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Đk : x >= 9
pt <=> \(\sqrt{\left(x-9\right)+6\sqrt{x-9}+9}\)+ \(\sqrt{\left(x-9\right)-6\sqrt{x-9}+9}\)- 1 = 0
<=> \(\sqrt{\left(\sqrt{x-9}+3\right)^2}\)+ \(\sqrt{\left(\sqrt{x-9}-3\right)^2}\)- 1 = 0
<=> \(\sqrt{x-9}+3\)+ |\(\sqrt{x-9}\)- 3| - 1 = 0
Đến đó bạn xét 2 trường hợp đề loại dấu "| |" để giải pt nha
Tk mk
a) pt<=> \(\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
<=>\(\left|x-2\right|+\left|x-3\right|=1\)
đến đây chia 3 trường hợp để phá trị tuyệt đối là ra
b) \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\)
câu này cũng tương tự câu a nha
\(x+y+z-6046=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
\(\left(x-2019\right)+\left(x-2020\right)+\left(x-2021\right)+1+4+9\)\(=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
đặt :\(\hept{\begin{cases}\sqrt{x-2019}=a\\\sqrt{y-2020}=b\\\sqrt{z-2021}=c\end{cases}\left(đk:a,b,c\ge0\right)}\)
PT <=> \(a^2+b^2+c^2+1+4+9=2a+4b+6c\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}\left(tm\right)}}\)
\(\Rightarrow\hept{\begin{cases}x=2020\\y=2024\\z=2030\end{cases}}\)
1.
đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)
có \(a^2+b^2=4\)
pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)
vì a,b>o nên \(a-b=\sqrt{2}\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Bình phương 2 vế:
\(4-2\sqrt{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
\(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+...+\frac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+2}-\sqrt{x+1}\right)}+\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+2}+\sqrt{x+3}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}\)
\(+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{\left(\sqrt{x+2019}+\sqrt{x+2020}\right)\left(\sqrt{x+2020}-\sqrt{x+2019}\right)}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{x+2020-x-2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+3}-\sqrt{x+2}+...+\sqrt{x+2020}-\sqrt{x+2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}=11+\sqrt{x+1}\)
\(\Leftrightarrow\)\(x+2020=121+22\sqrt{x+1}+x+1\)
\(\Leftrightarrow\)\(22\sqrt{x+1}=1898\)
\(\Leftrightarrow\)\(\sqrt{x+1}=\frac{949}{11}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=\frac{900601}{121}\\x+1=\frac{-900601}{121}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{900480}{121}\\x=\frac{-900722}{121}\end{cases}}\)
Chúc bạn học tốt ~
PS : sai thì thui nhá
\(\sqrt{2020-x}+\sqrt{2023-x}+\sqrt{2028-x}=6\)\(\left(x\le2020\right)\)
\(\Leftrightarrow\sqrt{2020-x}-1+\sqrt{2023-x}-2+\sqrt{2020-x}-3=0\)
\(\Leftrightarrow\frac{\left(\sqrt{2020-x}-1\right)\left(\sqrt{2020-x}+1\right)}{\sqrt{2020-x}+1}\) \(+\frac{\left(\sqrt{2023-x}-2\right)\left(\sqrt{2023-x}+2\right)}{\sqrt{2023-x}+2}\)\(+\frac{\left(\sqrt{2028-x}-3\right)\left(\sqrt{2028-x}+3\right)}{\left(\sqrt{2028-x}+3\right)}\)=0
\(\Leftrightarrow\frac{2019-x}{\sqrt{2020-x}+1}+\frac{2019-x}{\sqrt{2023-x}+2}+\frac{2019-x}{\left(\sqrt{2028-x}+3\right)}\)=0
\(\Leftrightarrow\left(2019-x\right)\left(\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}\right)\)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=2019\left(tm\right)\\\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}=0\left(2\right)\end{matrix}\right.\)
vì \(\sqrt{2020-x}\ge0\Rightarrow\frac{1}{\sqrt{2020-x}+1}>0\)
cmtt: \(\frac{1}{\sqrt[]{2023-x}+2}>0\)
\(\frac{1}{\sqrt{2028-x}+3}>0\)
=>\(\frac{1}{\sqrt{2020-x}+1}+\frac{1}{\sqrt{2023-x}+2}+\frac{1}{\sqrt{2028-x}+3}>0\)(3)
từ (2) và (3)=> vô lý
vậy x=2019 là nghiệm của phương trình