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\(\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4}=\dfrac{a^2+2ab+b^2-a^2+2ab-b^2}{4}=\dfrac{4ab}{4}=ab\left(đpcm\right)\)
\(\left(x+y\right)^2+\left(x-y\right)^2=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2=2\left(x^2+y^2\right)\left(dpcm\right)\)
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a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
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\(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+bx+ax+ab\right)\left(x+c\right)\)
\(=x^3+bx^2+ax^2+abx+cx^2+bcx+acx+abc\)
\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+cax\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
\(\Rightarrowđpcm\)
Ta có: (x+a)(x+b)(x+c) = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
VT = (x2+ax+bx+ab)(x+c)
= x3 + ax2 + bx2 + abx + cx2 + cax + bcx + abc (1)
VP = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
= x3 + ax2 + bx2 + abx + cx2 + cax + bcx + abc (2)
Từ (1) và (2), suy ra:
(x+a)(x+b)(x+c) = x3 + (a+b+c)x2 +(ab+bc+ca)x + abc
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từ A=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)
=>A=x2-ax-bx+ab+x2-bx-cx+bc+x2-cx-ax+ac
=>A=3x2-2ax-2bx-2cx+ab+bc+ac
=>A=3x2-2x(a+b+c)+ab+bc+ac
mà a+b+c=2x(gt)
=>A=3x2-2x.2x+ab+bc+ac
=>A=3x2-4x2+ab+bc+ac
=>A=ab+bc+ac-x2=VP
Vậy ...........................................
Với a = 1, b = 4, c = 2, d = 3 thì a + b = 5 =c + d.
Biến đổi: P(x) = (x + 1)(x + 4)( x + 2)( x + 3) – 15
= (x2 + 5x + 4)(x2 + 5x + 6) – 15
Đặt y = x2 + 5x + 4 thì P(x) trở thành
Q(y) = y(y + 2) – 1
= y2 +2y – 15
= y2 – 3y + 5y – 15
= y(y – 3) + 5( y – 3)
= (y – 3)(y + 5)
Do đó: P(x) = (x2 +5x + 1)(x2 + 5x + 9)
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a) Ta có:
\(VT=\left(a-b\right)^2\)
\(=a^2-2\cdot a\cdot b+b^2\)
\(=a^2-2ab+b^2\)
\(=a^2-4ab+2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab=VP\)
⇒ Đpcm
b) Ta có:
\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+2\cdot x\cdot y+y^2+x^2-2\cdot x\cdot y+y^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2\)
\(=\left(x^2+x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)
\(=2x^2+0+2y^2\)
\(=2x^2+2y^2\)
\(=2\left(x^2+y^2\right)=VP\)
⇒ Đpcm
a: (a-b)^2
=a^2-2ab+b^2
=a^2+2ab+b^2-4ab
=(a+b)^2-4ab
b: (x+y)^2+(x-y)^2
=x^2+2xy+y^2+x^2-2xy+y^2
=2x^2+2y^2
=2(x^2+y^2)
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a: (a+b+c)^2+a^2+b^2+c^2
=a^2+b^2+c^2+a^2+b^2+c^2+2ab+2ac+2bc
=(a^2+2ab+b^2)+(b^2+2bc+c^2)+(a^2+2ac+c^2)
=(a+b)^2+(b+c)^2+(c+a)^2
b: (x+y)^4-2(x^2+xy+y^2)^2
=(x^2+2xy+y^2)^2-2(x^2+xy+y^2)^2
=x^4+4x^2y^2+y^4+4x^3y+2x^2y^2+4xy^3-2(x^4+x^2y^2+y^4+2x^3y+2x^2y^2+2xy^3)
=-x^4-y^4
=>ĐPCM
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\(\left(x+a\right)\left(x+b\right)\)
=>\(x^2+bx+ax+ab\)
=>\(x^2+\left(a+b\right)x+ab\)(ĐPCM)
Nhớ H cho mik nhé, các bạn.