1 (x-3) (x^2+3x+9)
2 (x+4) (x^2+4x+16)
3. (x-3y) (x^2+3xy+9y^2)
hằng đẳng thức
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\(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^2+3x^2y+9xy^2-3x^2y-9xy^2-27y^3\)
\(=\)\(x^2-27y^3\)
\(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3xy}+4y^2\right)\)
\(=\)\(\frac{x^3}{27}-\frac{2}{9xy}+\frac{4xy^2}{3}+\frac{2x^2y}{9}-\frac{4y}{3xy}+8y^3\)
làm nốt nha
a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
Gọi diện tích hình vuông là Shv.Khi đó mỗi ô vuông nhỏ có diện tích là Shv9 . Ta thấy ngay diện tích tam giác ABK bằng một nửa diện tích hình chữ nhật AKBH và bằng Shv9 .
Tương tự SAID=SDNC=SBMC=SABK=Shv9 và SIKMN=Shv9
Vậy thì SABCD=4.Shv9 +Shv9 =59 Shv
Vậy diện tích phần còn lại bằng 49 Shv
Suy ra diện tích hình vuông ABCD bằng 54 diện tích phần còn lại.
k mình nha
\(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\)
\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)
\(=x^4-\frac{4}{25}y^2\)
\(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}xy\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Giải:
+) \(\left(x-4\right)\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4.x+4^2\right)\)
\(=x^3-4^3\)
\(=x^3-64\)
+) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
+) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
+) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+\dfrac{1}{3}.x^2+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
Vậy ...
a) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=x^3-3^3\)
\(=x^3-27.\)
a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)
b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)
\(=\left(x^2\right)^3-3^3=x^6-27\)
\(a,\left(x+4\right).\left(x^2-4x+16\right)=x^3-4x^2+16x+4x^2-16x+64\) \(64\)
\(=x^3+64\)
hoặc \(\left(x+4\right).\left(x^2-4x+16\right)=x^3+64\) ÁP Dụng hằng đẳng thức
\(b,\left(x-3y\right).\left(x^2+3xy+9y^2\right)=x^3-27\)
a) (x+4).(x^2-4x+16)
= (x+4).(x^2-x.4+4^2)
= x^3+4^3
= x^3+64
b) (x-3y).(x^2+3xy+9y^2)
= (x-3y).(x^2+x.3y+(3y)^2)
= x^3-(3y)^3
= x^3-27y^3
`(x-3).(x^2+3x+9)`
`=x.(x^2+3x+9)-3.(x^2+3x+9)`
`=x^3 - 27`
______________________
`(x+4).(x^2+4x+16)`
`=x.(x^2+4x+16) + 4.(x^2+4x+16)`
`=x^3 +8x^2 +32x +64`
______________________
`(x-3y).(x^2+3xy+9y^2)`
`=x.(x^2+3xy+9y^2) - 3y.(x^2+3xy+9y^2)`
`=x^3 -27y^3`