\(\sqrt{2028-x}+\sqrt{2093-x}+\sqrt{2268-x}=29\)
giúp t với
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\(4\sqrt{x+3}-\sqrt{x-1}=7\left(1\right)\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow4\sqrt{x+3}=7+\sqrt{x-1}\)
\(\Leftrightarrow16\left(x+3\right)=49+x-1+14\sqrt{x+1}\)
\(\Leftrightarrow15x=14\sqrt{x-1}\)\(\left(x\ge1\right)\)
\(\Leftrightarrow225x^2=196\left(x-1\right)\)
\(\Leftrightarrow225x^2-196x+196=0\)
\(\Delta=196^2-4.225.196< 0\)
\(\Rightarrow pt\)vô nghiệm
Vậy pt vô nghiệm.
\(9x^2-\left(3x+2\right)\sqrt{3x-1}+2=3x\left(1\right)\)\(\left(x\ge\frac{1}{3}\right)\)
Đặt \(\sqrt{3x-1}=a\ge0\)
\(\Rightarrow\hept{\begin{cases}3x=a^2+1\\3x+2=a^2+3\\3x-1=a^2\end{cases}}\)
Pt (1) \(\Leftrightarrow3x\left(3x-1\right)-\left(3x+2\right)\sqrt{3x-1}+2=0\)
\(\Leftrightarrow\left(a^2+1\right)a^2-\left(a^2+3\right)a+2=0\)
\(\Leftrightarrow a^3+a^2-a^3-3a+2=0\)
\(\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=2\end{cases}}\)
TH1: a=1
\(\Rightarrow\sqrt{3x-1}=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow x=\frac{2}{3}\left(tm\right)\)
TH2: a=2
\(\Rightarrow\sqrt{3x-1}=2\)
\(\Leftrightarrow3x-1=4\)
\(\Leftrightarrow x=\frac{5}{3}\left(tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{\frac{2}{3};\frac{5}{3}\right\}\)
ĐK: \(x\ge1\).
Đặt \(\sqrt{x-1}=a,2x-5=b\).
Phương trình ban đầu tương đương với:
\(\sqrt{b^2+5a^2}=2b+7a\)
\(\Rightarrow b^2+5a^2=4b^2+49a^2+28ab\)
\(\Leftrightarrow\left(2a+b\right)\left(22a+3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=-b\\22a=-3b\end{cases}}\)
Với \(2a=-b\Rightarrow2\sqrt{x-1}=5-2x\)
\(\Rightarrow4\left(x-1\right)=25-20x+4x^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=3-\frac{\sqrt{7}}{2}\\x=3+\frac{\sqrt{7}}{2}\end{cases}}\)
Thử lại chỉ có \(x=3-\frac{\sqrt{7}}{2}\)thỏa mãn.
Với \(22a=-3b\Rightarrow22\sqrt{x-1}=-3\left(2x-5\right)\)
\(\Rightarrow484\left(x-1\right)=9\left(2x-5\right)^2\)
\(\Leftrightarrow x=\frac{83}{9}\pm\frac{55\sqrt{7}}{18}\)
Thử lại đều không thỏa mãn.
ĐK: x \(\ge\)-4
Ta có: \(x^2+\sqrt{x+4}+\sqrt{x+1}=x+27\)
<=> \(x^2-x-27+\sqrt{x+4}+\sqrt{x+11}=0\)
<=> \(x^2-x-20+\sqrt{x+4}-3+\sqrt{x+11}-4=0\)
<=> \(\left(x-5\right)\left(x+4\right)+\frac{x+4-9}{\sqrt{x+4}+3}+\frac{x+11-16}{\sqrt{x+4}+4}=0\)
<=> \(\left(x-5\right)\left(x+4\right)+\frac{x-5}{\sqrt{x+4}+3}+\frac{x-5}{\sqrt{x+4}+4}=0\)
<=> \(\left(x-5\right)\left(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}\right)=0\)
<=> \(x=5\)
(vì x \(\ge\)-4 => \(x+4\ge0\); \(\frac{1}{\sqrt{x+4}+3}>0\); \(\frac{1}{\sqrt{x+11}+4}>0\)
=> \(x+4+\frac{1}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x+11}+4}>0\))
Đk: \(-2\le x\le3\)
Ta có: \(\sqrt{x+2}-\sqrt{3-x}=x^2-6x+9\)
<=> \(\sqrt{x+2}-2-\sqrt{3-x}+1-x^2+6x-8=0\)
<=> \(\frac{x+2-4}{\sqrt{x+2}+2}-\frac{3-x-1}{\sqrt{3-x}+1}-\left(x-4\right)\left(x-2\right)=0\)
<=> \(\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{3-x}+1}-\left(x-4\right)\left(x-2\right)=0\)
<=> \(\left(x-2\right)\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4=0\left(1\right)\end{cases}}\)
Vì \(-2\le x\le3\)
Do đó: \(\frac{1}{\sqrt{x+2}+2}>0\); \(\frac{1}{\sqrt{3-x}+1}>0\); 4 - x > 0
=> \(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4>0\)
=> pt (1) vô nghiệm
Vậy S = {2}
Đk: x \(\le\)2028
Ta có: \(\sqrt{2028-x}+\sqrt{2093-x}+\sqrt{2268-x}=29\)
<=> \(\sqrt{2028-x}-4+\sqrt{2093-x}-9+\sqrt{2268-x}-16=0\)
<=> \(\frac{2028-x-16}{\sqrt{2028-x}+4}+\frac{2093-x-81}{\sqrt{2093-x}+9}+\frac{2268-x-256}{\sqrt{2268-x}+16}=0\)
<=> \(\left(2012-x\right).\left(\frac{1}{\sqrt{2028-x}+4}+\frac{1}{\sqrt{2093-x}+9}+\frac{1}{\sqrt{2268-x}+16}\right)=0\)
<=> x = 2012 (tm)