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Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
a) Ta có: \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2bc-2ab-2ac+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ca\)
\(=a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
b) Đặt x = b + c - a
y = c + a - b
z = a + b - c
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y+z\)
Ta có: \(\left(a+b+c\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left[xy+\left(x+y\right)z+z^2\right]\)
\(=3\left(x+y\right)\left[z^2+xy+xz+yz\right]\)
\(=3\left(x+y\right)\left[z\left(x+y\right)+y\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=3.2a.2b.2c\)
\(=24abc\) (đpcm)
a) \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)\left(ac+bc+c^2\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
b) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
a,\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)
Tương tự :
\(\left(b+c-a\right)^3=b^3+c^3-a^3+3\left(a^2b-b^2a+ca^2-ac^2+b^2c+c^2b\right)\)
\(\left(b+a-c\right)^3=b^3-c^3+a^3+3\left(a^2b+b^2a-ca^2+ac^2-b^2c+c^2b\right)\)
\(\left(a+c-b\right)^3=c^3+a^3-b^3+3\left(-a^2b+b^2a+ca^2+ac^2+b^2c-c^2b\right)\)
Biểu thức sau khi rút gọn ta được
24abc
b,\(\left(a+b\right)^3=a^3+b^3+3\left(a^2b+b^2a\right)\)
\(\left(c+b\right)^3=c^3+b^3+3\left(c^2b+b^2c\right)\)
\(\left(a+c\right)^3=a^3+c^3+3\left(a^2c+b^2c\right)\)
=>\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3=\)\(2\left(a^2+b^2+c^2\right)+3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b\right)\)
Lại có
\(3\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(3\left(a^2b+b^2a+c^2a+ca^2+b^2c+c^2b+2abc\right)\right)\)
Biểu thức khi đó trở thành
\(2\left(a^2+b^2+c^2\right)-6abc=2\left(a^2+b^2+c^2-3abc\right)\)
Tặng vk iu
b) \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Biến đổi VT ta có :
+) \(a^3+b^3+c^3=ab+bc+ca\)
\(\Leftrightarrow3a^3+3b^3+3c^3=3ab+3bc+3ca\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0\)
\(\Rightarrow a=b=c\)
< => VT = VP
=> đpcm
\(VP=\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3=VT\)