Lời giải:

a. $=(2x)^2-2.2x.5y+(5y)^2=4x^2-20xy+25y^2$
b. $=(3x)^2+2.3x.2y+(2y)^2=9x^2+12xy+4y^2$

c. $=(4y+3x)(4y-3x)=(4y)^2-(3x)^2=16y^2-9x^2$

\(a.4x^2-10xy+25y^2\)

\(b.9x^2+6xy+4y^2\)

\(c.16y^2-9x^2\)

a: \(\left(x^2+2y\right)^3\)

\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2y+3\cdot x^2\cdot\left(2y\right)^2+\left(2y\right)^3\)

\(=x^6+6x^4y+12x^2y^2+8y^3\)

b: \(\left(\dfrac{1}{2}x-1\right)^3\)

\(=\left(\dfrac{1}{2}x\right)^3-3\cdot\left(\dfrac{1}{2}x\right)^2\cdot1+3\cdot\dfrac{1}{2}x\cdot1^2-1^3\)

\(=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1\)

a) Sử dụng công thức bình phương của tổng với số hạng thứ nhất là a + b và số hạng thứ hai là c.

Biến đổi thu được A = a 2   +   b 2   +   c 2  + 2ab + 2bc + 2 ac;

b)  a 2   +   b 2   +   c 2  - 2ab + 2bc - 2 ac.

a, \(\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ca\)

b, \(\left(a+2b-c\right)^2=a^2+4b^2+c^2+4ab-4bc-2ca\)

c, \(\left(2a-b-c\right)^2=4a^2+b^2+c^2-4ab+2bc-4ca\)

a,

b,

a,

b,

a) \((a+b+c)^2\) \(= a^2 + b^2 +c ^2 +2ab+2bc+2ca\)

b) \((a-b-c)^2 = a^2 + b^2 +c^2 - 2ab + 2bc -2ca\)

học tốt !

a) ( a - b + c ) ² = a² + b² + c² + 2ab - 2ac - 2bc

b ) ( a - b - c )² = a² + b² + c² - 2ab + 2bc - 2ca

Giải

a/\(\left(a-b+c\right)^2=a^2+b^2+c^2+2ab-2ac-2bc\)

b/\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ca\)

\(\left(a-b\right)\left(a-b\right)\)

\(=a\left(a-b\right)-b\left(a-b\right)\)

\(=a^2-ab-ab+b^2\)

\(=a^2-2ab+b^2\)

\(\left(a+b\right)\left(a-b\right)\)

\(=a\left(a-b\right)+b\left(a-b\right)\)

\(=a^2-ab+ab-b^2\)

\(=a^2-b^2\)

\(\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a+b\right)\left(a-b\right)^2\)

\(=\left(a+b\right)\left(a-b\right)\left(a-b\right)\)

\(=\left(a^2-b^2\right)\left(a-b\right)\)

\(=a^2\left(a-b\right)-b^2\left(a-b\right)\)

\(=a^3-a^2b-b^2a+b^3\)

p/s thật ra cái này áp dụng hđt là ra ồi

a) a 2   +   b 2   +   c 2  + 2ab - 2bc - 2 ac.

b) 1 – 2x + x 2 .

\(a.\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ca\)

\(b.\left(a+b-c-d\right)^2\)

\(=\left[a+b-\left(c+d\right)\right]\)

\(=a^2+b^2+\left(c+d\right)^2+2ab-2b\left(c+d\right)-2a\left(c+d\right)\)

\(=a^2+b^2+c^2+2cd+d^2+2ab-2bc-2bd-2ac-2ad\)

\(c.\left(2x-y+3z\right)^2=4x^2+y^2+9z^2-4xy-6yz+12xz\).

a) \(\left(2x-3y\right)^2=4x^2-12xy+9y^2\)

b) \(\left(5p-q\right)^2=25p^2-10pq+q^2\)

c) \(\left(-a-b\right)^2=-a^2-2ab-b^2\)

d) \(\left(1+3s\right)^2=1+6s+9s^2\)

e) \(\left(a^2b+2b\right)^2=a^4b^2+4a^2b^2+4b^2\)

f) \(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)

a,\(\left(2x-3y\right)=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\)

=\(4x^2-12xy+6y^2\)

b,\(\left(5p-q\right)^2=\left(5p\right)^2-2.5p.q+q^2\)

=\(25p^2-10pq+q^2\)

c,(-a-b)\(^2=\left(-a\right)^2-2.\left(-a\right).b+b^2\)

=\(a^2+2ab+b^2\)

d,\(\left(1+3s\right)^2=1+6s+9s^2\)

e,(a\(^2b+2b)^2=(a^2b)^2+2.a^2b.2b^2+\left(2b\right)^2\)

=\(a^4b^2+4a^2b^2+4b^2\)

f,\(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)