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Bài 1:
b) Ta có: \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Leftrightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)
\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)
nên x-89=0
hay x=89
Vậy: S={89}
Bài 1:
a)ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}\)
Suy ra: \(x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhân\right)\\x=6\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;6}
\(45.\)
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
\(=a^2-ab+b^2+3ab-6a^2b^2+6a^2b^2\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
\(=1^2\)
\(=1\).
42:
a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3ab(a+b)-3bac
=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)
=0
=>a^3+b^3+c^3=3abc
44:
a: x^3+y^3+3xy
=(x+y)^3-3xy(x+y)+3xy
=1^3-3xy+3xy=1
b: x^3-y^3-3xy
=(x-y)^3+3xy(x-y)-3xy
=1^3+3xy-3xy=1
