\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)

\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

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

\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

b) Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: Để B=0 thì x=1

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:

\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Vậy: Khi x=3 thì B=1

d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: 

\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Để B>0 thì \(\dfrac{x-1}{2}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để B>0 thì x>1

\(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
a) ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
b) \(P=0\Leftrightarrow x^3+4x^2-5x=0\)
\(\Leftrightarrow\)x=0 ( ko tm đkxđ) hoặc x=1(tm đkxđ) hoặc x=-5(ktmdkxd)=> x=1
c)\(P=\frac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\frac{\left(x-1\right)}{2}\)
P>0 => x>1
P<0=> x<1
Chúc bạn học tốt :)

a,Tìm ĐKXĐ

\(2x+10\ne0\Rightarrow2\left(x+5\right)\ne0\Rightarrow x\ne-5\)

\(x\ne0\)

\(2x\left(x+5\right)\ne0\Rightarrow x\ne0;x\ne-5\)

a) để phân thức P được xác định thì x\(\ne\pm2\)

b)rút gon P=3x(x-1)

Khi p=0 thì p=3x(x-1)=0

                     \(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\:\:\:\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Để phân thức p=0 thì x=1

a: Ta có: \(x^2+3x-10=0\)

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

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

b: Ta có: \(x^2-5x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

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

<=> x² - 2x + 1 - ( x² - 1 ) = 0

<=> x² - 2x + 1 - x² + 1 = 0

<=> 2 - 2x = 0

<=> 2x = 2 

<=> x = 1

b) ( 2x - 1 )² - ( 2x + 1 )² = 0

<=> [ ( 2x - 1 ) - ( 2x + 1 ) ][ ( 2x - 1 ) + ( 2x + 1 ) ] = 0

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

<=> -2.4x = 0

<=> -8x = 0

<=> x = 0

c) 25( x + 3 )² + ( 1 - 5x )( 1 + 5x ) = 8

<=> 5²( x + 3 )² + 1² - 25x² = 8

<=> [ 5( x + 3 ) ]² + 1 - 25x² = 8

<=> ( 5x + 15 )² + 1 - 25x² = 8

<=> 25x² + 150x + 225 + 1 - 25x² = 8

<=> 150x + 226 = 8

<=> 150x = -218

<=> x = -218/150 = -109/75

d) 9( x + 1 )² - ( 3x - 2 )( 3x + 2 ) = 10

<=> 3²( x + 1 )² - ( 9x² - 4 ) = 10

<=> [ 3( x + 1 ) ]² - 9x² + 4 = 10

<=> ( 3x + 3 )² - 9x² + 4 = 10

<=> 9x² + 18x + 9 - 9x² + 4 = 10

<=> 18x + 13 = 10

<=> 18x = -3

<=> x = -3/18 = -1/6

a) (x - 1)² - (x - 1)(x + 1) = 0

=> (x - 1)² - (x² - 1²) = 0

=> x² - 2.x.1 + 1² - x² + 1 = 0

=> x² - 2x + 1 - x² + 1 = 0

=> -2x + 1 + 1 = 0

=> -2x + 2 = 0

=> -2x = -2 => x = 1

b) (2x - 1)² - (2x + 1)² = 0

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

=> 0 = 0(đúng)

c) 25(x + 3)² + (1 - 5x)(1 + 5x) = 8

=> 25(x² + 2.x.3 + 3²) + (1² - (5x)²) = 8

=> 25x² + 150x + 225 + 1 - 25x² = 8

=> 150x +225 + 1 = 8

=> 150x = -218

=> x = -109/75

d) 9(x + 1)² - (3x - 2)(3x + 2) = 10

=> 9(x² + 2x + 1) - [(3x)² - 2²  ] = 10

=> 9x² + 18x + 9 - (9x² - 4) = 10

=> 9x² + 18x + 9 - 9x² + 4 = 10

=> 18x + 9 + 4 = 10

=> 18x = -3

=> x = -1/6

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

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

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

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\)

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

\(\Leftrightarrow-2.4x=0\)

\(\Rightarrow x=0\)

c) \(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=8\)

\(\Leftrightarrow25x^2+150x+225+1-25x^2-8=0\)

\(\Leftrightarrow150x=-218\)

\(\Rightarrow x=-\frac{109}{75}\)

d) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)

\(\Leftrightarrow9x^2+18x+9-9x^2+4=10\)

\(\Leftrightarrow18x=-3\)

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

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p