Căn X^2-1 – căn 3x^2+4x+1 = (8-2x).căn x+1 16/07/2021 Bởi Delilah Căn X^2-1 – căn 3x^2+4x+1 = (8-2x).căn x+1
$\sqrt[]{x^2-1}-\sqrt[]{3x^2+4x+1}=(8-2x).\sqrt[]{x+1}$ <=>$\sqrt[]{(x-1)(x+1)}$- $\sqrt[]{3.(x+1/3)(x+1)}$=(8-2x). $\sqrt[]{x+1}$ <=>$\sqrt[]{x+1}.(\sqrt[]{x-1}-\sqrt[]{3x+1})=(8-2x).\sqrt[]{x+1}$ <=>$\sqrt[]{x-1}-\sqrt[]{3x+1}=(8-2x)$ mk thu gọn xong nhé Bình luận
`\sqrt(x^2-1)-\sqrt(3x^2+4x+1)=(8-2x)\sqrt(x+1)` `⇔\sqrt((x-1)(x+1))-3\sqrt((x+1/3)(x+1))-(8-2x)\sqrt(x+1)=0` `⇔\sqrt(x+1)(\sqrt(x-1)-3\sqrt(x+1/3)-8+2x)=0` ⇔\(\left[ \begin{array}{l}\sqrt(x+1)=0\\\sqrt(x-1)-3\sqrt(x+1/3)-8+2x=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x+1=0\\(x-5)(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4)=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-1\\x-5=0(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4>0)\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\) Bình luận
$\sqrt[]{x^2-1}-\sqrt[]{3x^2+4x+1}=(8-2x).\sqrt[]{x+1}$
<=>$\sqrt[]{(x-1)(x+1)}$- $\sqrt[]{3.(x+1/3)(x+1)}$=(8-2x). $\sqrt[]{x+1}$
<=>$\sqrt[]{x+1}.(\sqrt[]{x-1}-\sqrt[]{3x+1})=(8-2x).\sqrt[]{x+1}$
<=>$\sqrt[]{x-1}-\sqrt[]{3x+1}=(8-2x)$
mk thu gọn xong nhé
`\sqrt(x^2-1)-\sqrt(3x^2+4x+1)=(8-2x)\sqrt(x+1)`
`⇔\sqrt((x-1)(x+1))-3\sqrt((x+1/3)(x+1))-(8-2x)\sqrt(x+1)=0`
`⇔\sqrt(x+1)(\sqrt(x-1)-3\sqrt(x+1/3)-8+2x)=0`
⇔\(\left[ \begin{array}{l}\sqrt(x+1)=0\\\sqrt(x-1)-3\sqrt(x+1/3)-8+2x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x+1=0\\(x-5)(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4)=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x-5=0(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4>0)\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\)