Giải pt:
\(x^2+\sqrt{x+2011}=2011\)
\(x+\sqrt{x+2}=x^2-\sqrt{4-x}+\frac{9}{4}\)
Những câu hỏi liên quan
Giải pt :\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
giải pt:\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=\frac{1}{2}\)
Thay vào tìm x;y;z
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Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)
Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1−a21+b1−b21+c1−c21−43=0
\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21−a1+b21−b1+c21−c1+43=0
\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21−a1+41)+(b21−b1+41)+(c21−c1+41)=0
\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0⇔(a1−21)2+(b1−21)2+(c1−21)2=0
\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21
Thay vào tìm x;y;z
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Tìm giá trị nhỏ nhất của biểu thức \(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
Giai phương trình a, \(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
b,\(\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}+\frac{\sqrt{z-2012}-1}{2012}=\frac{3}{4}\)
\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\)
\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)
\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)
Làm nốt
giải pt
\(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
Đặt \(a=\sqrt{x-2009};b=\sqrt{y-2010};c=\sqrt{z-2011};a>0;b>0;c>0\)
\(Pt\Leftrightarrow\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)
\(\Leftrightarrow\frac{\left(4a^2-a+1\right)}{a^2}+\frac{\left(4b^2-b+1\right)}{b^2}+\frac{\left(4c^2-c+1\right)}{c^2}=0\)
\(\Leftrightarrow\left(\frac{2a-1}{a}\right)^2+\left(\frac{2b-1}{b}\right)^2+\left(\frac{2c-1}{c}\right)^2=0\)
\(\Rightarrow a=b=c=\frac{1}{2}\Rightarrow\sqrt{x-2009}=\frac{1}{2}\Rightarrow x=2009\frac{1}{4}\)
\(\Rightarrow b=\frac{1}{2}\Rightarrow\sqrt{y-2010}=\frac{1}{2}\Rightarrow y=2010\frac{1}{4}\)
\(\Rightarrow c=\frac{1}{2}\Rightarrow\sqrt{z-2011}=\frac{1}{2}\Rightarrow z=2011\frac{1}{4}\)
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a) Giải Phương trình: \(\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4}\)
b) Giải Phương Trình: \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
Giúp mình nha.......
a) ĐK: \(x>2009;y>2010;z>2011\)
\(\Leftrightarrow\frac{\sqrt{x-2009}-1}{x-2009}-\frac{1}{4}+\frac{\sqrt{y-2010}-1}{y-2010}-\frac{1}{4}+\frac{\sqrt{z-2011}-1}{z-2011}-\frac{1}{4}=0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{-\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{-\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\left(1\right)\)
Dễ thấy với đkxđ thì \(VT\left(1\right)\le0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}\left(tm\right)}}\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)(*)
\(ĐK:\orbr{\begin{cases}x\ge3\\x\le-3\end{cases}}\)
(*)\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)
Xét phương trình\(\sqrt{x+3}+\sqrt{x-3}=0\)(**) có \(\sqrt{x+3}\ge0;\sqrt{x-3}\ge0\)nên (**) xảy ra khi \(\hept{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\x=3\end{cases}}\left(L\right)\)
Vậy phương trình có một nghiệm duy nhất là 3
a. ĐK : x > 2009 ; y > 2010 ; z > 2011
Pt <=> \(\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)\)
\(\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2=0\\\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2=0\\\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)( tmđk )
b. ĐK : x2 - 9 \(\ge\)0 <=> x2\(\ge\)9 <=> - 3\(\le\)x\(\le\)3
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(tmdk\right)\\\sqrt{x+3}+\sqrt{x-3}=0\end{cases}}\)
TH :\(\sqrt{x+3}+\sqrt{x-3}=0\)
Vì \(\sqrt{x+3}+\sqrt{x-3}\ge0\forall x\). Dấu "=" xảy ra <=> \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x-3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)( mâu thuẫn )
Vậy pt có nghiệm duy nhất là x = 3
Tính giá trị biểu thức :
\(A=\frac{\left(x^2+x-3\right)^{2011}}{\left(x^5+x^4-x^3-2\right)^{2011}}+\left(x^5+x^4-x^3+1\right)^{2011}\)
Với \(x=\frac{\sqrt{5}-1}{2}\)
Giải pt sau :
1, sqrt{x+1}+sqrt{4-x}+sqrt{left(x+1right)left(4-xright)}5
2, sqrt{x+4}+sqrt{x-4}2x-12+2sqrt{x^2-16}
3, sqrt{x+sqrt{6x-9}}+sqrt{x-sqrt{6x-9}}sqrt{6}
4, frac{4}{x+sqrt{x^2+x}}-frac{1}{x-sqrt{x^2+x}}frac{3}{x}
5, sqrt{x^2+x+4}+sqrt{x^2+x+1}sqrt{2x^2+2x+9}
Đọc tiếp
Giải pt sau :
1, \(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\)
2, \(\sqrt{x+4}+\sqrt{x-4}=2x-12+2\sqrt{x^2-16}\)
3, \(\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)
4, \(\frac{4}{x+\sqrt{x^2+x}}-\frac{1}{x-\sqrt{x^2+x}}=\frac{3}{x}\)
5, \(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
1.
ĐK: \(-1\le x\le4\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)
\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)
2.
ĐK:\(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)
\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)
\(PT\Leftrightarrow t=2x-12+t^2-2x\)
\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.
Cho \(x=\frac{2}{\frac{1}{\sqrt{\sqrt{2}+1}-1}-\frac{1}{\sqrt{\sqrt{2}+1}+1}}\)
Tính giá trị biểu thức \(B=\left(x^4-x^3-x^2+2x-1\right)^{2011}\)
Tìm x, y, z biết \(\frac{\sqrt{x-2009}-1}{x-2009x}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{x-2011}=\frac{3}{4}\)
\(pt\Leftrightarrow\frac{1-\sqrt{x-2009}}{x-2009}+\frac{1-\sqrt{y-2010}}{y-2010}+\frac{1-\sqrt{z-2011}}{z-2011}=-\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{\sqrt{x-2009}}{x-2009}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{\sqrt{y-2010}}{y-2010}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{\sqrt{z-2011}}{z-2011}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x-2009}-\frac{1}{\sqrt{x-2009}}+\frac{1}{4}\right)+\left(\frac{1}{y-2010}-\frac{1}{\sqrt{y-2010}}+\frac{1}{4}\right)+\left(\frac{1}{z-2011}-\frac{1}{\sqrt{z-2011}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2009}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{y-2010}}-\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{z-2011}}-\frac{1}{2}\right)^2=0\)
Xảy ra khi \(\hept{\begin{cases}\frac{1}{\sqrt{x-2009}}=\frac{1}{2}\\\frac{1}{\sqrt{y-2010}}=\frac{1}{2}\\\frac{1}{\sqrt{z-2011}}=\frac{1}{2}\end{cases}}\Rightarrow\hept{\begin{cases}\sqrt{x-2009}=2\\\sqrt{y-2010}=2\\\sqrt{z-2011}=2\end{cases}}\Rightarrow\hept{\begin{cases}x=2013\\y=2014\\z=2015\end{cases}}\)
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