Tính
3x-2 + 3x = 90
x =
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\(90x-6750=75x-x^2\)
\(\Leftrightarrow180x-6750=75x-x^2\)
\(\Leftrightarrow x^2+105x-6750=0\)
\(\Leftrightarrow x^2-45x+150x-6750=0\)
\(\Leftrightarrow\left(x-45\right)\left(x+150\right)=0
\)
\(\Leftrightarrow\left[{}\begin{matrix}x-45=0\\x+150=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=45\\x=-150\end{matrix}\right.\)
Vậy x=45 hoặc x=-150
x=89 nên x+1=90
\(f\left(x\right)=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-x^4\left(x+1\right)+...+x\left(x+1\right)+1928\)
\(=x^7-x^7-x^6+x^6-...+x^2+x+1928\)
=x+1928=89+1928=2017
\(f\left(x\right)=x^7-90x^6+90x^5-90x^4+...+90x+1928\)\(\Rightarrow f\left(89\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5\)\(-\left(x+1\right)x^4+...+\left(x+1\right)x+1928\)
\(\Rightarrow f\left(89\right)=x^7-x^7-x^6+x^6+x^5-x^5\)\(-x^4+...+\)\(x^2+x+1928\)\(=89+1928=2017\)
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
\(3^{x-2}+3^x=90\)
\(3^{x-2}.1+3^{x-2}.3^2=90\)
\(3^{x-2}.\left(1+3^2\right)=90\)
\(3^{x-2}.10=90\)
\(3^{x-2}=90:10\)
\(3^{x-2}=9\)
\(3^{x-2}=3^2\)
\(x-2=2\)
\(x=4\)