\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)

a: =(x-y)(5+3x)

c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)

hay x=-1/3

`a, = 5^3 xx 3^3 = 15^3`

`b, = 12^3 xx 4^3 = 48^3`

`c, = 625^3 xx 8^3 = 5000^3`

`d, = 625 ^3 xx 125^3 = 78125^3`

`e, = 5^20 : 5^14 = 5^6`

1,\(125\cdot27=5^3\cdot3^3=\left(5\cdot3\right)^3=15^3\)

2, \(12^3\cdot64=12^3\cdot4^3=\left(12\cdot4\right)^3=48^3\)

3, \(25^6\cdot8^3=\left(5^2\right)^6\cdot\left(2^3\right)^3=5^8\cdot2^9\)

4, \(25^6\cdot125^3=\left(5^2\right)^3\cdot\left(5^3\right)^3=5^6\cdot5^9=5^{15}\)

5,\(625^5:25^7=\left(5^4\right)^5:\left(5^2\right)^7=5^{20}:5^{14}=5^6\)

a, 0,5 = \(\dfrac{5}{10}\);    0,71 = \(\dfrac{71}{100}\);    1,348 = \(\dfrac{1348}{1000}\);  49,16 = \(\dfrac{4916}{100}\)

b, \(\dfrac{3}{5}\) = \(\dfrac{3\times2}{5\times2}\) = \(\dfrac{6}{10}\);     \(\dfrac{7}{25}\) = \(\dfrac{7\times4}{25\times4}\) = \(\dfrac{28}{100}\); \(\dfrac{6}{125}\) = \(\dfrac{6\times8}{125\times8}\)= \(\dfrac{48}{1000}\)

\(\dfrac{7}{4}\) = \(\dfrac{7\times25}{4\times25}\) = \(\dfrac{175}{100}\)

ủa sao bên trên là số thập phân rồi mà

a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)

Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))

\(\frac{1}{4}a^2+2ab+4b^2\)

\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)

\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)

\(=\left(\frac{1}{2}a+2b\right)^2\)

b. \(25+10x+x^2\)

\(=x^2+2.x.5+5^2\)

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

c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)

\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)

\(=\left(y^4-\frac{1}{3}\right)^2\)

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)

2)
a)
\(=x^3+125\)
\(\)b)\(=1-x^3\)
c)
=\(y^3+27t^3\)
 

a, \(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
b, \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
c, \(x^2-10x+25=x^2-2.5x+5^2=\left(x-5\right)^2\)

1. \(25+10x+x^2\\ \Leftrightarrow5^2+2\cdot5\cdot x+x^2\\ \Leftrightarrow\left(5+x\right)^2\\ \Leftrightarrow\left(5+x\right)\left(5+x\right)\)

2. \(8x^3-\dfrac{1}{8}\\ \Leftrightarrow\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]\\ \Leftrightarrow\left(2x-\dfrac{1}{2}\right)\left[4x^2+x+\dfrac{1}{4}\right]\)

3. \(x^2-10x+25\\ \Leftrightarrow x^2-2\cdot5\cdot x+5^2\\ \Leftrightarrow\left(x-5\right)^2\\ \Leftrightarrow\left(x-5\right)\left(x-5\right)\)