Giúp mình nhé mình đang cần gấp

7853 x 9 =70677

666 x 6 = 3996

nhớ **** cho mình nhé

a: \(A=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)

\(=-x-15\)

\(=-\left(-1\right)-15=1-15=-14\)

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

   aaa : ( ax3 ) - 360 : ( 5x2 )

= aaa : a x 3 - 360 : 10

= (aaa : a) x 3 - 360 : 10

= 111 x 3 - 36

= 333 - 36= 297

a: Ta có: \(2x\left(x-3\right)+x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

 b. a0a0a : a x 3 - 20202 = 10101

Em xin tham khảo:

a, aaa : a : 3 - 360 : 10

= 111 : 3 - 36

= 37 - 36

= 1

b, b. a0a0a : a x 3 - 20202 = 10101

a: Ta có: \(\left(x-3\right)^2-x\left(x+5\right)=9\)

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

\(\Leftrightarrow x=0\)

b: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)