\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

\(a^3+b^3+c^3=3abc\\ \Leftrightarrow a^3+b^3+c^3-3abc=0\\ \Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\\ \Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

Vậy \(a^3+b^3+c^3=3abc\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Lời giải:

Từ \(10x^2=10y^2+z^2\Rightarrow 10x^2-10y^2=z^2\)

Theo hằng đẳng thức đáng nhớ ta có:

\((7x-3y+2z)(7x-3y-2z)=(7x-3y)^2-(2z)^2\)

\(=(7x-3y)^2-4z^2=(49x^2-42xy+9y^2)-4(10x^2-10y^2)\)

\(=9x^2-42xy+49y^2=(3x)^2-2.(3x).(7y)+(7y)^2=(3x-7y)^2\)

Ta có đpcm.

\(VT=\left(7x-3y+2z\right)\left(7x-3y-2z\right)\)

\(=\left(7x-3y\right)^2-4z^2\)

\(=49x^2-42xy+9y^2-4z^2\)

\(=4\cdot10x^2+9x^2-42xy+9y^2-4z^2\)

mà 10x² = 10y² + z²

\(\Rightarrow VT=4\left(10y^2+z^2\right)+9x^2-42xy+9y^2-4z^2\)

\(=40y^2+4x^2+9x^2-42xy+9y^2-4z^2\)

\(=9x^2-42xy+49y^2\)

\(=\left(3x-7y\right)^2=VP\)

Ta có :

10x²=10y²+z²

=>40x²=40y²+4z²

=>49x²-9x²-49y²+9y²-4z²=0

=>49x²+9y²-4z²=9x²+49y²

=>49x²-2.7x.3y+9y²-4z²=9x²-2.3x.7y+49y²

=>(7x-3y)²-4z²=(3x-7y)²

=>(7x-3y+2z)(7x-3y-2z)=(3x-7y)²

^kbnha

Bài 1:

Bạn tham khảo tại link sau:

Câu hỏi của hậuu đậuu - Toán lớp 8 | Học trực tuyến

Bài 2:

Ta có:

\(a^3+b^3+c^2-3abc=0\)

\(\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0\)

\(\Leftrightarrow [(a+b)^3+c^3]-3ab(a+b+c)=0\)

\(=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)=0\)

\(\Leftrightarrow (a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0\)

\(\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\)

Vì $a,b,c$ là 3 số dương nên $a+b+c>0$ . Suy ra $a+b+c\neq 0$

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)

Vì \((a-b)^2; (b-c)^2; (c-a)^2\geq 0, \forall a,b,c>0\). Do đó để tổng của chúng bằng $0$ thì \((a-b)^2=(b-c)^2=(c-a)^2=0\)

\(\Rightarrow a=b=c\)

Ta có đpcm.

Bài 3:

Áp dụng công thức \((a-b)(a+b)=a^2-b^2\):

\(C=(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8)(3^{16}+2^{16})\)

\(=(3-2)(3+2)(3^2+2^2)(3^4+2^4)(3^8+2^8)(3^{16}+2^{16})\)

\(=(3^2-2^2)(3^2+2^2)(3^4+2^4)(3^8+2^8)(3^{16}+2^{16})\)

\(=(3^4-2^4)(3^4+2^4)(3^8+2^8)(3^{16}+2^{16})\)

\(=(3^8-2^8)(3^8+2^8)(3^{16}+2^{16})\)

\(=(3^{16}-2^{16})(3^{16}+2^{16})=3^{32}-2^{32}\)