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\(\begin{array}{l}{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\\{a^3} + {b^3} = {\left( {a + b} \right)^3} - 3{a^2}b - 3a{b^2}\\\;\;\;\;\;\;\;\;\;\; = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\\\;\;\;\;\;\;\;\;\;\; = \left( {a + b} \right)\left[ {{{\left( {a + b} \right)}^2} - 3ab} \right]\\\;\;\;\;\;\;\;\;\;\; = \left( {a + b} \right)\left[ {{a^2} + 2ab + {b^2} - 3ab} \right]\\\;\;\;\;\;\;\;\;\;\; = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\end{array}\) \(\begin{array}{l}{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\\{a^3} - {b^3} = {\left( {a - b} \right)^3} + 3{a^2}b - 3a{b^2}\\\;\;\;\;\;\;\;\;\;\; = {\left( {a - b} \right)^3} + 3ab\left( {a - b} \right)\\\;\;\;\;\;\;\;\;\;\; = \left( {a - b} \right)\left[ {{{\left( {a - b} \right)}^2} + 3ab} \right]\\\;\;\;\;\;\;\;\;\;\; = \left( {a - b} \right)\left[ {{a^2} - 2ab + {b^2} + 3ab} \right]\\\;\;\;\;\;\;\;\;\;\; = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\end{array}\)
Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)
\(=\left(ac+bd\right)^2+\left(ad+bc\right)^2\)
( 2x - 3y )2 = 4x2 - 12xy + 9y2
( 3√x - y )2 = 9x - 6y√x + y2 ( x ≥ 0 )
1. \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(VP=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2\)
\(\Rightarrow VT=VP\)
2. \(a^4-b^4=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)
\(VP=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)=\left(a^2-b^2\right)\left(a^2+b^2\right)=a^4+a^2b^2-b^2a^2-b^4=a^4-b^4\)
\(\Rightarrow VT=VP\)
3. \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(bx+ay\right)^2\)
\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(VP=\left(ax-by\right)^2+\left(bx+ay\right)^2=a^2x^2-2axby+b^2y^2+b^2x^2+2bxay+a^2y^2=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\Rightarrow VT=VP\)
a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\rightarrow a=b=c=1\)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\rightarrow a=b=c\)
\(a+b+c=0\)
=>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\)
=>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
=>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\)
=>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)
các bạn giúp mình nha
các bạn ơi giúp mình đi mình đang cần gấp lắm. cảm ơn trước nha