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

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

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,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)

\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)

\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+zx\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=5\left(x+y+z\right)^2+\left(x^2+y^2+z^2\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(\ge5.\left(\frac{3}{4}\right)^2+\frac{\left(x+y+z\right)^2}{3}+\frac{2.9}{4\left(x+y+z\right)}\)

\(=5.\left(\frac{3}{4}\right)^2+\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{2.9}{\frac{4.3}{4}}=9\)

Tích mình đi

Ai tích sẽ có lợi

vì khi có lợi bạn sẽ được người khác tích lại.

THANKS

\(\frac{9}{16}x^{2m-2}y^2-2x^my^m+\frac{16}{9}x^2y^{2m-2}\)

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

p/s: chúc bạn học tốt

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

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