mng giúp với

\(\Leftrightarrow7x\left(x+5\right)+\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(7x+x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(8x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{5}{8}\end{matrix}\right.\)

\(\Leftrightarrow x^2+x-5x-5-x^2-6x-9-6=0\\ \Leftrightarrow-10x-20=0\\ \Leftrightarrow x=-2\)

\(x^2-x+8=4\sqrt{x+3}\)đk : x >= -3

\(\Leftrightarrow x\left(x-1\right)+8-4\sqrt{x+3}=0\)

Đặt \(\sqrt{x+3}=t;\Rightarrow x+3=t^2\Leftrightarrow x=t^2-3;x-1=t^2-4\)

khi đó : \(\left(t^2-3\right)\left(t^2-4\right)+8-4t=0\)

\(\Leftrightarrow t^4-7t^2+20-4t=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^3+2t^2-3t-10\right)=0\)

\(\Leftrightarrow t=2;t=\frac{-4+2i}{2}\left(loại\right);\frac{-4-2i}{2}\left(loại\right)\)

Theo cách đặt \(\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)

xin vui lòng giúp em, em rất rất gấp!!

ai chơi dragonboy ko cho mình xin 1 nick

x+17 / x+7= x+5 / x       (ĐKXĐ: x khác -7; x khác 0)

<=> x(x+17)=(x+7)(x+5)

<=>x^2 + 17x = x^2 + 12x + 35

<=>5x=35

<=>x=7(Tm ĐKXĐ)

Vậy ...............

x -x + 17/x - 5/x = -7        <=>   12/x = -7  

                                       <=>     12 = -7x   =>   x =  -12/7

a/

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

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

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

a.(x+2)²-x(x+2)=0

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

\(\Leftrightarrow\)(x+2)*2=0

\(\Leftrightarrow\)x+2=0

\(\Leftrightarrow\)x=-2

vay s={-2}

b.\(\frac{2x+7}{3}\)-\(\frac{x-2}{4}\)=2

\(\Leftrightarrow\)\(\frac{4\left(2x+7\right)}{12}\)+\(\frac{-3\left(x-2\right)}{12}\)=\(\frac{24}{12}\)

\(\Leftrightarrow\)8x+28-3x+6=24

\(\Leftrightarrow\)5x=-10

\(\Leftrightarrow\)x=-2

vay s={-2}

c.|x+5|=3x+1

neu x+5\(\ge\)0 thi |x+5|=x+5

\(\Leftrightarrow\)x\(\ge\)-5

ta co phuong trinh

x+5=3x+1

\(\Leftrightarrow\)-2x=-4

\(\Leftrightarrow\)x=2( thoa man dieu kien x\(\ge\)-5)

neu x+5<0 thi |x+5|=5-x

\(\Leftrightarrow\)x<-5

ta co phuong trinh

5-x=3x+1

\(\Leftrightarrow\)-4x=-4

\(\Leftrightarrow\)x=1 (k thoa man dieu kien x<5)

vay s={2}

chuc bn hoc tot

a, -2

b, -2

c, 2

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

(x+2). 2 = 0

<=> x +2 = 0

<=> x = -2

b , \(\frac{2x+7}{3}-\frac{x-2}{4}=2\)

\(\frac{4\left(2x+7\right)}{12}-\frac{3\left(x-2\right)}{12}=\frac{24}{12}\)

=> 8x + 28 - 3x + 6 = 24

<=> 5x = -10

=> x = -2

c , \(|x+5|=3\)x +1

\(|x+5|=x+5\) <=> x + 5 \(\ge0\)

<=> x \(\ge\) -5

\(|x+5|=3\)

\(x+5=3\)

x =-2 (tm)

+ \(|x+5|=-3\Leftrightarrow x+5< 0=>x< -5\)

\(|x+5|=-3\)