\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2\right)^2+2.5xy^2.3z+\left(3z\right)^2=\left(5xy^2+3z\right)^2\)

\(\frac{16}{9}x^2+4xyz^2+\frac{9}{4}y^2z^4=\left(\frac{4}{3}x\right)^2+2.\frac{4}{3}x.\frac{3}{2}yz^2+\left(\frac{3}{2}yz^2\right)^2=\left(\frac{4}{3}x+\frac{3}{2}yz^2\right)^2\)

\(\frac{9}{25}x^2+\frac{12}{35}xy+\frac{4}{49}y^2=\left(\frac{3}{5}x\right)^2+2.\frac{3}{5}x.\frac{2}{7}y+\left(\frac{2}{7}y\right)^2=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2\)( tự thay vào tính nhé )

\(\frac{25}{16}u^4y^2+\frac{1}{5}u^2+y^3+\frac{4}{625}y^4=\left(\frac{5}{4}u^2y\right)^2+2.\frac{5}{4}u^2y.\frac{2}{25}.y^2+\left(\frac{2}{25}y^2\right)^2=\left(\frac{5}{4}u^2y+\frac{2}{25}y^2\right)^2\)( tự thay vào tính nhé )

Tham khảo nhé~

a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)

b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)

1)\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2+3z\right)^2\)

\(\dfrac{16}{9}x^2+4xyz^2+\dfrac{9}{4}y^2z^4=\left(\dfrac{4}{3}x+\dfrac{3}{2}yz^2\right)^2\)

2)

a)\(\dfrac{9}{25}x^2+\dfrac{12}{35}xy+\dfrac{4}{49}y^2=\left(\dfrac{3}{5}x+\dfrac{2}{7}y\right)^2=\left(\dfrac{3}{5}.5+\dfrac{2}{7}.\left(-7\right)\right)^2=\left(3-2\right)^2=1\)b)\(\dfrac{25}{16}u^4v^2+\dfrac{1}{5}u^2v^3+\dfrac{4}{625}v^4\)

\(=\left(\dfrac{5}{4}u^2v+\dfrac{2}{25}v^2\right)^2=\left(\dfrac{5}{4}.\dfrac{4}{25}.\left(-5\right)+\dfrac{2}{25}.\left(-5\right)^2\right)^2\)

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

A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.........\frac{899}{900}\)

A=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}..........\frac{29.31}{30.30}\)

A=\(\frac{1.2.3.......29}{2.3.4.......30}.\frac{3.4.5........31}{2.3.4.......30}\)

A=\(\frac{1}{30}.\frac{2}{31}=\frac{1}{465}\)

Bài 50:

\((5x+3y)^2=25x^2+30xy+9y^2\)

Bài 51:

\((\frac{1}{3}xy^m+4x^2y)^2=\frac{1}{9}x^2y^{2m}+2.\frac{1}{3}xy^m.4x^2y+16x^4y^2\)

\(=\frac{1}{9}x^2y^{2m}+\frac{}{3}x^3y^{m+1}+16x^4y^2\)

Bài 54:

\(25x^2y^4+30xy^2z+9z^2=(5xy^2)^2+2.(5xy^2).(3z)+(3z)^2\)

\(=(5xy^2+3z)^2\)

Bài 55:

\(\frac{16}{9}x^2+4xyz^2+\frac{9}{4}y^2z^4=(\frac{4}{3}x)^2+2.(\frac{4}{3}x).(\frac{3}{2}yz^2)+(\frac{3}{2}yz^2)^2\)

\(=(\frac{4}{3}x+\frac{3}{2}yz^2)^2\)

Bạn chỉ cần nhớ rõ hằng đẳng thức đáng nhớ số 1 là được.

Bài 58:

Nếu số tự nhiên $n$ chia $11$ dư $4$, đặt $n=11k+4$ ($k\in\mathbb{N}$)

Khi đó:

\(n^2=(11k+4)^2=11^2k^2+88k+4^2=(11^2k^2+88k+11)+5\)

\(=11(11k^2+8k+1)+5\) chia $11$ dư $5$

Vậy $n^2$ chia $11$ dư $5$ (đpcm)

a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)

\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)

\(=x\left(x+2y+3z\right)\)

b: \(\left(x+3\right)^2+4\left(x+3\right)+4\)

\(=\left(x+3+2\right)^2\)

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

c: \(25+10\left(x+1\right)+\left(x+1\right)^2\)

\(=\left(x+1+5\right)^2\)

\(=\left(x+6\right)\left(x+6\right)\)

giúp mình nha!!!!!!!!!!!!!!!!!!!!!!!!!!!

Dấu. là dấu nhân

\(A=\left(9\frac{3}{4}:5,2+x.2\frac{7}{24}\right):1\frac{9}{16}=6\)

\(A=\left(9\frac{3}{4}:5,2+x.2\frac{7}{24}\right)=\frac{75}{8}\)

 \(A=\left(2\frac{7}{24}.\frac{9}{11}+x.2\frac{7}{24}\right)=\frac{75}{8}\)

\(A=\left[2\frac{7}{24}.\left(\frac{9}{11}+x\right)\right]=\frac{75}{8}\)

\(A=\frac{9}{11}+x=\frac{45}{11}\)

\(x=\frac{45}{11}-\frac{9}{11}\)

\(x=\frac{36}{11}\)

36/11