Tìm `x` biết: `| x+1| + | x+7| = 6- ( x+2) ^2` Giúp iem với 29/08/2021 Bởi Reese Tìm `x` biết: `| x+1| + | x+7| = 6- ( x+2) ^2` Giúp iem với
với x≥-1 ta có: x+1+x+7=6-(x+2)² ⇔2x+8=6-x²-4x-4 ⇔x²+6x+6=0 với x≤-7 ta có -x-1-x-7=6-(x+2)² ⇔-2x-8=6-x²-4x-4 ⇒x²+2x-10=0 ⇒x=-1±√11 với -7<x<-1 ta có -x-1+x+7=6-(x+2)² ⇔6=6-(x+2)² ⇒x=-2 Chúc học tốt ah. Cố lên Bình luận
\(BXD:\begin{array}{|c|cc|}\hline x&-\infty&&-7&&-1&&+\infty\\\hline x+1&&-&|&-&0&+&\quad\\\hline x+7&&-&0&+&|&+\\\hline\end{array}\\Xét\,\, x<-7\\→-(x+1)-(x+7)=6-(x+2)^2\\↔-x-1-x-7=6-x^2-4x-4\\↔-2x-8=2-x^2-4x\\↔x^2+2x-10=0\\↔(x+1)^2-11=0\\↔(x+1-\sqrt {11})(x+1+\sqrt{11})=0\\↔x+1-\sqrt{11}=0\quad or\quad x+1+\sqrt{11}=0\\↔x=-1+\sqrt{11}(KTM)\quad or\quad x=-1-\sqrt{11}(KTM)\\Xét\,\, -7\le x\le -1\\→-(x+1)+(x+7)=6-(x+2)^2\\↔-x-1+x+7=6-x^2-4x-4\\↔6=6-x^2-4x-4\\↔x^2+4x+4=0\\↔(x+2)^2=0\\↔x+2=0\\↔x=-2(TM)\\Xét\,\, x>-1\\→ (x+1)+(x+7)=6-(x+2)^2\\↔2x+8=6-x^2-4x-4\\↔x^2+6x+6=0\\↔(x+3)^2-3=0\\↔(x+3-\sqrt 3)(x+3+\sqrt 3)=0\\↔x+3-\sqrt 3=0\quad or\quad x+3+\sqrt 3=0\\↔x=-3+\sqrt 3(KTM)\quad or\quad x=-3-\sqrt 3(KTM)\\Suy\,\,ra\,\,:\,\,x=2\\Vậy\,\,S=\{-2\}\) Bình luận
với x≥-1 ta có:
x+1+x+7=6-(x+2)²
⇔2x+8=6-x²-4x-4
⇔x²+6x+6=0
với x≤-7 ta có
-x-1-x-7=6-(x+2)²
⇔-2x-8=6-x²-4x-4
⇒x²+2x-10=0
⇒x=-1±√11
với -7<x<-1 ta có
-x-1+x+7=6-(x+2)²
⇔6=6-(x+2)²
⇒x=-2
Chúc học tốt ah. Cố lên
\(BXD:\begin{array}{|c|cc|}\hline x&-\infty&&-7&&-1&&+\infty\\\hline x+1&&-&|&-&0&+&\quad\\\hline x+7&&-&0&+&|&+\\\hline\end{array}\\Xét\,\, x<-7\\→-(x+1)-(x+7)=6-(x+2)^2\\↔-x-1-x-7=6-x^2-4x-4\\↔-2x-8=2-x^2-4x\\↔x^2+2x-10=0\\↔(x+1)^2-11=0\\↔(x+1-\sqrt {11})(x+1+\sqrt{11})=0\\↔x+1-\sqrt{11}=0\quad or\quad x+1+\sqrt{11}=0\\↔x=-1+\sqrt{11}(KTM)\quad or\quad x=-1-\sqrt{11}(KTM)\\Xét\,\, -7\le x\le -1\\→-(x+1)+(x+7)=6-(x+2)^2\\↔-x-1+x+7=6-x^2-4x-4\\↔6=6-x^2-4x-4\\↔x^2+4x+4=0\\↔(x+2)^2=0\\↔x+2=0\\↔x=-2(TM)\\Xét\,\, x>-1\\→ (x+1)+(x+7)=6-(x+2)^2\\↔2x+8=6-x^2-4x-4\\↔x^2+6x+6=0\\↔(x+3)^2-3=0\\↔(x+3-\sqrt 3)(x+3+\sqrt 3)=0\\↔x+3-\sqrt 3=0\quad or\quad x+3+\sqrt 3=0\\↔x=-3+\sqrt 3(KTM)\quad or\quad x=-3-\sqrt 3(KTM)\\Suy\,\,ra\,\,:\,\,x=2\\Vậy\,\,S=\{-2\}\)