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\(\left(8x+5\right)\left(8x+7\right)\left(8x+6\right)^2=72\)
Đặt \(8x+5=t\left(t\ge0\right)\)
\(t\left(t+2\right)\left(t+1\right)^2-72=0\)
\(\Leftrightarrow t\left(t+1\right)\left(t+2\right)\left(t+1\right)-72=0\)
\(\Leftrightarrow\left(t^2+t\right)\left(t^2+3t+2\right)-72=0\)
\(\Leftrightarrow t^4+3t^3+2t^2+t^3+3t^2+2t-72=0\)
\(\Leftrightarrow t^4+4t^3+5t^2+2t-72=0\)
\(\Leftrightarrow\left(t^2+2t+9\ne0\right)\left(t+4\right)\left(t-2\right)=0\Leftrightarrow t=-4;2\)
hay \(8x+5=-4\Leftrightarrow x=-\frac{9}{8}\)( trường hợp 1 )
\(8x+5=2\Leftrightarrow x=-\frac{3}{8}\)( trưởng hợp 2 )
Vậy tập nghiệm của phương trình là S = { -9/8 ; -3/8 }
\(\left(8x+5\right)\cdot\left(8x+7\right)\cdot\left(8x+6\right)^2=72\)
Đặt \(t=8x+6\)
\(Pt\Leftrightarrow\left(t-1\right)\left(t+1\right)t^2-72=0\)
\(\Leftrightarrow\left(t^2-1\right)t^2-72=0\Leftrightarrow t^4-t^2-72=0\)
\(\Leftrightarrow\left(t^2-9\right)\left(t^2+8\right)=0\Leftrightarrow\orbr{\begin{cases}t^2=9\\t^2=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}t=3\\t=-3\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}8x+6=3\\8x+6=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy....
Đặt \(A=x^{13}-\left(8x^{12}-8x^{11}+8x^{10}-8x^9+.....+8x^2-8x^1\right)+8\)
Đặt \(B=8x^{12}-8x^{11}+8x^{10}-....+8x^2-8x^1\)
\(B=8.\left(x^{12}-x^{11}+x^{10}-x^9+....+x^2-x^1\right)\)
Đặt \(C=x^{12}-x^{11}+x^{10}-x^9+...+x^2-x\)
Suy ra \(C.x=x^{13}-x^{12}+x^{11}-x^{10}+.....+x^3-x^2\)
Nên \(C.x-C=x^{13}-x\)hay \(C.\left(x-1\right)=x^{13}-x\)
Khi đó \(C=\frac{x^{13}-x}{x-1}\)nên\(B=8.\frac{x^{13}-x}{x-1}\)
Từ đó tính tương tự nha , cách làm thì có thể sai những em vẫn cố gắng giúp , ai có cách hay hơn thì giải nhé
\(A=x^5-5x^4+5x^3-5x^2+5x-6\)
\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x-2\)
\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-2\)
\(=-2\)
\(a,x+\frac{4}{5}-x+4=\frac{x}{3}-x-1\)
\(x+\frac{24}{5}-x=\frac{x}{3}-x-1\)
\(x+\frac{24}{5}-x-\frac{x}{3}+x+1=0\)
\(x+\frac{29}{5}-\frac{x}{3}=0\)
\(x-\frac{1}{3}x=-\frac{29}{5}\)
\(\frac{2}{3}x=-\frac{29}{5}\)
\(x=-\frac{87}{10}\)
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)