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\(a,\left(2x+y+3\right)^2=4x^2+y^2+9+4xy+12x+6y\)
\(b,\left(x-2y+1\right)^2=x^2+4y^2+1-4xy+2x-4y\)
\(c,\left(x^2-2xy^2-3\right)^2=x^4+2x^2y^4+9-4x^3y^2-6x^2+12xy^2\)
Giải:
a) \(\left(2x+y+3\right)^2\)
\(=\left(2x+y\right)^2+2.3\left(2x+y\right)+3^2\)
\(=\left(2x\right)^2+2.2x.y+y^2+2.3\left(2x+y\right)+3^2\)
\(=4x^2+4xy+y^2+12x+6y+9\)
Vậy ...
b) \(\left(x-2y+1\right)^2\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1^2\)
\(=x^2-2.x.2y+\left(2y\right)^2+2x-4y+1^2\)
\(=x^2-4xy+4y^2+2x-4y+1\)
Vậy ...
c) \(\left(x^2-2xy^2-3\right)^2\)
\(=\left(x^2-2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=\left(x^2\right)^2-2.x^2.2xy^2+\left(2xy^2\right)^2+2.3.\left(x^2-2xy^2\right)-3^2\)
\(=x^4-4x^3y^2+4x^2y^4+6x^2-12xy^2-9\)
Vậy ...
Câu 1: Đặt a/x là m; b/y là n; c/z là p, ta có: m + n + p = 2; 1/m + 1/n + 1/p = 0. Tìm m2 + n2 + p2 ?
Từ 1/m + 1/n + 1/p = 0
=> mnp(1/m + 1/n + 1/p) = 0
<=> mn + np + mp = 0
Mặt khác, ta có (m + n + p)2 = m2 + n2 + p2 + 2(mp + np + mp) = 4
Mà mn + np + mp = 0 => m2 + n2 + p2 + 0 = 4
Trả lời: Vậy a2/x2 + b2/y2 + c2/z2 = 4
Giúp bạn câu 1 thôi (Mình lười lắm)
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Chúc bn học tốt!!
1) \(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}\right)^2=\dfrac{1}{4}x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\)
\(\left(a+b\sqrt{3}\right)^2=a^2+2\sqrt{3}ab+3b^2\)
2) \(4a^2+4a+1=\left(2a+1\right)^2\)
\(9x^2-6x+1=\left(3x-1\right)^2\)
\(\dfrac{1}{4}x^2-\dfrac{1}{3}xy+\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right)^2\)
\(a,\left(-4xy-5\right)\left(5-4xy\right)=\left(4xy+5\right)\left(4xy-5\right).\)
\(=\left(4xy\right)^2-5^2=16x^2y^2-25\)
\(b,\left(a^2b+ab^2\right)\left(ab^2-a^2b\right)=\left(ab^2+a^2b\right)\left(ab^2-a^2b\right)\)
\(=\left(ab^2\right)^2-\left(a^2b\right)^2=a^2b^4-a^4b^2\)
\(c,\left(3x-4\right)^2+2\left(3x-4\right)\left(4-x\right)+\left(4-x\right)^2\)
\(=\left[\left(3x-4\right)+\left(4-x\right)\right]^2\)
\(=\left(3x-4+4-x\right)^2=\left(2x\right)^2=4x^2\)
\(d,\left(a^2+ab+b^2\right)\left(a^2-ab+b^2\right)-\left(a^4+b^4\right)\)
\(=\left[\left(a^2+b^2\right)+ab\right]\left[\left(a^2+b^2\right)-ab\right]-\left(a^4+b^4\right)\)
\(=\left(a^2+b^2\right)^2-\left(ab\right)^2-a^4-b^4\)
\(=a^4+2a^2b^2+b^4-a^2b^2-a^4-b^4=a^2b^2\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a) \(\left(\frac{1}{3}u+3v\right)^2=\frac{1}{9}u^2+2uv+9v^2\)
b) \(\left(\frac{1}{2}x^2-6x\right)^2=\frac{1}{4}x^4-6x^3+36x^2\)
c) \(\left(-\frac{1}{2}a+b\right)^2=\frac{1}{4}a^2-ab+b^2\)
d) \(\left(-\frac{4}{3}a-\frac{1}{3}b\right)^2=\frac{16}{9}a^2+\frac{8}{9}ab+\frac{1}{9}b^2\)
e) \(\left(\frac{2}{3}x-\frac{3}{2}y\right)\left(\frac{2}{3}x+\frac{3}{2}y\right)=\left(\frac{2}{3}x\right)^2-\left(\frac{3}{2}y\right)^2=\frac{4}{9}x^2-\frac{9}{4}y^2\)
a,\(\left(x^2+2xy\right)^3=\left(x^2\right)^3+3.\left(x^2\right)^2.2xy+3.\left(2xy\right)^2.x^2+\left(2xy\right)^3\)
\(=x^6+6x^5y+12x^4y^2+8x^3y^3\)
b,\(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.\left(2y\right)^2.3x^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36y^2x^2-8y^3\)
c,\(\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)
a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
a) a 2 + b 2 + c 2 + 2ab - 2bc - 2 ac.
b) 1 – 2x + x 2 .