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

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

\(\frac{9}{25}x^2+\frac{12}{35}xy+\frac{4}{49}y^2=\left(\frac{3}{5}x\right)^2+2.\frac{3}{5}x.\frac{2}{7}y+\left(\frac{2}{7}y\right)^2=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2\)( 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. \(9x^2+30x+25=\left(3x+5\right)^2\)

b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)

c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)

d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)

e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

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

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

x^2y/2;x/-5^2;-4/5

a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)

b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)

a,(4x+1)² e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)

b,(x-\(\dfrac{1}{2}\))² g,\(\left(xy+1\right)^2\)

c,(\(x+\dfrac{3}{2}\))² h,\(\left(x+5\right)^2\)

d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)

k,\(-\left(2x+3\right)^2\)

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)

Sao câu hỏi của bn giống của mình vậy ???

Bài này wá dễ!

Động não đi bn.