Giải pt: (đặt ẩn phụ) `(x+1)^2+((x+1)^2)/((x+2)^2)=8` nhanh ik mn ưi huhu=(((((((( 02/11/2021 Bởi Harper Giải pt: (đặt ẩn phụ) `(x+1)^2+((x+1)^2)/((x+2)^2)=8` nhanh ik mn ưi huhu=((((((((
`(x+1)^2+(x+1)^2/(x+2)^2=8` Đặt `x+1=t` Ta có pt ẩn phụ t là: `t^2+(t^2)/(t+1)^2=8` `<=> (t^2(t+1)^2+t^2)/(t+1)^2=8` `<=> t^2(t^2+2t+1)+t^2=8(t+1)^2` `<=> t^4+2t^3+t^2+t^2=8(t^2+2t+1)` `<=> t^4+2t^3+2t^2=8t^2+16t+8` `<=> t^4+2t^3-6t^2-16t-8=0` `<=> (t^4-2t^3-2t^2)+(4t^3-8t^2-8t)+(4t^2-8t-8)=0` `<=> t^2(t^2-2t-2)+4t(t^2-2t-2)+4(t^2-2t-2)=0` `<=> (t^2-2t-2)(t^2+4t+4)=0` `<=> (t^2-2t-2)(t+2)^2=0` `<=>`\(\left[ \begin{array}{l}t^2-2t-2=0\\t+2=0\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}(t-1)^2=3\\t=-2\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}t=1+\sqrt[]{3}\\t=1-\sqrt[]{3}\\t=-2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x+1=1+\sqrt[]{3}\\x+1=1-\sqrt[]{3}\\x+1=-2\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\sqrt[]{3}\\x=-\sqrt[]{3}\\x=-3\end{array} \right.\) Vậy `S={+-\sqrt{3};-3}` Bình luận
$\begin{array}{l}\text{Điều kiện:} \ x\ne-2\\\text{Đặt:} \ x+1=t \ (t\ne-1)\\\text{Phương trình đã cho trở thành:}\\t^2+\dfrac{t^2}{(t+1)^2}=8\\\to \dfrac{t^2(t+1)^2+t^2}{(t+1)^2}=8\\\to \dfrac{t^2(t^2+2t+1)+t^2}{t^2+2t+1}=8\\\to t^4+2t^3+t^2+t^2=8(t^2+2t+1)\\\to t^4+2t^3+2t^2=8t^2+16t+8\\\to t^4+2t^3-6t^2-16t-8=0\\\to t^4+2t^3-6t^2-12t-4t-8=0\\\to t^3(t+2)-6t(t+2)-4(t+2)=0\\\to (t+2)(t^3-6t-4)=0\\\to \left[\begin{array}{l}t+2=0\\t^3-6t-4=0\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\t^3+2t^2-2t^2-4t-2t-4=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}t=-2\\t^2(t+2)-2t(t+2)-2(t+2)=0\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\(t+2)(t^2-2t-2)=0\end{array}\right. \\\leftrightarrow\left[\begin{array}{l}t=-2\\t^2-2t-2=0\end{array}\right. \leftrightarrow\left[\begin{array}{l}t=-2\\t^2-2t+1=3\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}t=-2\\(t-1)^2=(\pm\sqrt3)^2\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\t-1=-\sqrt3\\t-1=\sqrt3\end{array}\right.\\\leftrightarrow\left[\begin{array}{l}t=-2\\t=1-\sqrt3\\t=1+\sqrt3\end{array}\right.\to \left[\begin{array}{l}x+1=-2\\x+1=1-\sqrt3\\x+1=1+\sqrt3\end{array}\right. \leftrightarrow \left[\begin{array}{l}x=-3\\x=-\sqrt3\\x=\sqrt3\end{array}\right.\\\text{Vậy phương trình có tập nghiệm} \ S=\{\pm \sqrt3;-3\}\end{array}$ Bình luận
`(x+1)^2+(x+1)^2/(x+2)^2=8`
Đặt `x+1=t`
Ta có pt ẩn phụ t là:
`t^2+(t^2)/(t+1)^2=8`
`<=> (t^2(t+1)^2+t^2)/(t+1)^2=8`
`<=> t^2(t^2+2t+1)+t^2=8(t+1)^2`
`<=> t^4+2t^3+t^2+t^2=8(t^2+2t+1)`
`<=> t^4+2t^3+2t^2=8t^2+16t+8`
`<=> t^4+2t^3-6t^2-16t-8=0`
`<=> (t^4-2t^3-2t^2)+(4t^3-8t^2-8t)+(4t^2-8t-8)=0`
`<=> t^2(t^2-2t-2)+4t(t^2-2t-2)+4(t^2-2t-2)=0`
`<=> (t^2-2t-2)(t^2+4t+4)=0`
`<=> (t^2-2t-2)(t+2)^2=0`
`<=>`\(\left[ \begin{array}{l}t^2-2t-2=0\\t+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(t-1)^2=3\\t=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}t=1+\sqrt[]{3}\\t=1-\sqrt[]{3}\\t=-2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x+1=1+\sqrt[]{3}\\x+1=1-\sqrt[]{3}\\x+1=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\sqrt[]{3}\\x=-\sqrt[]{3}\\x=-3\end{array} \right.\)
Vậy `S={+-\sqrt{3};-3}`
$\begin{array}{l}\text{Điều kiện:} \ x\ne-2\\\text{Đặt:} \ x+1=t \ (t\ne-1)\\\text{Phương trình đã cho trở thành:}\\t^2+\dfrac{t^2}{(t+1)^2}=8\\\to \dfrac{t^2(t+1)^2+t^2}{(t+1)^2}=8\\\to \dfrac{t^2(t^2+2t+1)+t^2}{t^2+2t+1}=8\\\to t^4+2t^3+t^2+t^2=8(t^2+2t+1)\\\to t^4+2t^3+2t^2=8t^2+16t+8\\\to t^4+2t^3-6t^2-16t-8=0\\\to t^4+2t^3-6t^2-12t-4t-8=0\\\to t^3(t+2)-6t(t+2)-4(t+2)=0\\\to (t+2)(t^3-6t-4)=0\\\to \left[\begin{array}{l}t+2=0\\t^3-6t-4=0\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\t^3+2t^2-2t^2-4t-2t-4=0\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}t=-2\\t^2(t+2)-2t(t+2)-2(t+2)=0\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\(t+2)(t^2-2t-2)=0\end{array}\right. \\\leftrightarrow\left[\begin{array}{l}t=-2\\t^2-2t-2=0\end{array}\right. \leftrightarrow\left[\begin{array}{l}t=-2\\t^2-2t+1=3\end{array}\right.\\\leftrightarrow \left[\begin{array}{l}t=-2\\(t-1)^2=(\pm\sqrt3)^2\end{array}\right. \leftrightarrow \left[\begin{array}{l}t=-2\\t-1=-\sqrt3\\t-1=\sqrt3\end{array}\right.\\\leftrightarrow\left[\begin{array}{l}t=-2\\t=1-\sqrt3\\t=1+\sqrt3\end{array}\right.\to \left[\begin{array}{l}x+1=-2\\x+1=1-\sqrt3\\x+1=1+\sqrt3\end{array}\right. \leftrightarrow \left[\begin{array}{l}x=-3\\x=-\sqrt3\\x=\sqrt3\end{array}\right.\\\text{Vậy phương trình có tập nghiệm} \ S=\{\pm \sqrt3;-3\}\end{array}$