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\(\frac{1}{4}x^6-0,01y^2=\left(\frac{1}{2}x^3\right)^2-\left(0,1y\right)^2\)
\(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)
Vậy \(\frac{1}{4}x^6-0,01y^2\)\(=\left(\frac{1}{2}x^3-0,1y\right).\left(\frac{1}{2}x^3+0,1y\right)\)
Tham khảo nhé ~
\(\frac{1}{4}x^6-0.01y^2\)
\(=\left(\frac{1}{2}x^3\right)^2-\left(0.1y\right)^2\)
\(=\left(\frac{1}{2}x^3-0.1y\right)\left(\frac{1}{2}x^3+0.1y\right)\)
Mong lần này không sai nữa ......
a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2\)
b,\(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)\)
\(=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2\)
Trả lời:
a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)\)\(=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2=\frac{x^2}{y^2}-\frac{4}{9}\)
b, \(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2=4x-\frac{4}{9}\)
\(4x^2-20xy^2+25y^4=\left(2x\right)^2-2.2x.5y^2+\left(5y^2\right)^2=\left(2x-5y^2\right)^2\)
Áp dụng hằng đẳng thức: \(\left(A-B\right)^2=A^2-2AB+B^2\)
\(4x^2-20xy^2+25y^4\)
\(=\left(2x\right)^2-2\cdot2x\cdot5y^2+\left(5y\right)^2\)
\(=\left(2x-5y\right)^2\)
b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
1) \(\left(\frac{1}{4}+k\right)^2=\frac{1}{16}+\frac{1}{2}k+k^2\)
2) \(\left(2x^2y+\frac{1}{2}xy^2\right)^2=4x^4y^2+2x^3y^3+\frac{1}{4}x^2y^4\) (hẳn đề là như thế này)
3) \(\left(x+\frac{1}{2}y\right)^2=x^2+xy+\frac{1}{4}y^2\)
x2-2.x.1/2 +(1/2)2-9/4
=(x-1/2)2-9/4
=(x-1/2)2-(3/2)2
=(x-1/2-3/2).(x-1/2+3/2)
=(x-2)(x+1)
\(\frac{4}{9}x^2-\frac{2}{3}xy+\frac{1}{4}y^2\)
\(=\left(\frac{2}{3}x\right)^2-2.\frac{2}{3}x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(\frac{2}{3}x-\frac{1}{2}y\right)^2\)
p/s: chúc bạn học tốt