2+2x+1+6x-2-(x^2+2x-3)}{2(x-1)(x+1)}\\&=\dfrac{6x+2}{2(x-1)(x+1)}=\dfrac{3x+1}{(x-1)(x+1)}.\end{aligned}\]

Suy ra $B=\dfrac{3x+1}{(x-1)(x+1)}\cdot\dfrac{x+1}{3}=\dfrac{3x+1}{3(x-1)}.$

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

Ta thấy x=0 ko tm bài toán => x khác 0

Có : x^2-3x+1 = 0

=> x^2+1 = 3x

=> x^2+1/x = 3x/x = 3

=> x+1/x = 3

=> (x+1/x)^2 = 9

=> x^2+1/x^2+2=9

=> x^2+1/x^2 = 9-2 = 7

=> (x^2+1/x^2)^2 = 49

=> x^4+1/x^4+2 = 49

=> x^4+1/x^4 = 49-2 = 47

Vậy x^4+1/x^4 = 47

Tk mk nha

hay quá