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Ta có: c + d = 4.
<=> (c+d)2 = 16.
<=> c2 + 2cd + d2 = 16.
<=> 4a2 + b2 + c2 + 2cd + d2 = 2 + 16 = 18. (1)
Áp dụng BĐT Cauchy, ta có:
4a2 + c2 ≥ 2*2a*c = 4ac. (2)
b2 + d2 ≥ 2bd. (3)
Từ (1), (2) và (3) suy ra:
18 ≥ 4ac + 2bd + 2cd.
<=> 9 ≥ 2ac + bd + cd.
max A = 9 <=> 2a=c ; b=d.
Ta có:
\(\left(a+b+c\right)^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)
\(=a^2+2ab+b^2+2ac+2bc+c^2\)
\(=a^2+b^2+c^2+2\left(ab+bc+ca\right)\) \(\Rightarrowđpcm\)
e)
$x^3+6x^2+12x+8=x^3+3.2.x^2+3.2^2.x+2^3=(x+2)^3$
f)
$a^3-2a^2-ab^2+2b^2=(a^3-ab^2)-(2a^2-2b^2)$
$=a(a^2-b^2)-2(a^2-b^2)=(a^2-b^2)(a-2)=(a-b)(a+b)(a-2)$
g)
$2a^2x-2a^2-2abx+4ab-2b^2=(2a^2x-2abx)-(2a^2-4ab+2b^2)$
$=2ax(a-b)-2(a-b)^2=2(a-b)(ax-a+b)$
h)
\(x^2-2xy+y^2-25=(x-y)^2-25=(x-y)^2-5^2=(x-y+5)(x-y-5)\)
a)
$4x^2-40x^4+100x^3=4x^2(1-10x^2+25x)$
b)
\(3xy(x-5)-7x+35=3xy(x-5)-7(x-5)\)
\(=(x-5)(3xy-7)\)
c)
\(a^2-am-b^2-bm=(a^2-b^2)-(am+bm)=(a-b)(a+b)-m(a+b)\)
\(=(a+b)(a-b-m)\)
d)
\(x^3-4x-x^2y+4y=(x^3-x^2y)-(4x-4y)\)
\(=x^2(x-y)-4(x-y)=(x^2-4)(x-y)=(x-2)(x+2)(x-y)\)
Bài 1
\(x+y+z=0\)
\(\Leftrightarrow x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^3=-z^3\)
\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=-z^3\)
\(\Leftrightarrow x^3+y^3-3xyz=-z^3\) (vì x+y=-z)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
Ta có:
\(\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=196\)
\(\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=169\)
Lại có:
\(\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=27\)
\(\Leftrightarrow a^6+6a^4b^2+9ab^4-b^6-6a^2b^4-9a^4b^2=27\)
\(\Leftrightarrow a^6-3a^4b^2+3a^2b^4-b^6=27\)
\(\Leftrightarrow\left(a^2-b^2\right)^3=27\)
\(\Leftrightarrow a^2-b^2=\sqrt[3]{27}=3\)
\(a^3+3ab^2+b^3+3a^2b=27=\left(a+b\right)^3\Rightarrow a+b=3\)
\(a^3+3ab^2-b^3-3a^2b=1\Rightarrow\left(a-b\right)^3=1\Rightarrow a-b=1\)
\(\Rightarrow a^2-b^2=\left(a-b\right).\left(a+b\right)=3\)