\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)Giải phương trình
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16 tháng 12 2020
ĐKXĐ: \(x\ge1\)
Ta có:
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+\left|\sqrt{x-1}-1\right|=\dfrac{x+1}{2}\left(1\right)\)
Ta xét 2 trường hợp sau:
TH1: \(x\ge2\)
Khi đó:
\(\left(1\right)\Leftrightarrow2\sqrt{x-1}-1=\dfrac{x+1}{2}\\ \Leftrightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\\ \Leftrightarrow16\left(x-1\right)=x^2+6x+9\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TMĐK\right)\)
TH2: \(1\le x< 2\)
Khi đó:
\(\left(1\right)\Leftrightarrow1=\dfrac{x+1}{2}\Leftrightarrow x=1\left(TMĐK\right)\)
Vậy x=1 hoặc x=5
3 tháng 11 2018
\(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+...+\frac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+2}-\sqrt{x+1}\right)}+\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+2}+\sqrt{x+3}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}\)
\(+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{\left(\sqrt{x+2019}+\sqrt{x+2020}\right)\left(\sqrt{x+2020}-\sqrt{x+2019}\right)}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{x+2020-x-2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+3}-\sqrt{x+2}+...+\sqrt{x+2020}-\sqrt{x+2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}=11+\sqrt{x+1}\)
\(\Leftrightarrow\)\(x+2020=121+22\sqrt{x+1}+x+1\)
\(\Leftrightarrow\)\(22\sqrt{x+1}=1898\)
\(\Leftrightarrow\)\(\sqrt{x+1}=\frac{949}{11}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=\frac{900601}{121}\\x+1=\frac{-900601}{121}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{900480}{121}\\x=\frac{-900722}{121}\end{cases}}\)
Chúc bạn học tốt ~
PS : sai thì thui nhá
10 tháng 6 2016
ĐKXĐ:x khác 0
Trục căn thức ở mẫu ta được:
\(\left(\sqrt{x+3}-\sqrt{x+2}\right)+\left(\sqrt{x+2}-\sqrt{x+1}\right)+\left(\sqrt{x+1}-\sqrt{x}\right)=1.\)
<=> \(\sqrt{x+3}=\sqrt{x}+1\)
<=> \(x+3=x+2\sqrt{x}+1\)
=> 2\(\sqrt{x}=2\)
=> x=1
10 tháng 6 2016
\(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}=1\left(DKXD:x\ge0\right)\)
\(\Rightarrow\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(x+3\right)-\left(x+2\right)}+\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(x+2\right)-\left(x+1\right)}+\frac{\sqrt{x+1}-\sqrt{x}}{\left(x+1\right)-x}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+1}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\Leftrightarrow x+3=\left(1+\sqrt{x}\right)^2\Leftrightarrow x+3=x+1+2\sqrt{x}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(TMDK\right)\)
Vậy tập nghiệm của phương trình : \(S=\left\{1\right\}\)
Ta có: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\) ( ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)
\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)
+ Với \(x=1\)\(\Rightarrow\)\(\frac{x+3}{2}=0\)\(\Leftrightarrow\)\(x=-3\)( loại )
+ Với \(x>1\)\(\Rightarrow\)\(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)
\(\Leftrightarrow\)\(2\sqrt{x-1}=\frac{x+3}{2}\)
\(\Leftrightarrow\)\(4\sqrt{x-1}=x+3\)
\(\Leftrightarrow\)\(16\left(x-1\right)=\left(x+3\right)^2\)
\(\Leftrightarrow\)\(16x-16=x^2+6x+9\)
\(\Leftrightarrow\)\(x^2-10x+25=0\)
\(\Leftrightarrow\)\(\left(x-5\right)^2=0\)
\(\Leftrightarrow\)\(x=5\)\(\left(TM\right)\)
Vậy \(S=\left\{5\right\}\)
ĐKXĐ: x≥1x≥1
Ta có:
√x+2√x−1+√x−2√x−1=x+32⇔√(√x−1+1)2+√(√x−1−1)2=x+32⇔√x−1+1+∣∣√x−1−1∣∣=x+32⇔√x−1+∣∣√x−1−1∣∣=x+12(1)x+2x−1+x−2x−1=x+32⇔(x−1+1)2+(x−1−1)2=x+32⇔x−1+1+|x−1−1|=x+32⇔x−1+|x−1−1|=x+12(1)
Ta xét 2 trường hợp sau:
TH1: x≥2x≥2
Khi đó:
(1)⇔2√x−1−1=x+12⇔2√x−1=x+32⇔16(x−1)=x2+6x+9⇔x2−10x+25=0⇔(x−5)2=0⇔x=5(TMĐK)(1)⇔2x−1−1=x+12⇔2x−1=x+32⇔16(x−1)=x2+6x+9⇔x2−10x+25=0⇔(x−5)2=0⇔x=5(TMĐK)
TH2: 1≤x<21≤x<2
Khi đó:
(1)⇔1=x+12⇔x=1(TMĐK)(1)⇔1=x+12⇔x=1(TMĐK)
Vậy x=1 hoặc x=5