Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

\(\hept{\begin{cases}x^2+4y^2=5\\4x^2y+8xy^2+5x+10y=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+2y\right)^2-4xy=5\\4xy\left(x+2y\right)+5\left(x+2y\right)=1\end{cases}}\)

Đặt \(\hept{\begin{cases}x+2y=a\\4xy=b\end{cases}}\)

Ta  thu được hệ \(\hept{\begin{cases}a^2-b=5\\ab+5a=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=a^2-5\\a\left(a^2-5\right)+5a=1\end{cases}}\)

Giải pt 2 tìm đc a -> b -> dễ

\(=2xy+5x+4y^2+10y\)

\(=x\left(2y+5\right)+2y\left(2y+5\right)\)

\(=\left(x+2y\right)\left(2y+5\right)\)

\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)

pt da thuc thanh nt ak bn

a,=x^2-[y-1]^2

=[x-y+1] [x+y-1]

b,=5[x-2y]-[x-2y].[x+2y]

=[x-2y].[5-x-2y]

câu c, d ,e lm tương tự nha

\(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy=5\\4xy\left(x+2y\right)+5\left(x+2y\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-4b=5\\4ab+5a=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}4b=a^2-5\\a\left(a^2-5\right)+5a=1\end{matrix}\right.\)

\(\Rightarrow a^3=1\)=> a=1 => 4b= 1 -5 =4=> b = -1

=>\(\left\{{}\begin{matrix}x+2y=1\\xy=-1\end{matrix}\right.\)

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

c: \(=x^2+6xy+9y^2\)

e: \(=x^4-4y^2\)

\(3x^2+4y^2-5x^2+6y^2-7x^2-10y^2-5\)\(=\left(3x^2-5x^2-7x^2\right)+\left(4y^2+6y^2-10y^2\right)-5\)

\(=-9x^2-5=-\left(9x^2+5\right)=-230\)

\(\Rightarrow9x^2+5=230\Rightarrow9x^2=225\Rightarrow x^2=25\)=> x = 5 hoặc x = -5

cho điểm đi rồi trả lời.không thì thôi. 

\(3x^2+4y^2-5x^2+6y^2-7x^2-10y^2-5\)

\(=\left(3-5-7\right)x^2+\left(4+6-10\right)y^2-5\)

\(=-9x^2-5\)

=> \(-9x^2-5=-230\)

=> \(-9x^2=-225\)

=> \(x^2=25\)

=> \(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)