rut gon bieu thuc \(\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3-\left(a+b-c\right)^3\)
Những câu hỏi liên quan
Rut gon phan thuc
\(M=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)
rut gon phan thuc \(\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2}\)
ĐK : \(a\ne b\ne c\)
\(\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca\right)-3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)
\(=\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)
\(=\dfrac{a+b+c}{2}\)
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rut gon bieu thuc :
a,\(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
b,\(\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)
a) \(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(\Leftrightarrow2b^3\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6ab^2\)
\(\Leftrightarrow2b^3+6a^2b-6ab^2\)
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rut gon \(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
rut gon
\(C=\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(c-a-b\right)^3\)
rut gon cac bieu thuc
a) \(4x^2\left(5x^2-2x+3\right)\)
b)\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
c)\(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2-x-2\right)\)
\(=\left(-4\right)\left(x+2\right)\)
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ai tra loi nhanh cau nay minh tich ngay nguoi do
de bai rut gon bieu thuc
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab-2ac-2bc-a^2+2ac-c^2-2ab+2bc\)
\(=b^2\)
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rut gon A=\(\frac{\left(a+b+c\right)^5-a^5-b^5-c^5}{\left(a+b+c\right)^3-a^3-b^3-c^3}\)iup minh voi di may ban
cho biểu thức p=\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
a)rut gon p
b) xet dau cua bieu thuc M = a. \(\left(P-\frac{1}{2}\right)\)