Giải pt
\(\sqrt[3]{x+11}-\sqrt[3]{x+2}=3\)
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19 tháng 2 2022
đk -3 =< x =< 10
\(\sqrt{x+3}-2+\sqrt{10-x}-3=x^2-7x+6\)
\(\Leftrightarrow\dfrac{x+3-4}{\sqrt{x+3}+2}+\dfrac{10-x-9}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{x+3}+2}+\dfrac{1-x}{\sqrt{10-x}+3}=\left(x-6\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x+3}+2}-\dfrac{1}{\sqrt{10-x}+3}-x+6\ne0\right)=0\Leftrightarrow x=1\)(tm)
3 tháng 11 2018
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
TA
11 tháng 12 2018
\(\left(x-1\right)+4.\left(\sqrt{x+3}-2\right)+2.\left(\sqrt{3-2x}-1\right)=0\)
\(x-1+\dfrac{4.\left(x+3-4\right)}{\sqrt{x+3}+2}+\dfrac{2.\left(3-2x-1\right)}{\sqrt{3-2x}+1}=0\)
=> x-1+\(\dfrac{4.\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{4.\left(1-x\right)}{\sqrt{3-2x}+1}=0\)
=> (x-1).\(\left(\dfrac{4}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}\right)=0\)
=> x=1 (do \(\dfrac{4}{\sqrt{x+3}+2}+\dfrac{4}{\sqrt{3-2x}+1}>0\)
8 tháng 8 2016
ta đặt: \(\sqrt[3]{x+5}=u\)
\(\sqrt[3]{x+6}=v\)
ta có \(u^3+v^3=2x+11\)
=> \(u+v=\sqrt[3]{u^3+v^3}\)
=>\(\left(u+v\right)^3=u^3+v^3+3uv\left(u+v\right)=u^3+v^3\)
=> \(3uv\left(u+v\right)=3uv\sqrt[3]{u^3+v^3}=0\)
<=> \(3\sqrt[3]{x+5}\sqrt[3]{x+6}\sqrt[3]{2x+11}=0\)
<=> x=-5 hoặc x=-6 hoặc x=-11/2
vậy pt có 3 nghiệm ....
EN
1
13 tháng 8 2016
Lập phương hai vế : \(\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)^3=\left(\sqrt[3]{2x+11}\right)^3\)
\(\Leftrightarrow2x+11+3.\sqrt[3]{x+5}.\sqrt[3]{x+6}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{x+5}.\sqrt[3]{x+6}\left(\sqrt[3]{x+6}+\sqrt[3]{x+5}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}+\sqrt[3]{x+6}=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-6\\x=-\frac{11}{2}\end{array}\right.\)
V
1
21 tháng 6 2021
`ĐK:x>=2`
`pt<=>sqrt{(x-1)(x-2)}+sqrt{x+3}=sqrt{x-2}+sqrt{(x-1)(x+3)}`
`<=>sqrt{x-1}(sqrt{x-2}-sqrt{x+3})-(sqrt{x-2}-sqrt{x+3})=0`
`<=>(sqrt{x-2}-sqrt{x+3})(sqrt{x-1}-1)=0`
`+)sqrt{x-2}=sqrt{x+3}`
`<=>x-2=x+3`
`<=>0=5` vô lý
`+)sqrt{x-1}-1=0`
`<=>x-1=1`
`<=>x=2(tm)`.
Vậy `x=2`.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+11}=a\\\sqrt[3]{x+2}=b\end{matrix}\right.\) . Ta có hệ phương trình :
\(\left\{{}\begin{matrix}a-b=3\\a^3-b^3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\\left(b+3\right)^3-b^3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\9b^2+27b+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b+3\\\left(b+1\right)\left(b+2\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\\\left\{{}\begin{matrix}a=1\\b=-2\end{matrix}\right.\end{matrix}\right.\)
Với \(a=2;b=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{x+11}=2\\\sqrt[3]{x+2}=-1\end{matrix}\right.\Rightarrow x=-3\)
Với \(a=1;b=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{x+11}=1\\\sqrt[3]{x+2}=-2\end{matrix}\right.\Rightarrow x=-10\)
Vậy \(S=\left\{-10;-3\right\}\)
Cảm ơn bạn nhiều nha <3