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\(\sqrt{x-4\sqrt{x-1}+3}+\sqrt{x-6\sqrt{x-1}+8}=1\\ < =>\sqrt{x-1-2\sqrt{x-1}.2+4}+\sqrt{x-1-2\sqrt{x-1}.3+9}=1\\ < =>\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)ĐK: x>=1
\(< =>|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ < =>\left(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\right)^2=1\\ < =>\sqrt{x-1}-2+2\left|\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-3\right)\right|+\sqrt{x-1}-3=1\\ < =>2\sqrt{x-1}-5+2\left|x+5-5\sqrt{x-1}\right|=1\\ < =>2\left|x+5-5\sqrt{x-1}\right|=6-2\sqrt{x-1}\\ < =>\left|x+5-5\sqrt{x-1}\right|=3-\sqrt{x-1}\)
\(< =>\left[{}\begin{matrix}x+5-5\sqrt{x-1}=3-\sqrt{x-1}\left(1\right)\\x+5-5\sqrt{x-1}=\sqrt{x-1}-3\left(2\right)\end{matrix}\right.\)
Giải (1): \(x+5-5\sqrt{x-1}=3-\sqrt{x-1}\\ < =>x+2-4\sqrt{x-1}=0\\ < =>x-1-2\sqrt{x-1}.2+4=1\\ < =>\left(\sqrt{x-1}-2\right)^2=1\\ < =>\left[{}\begin{matrix}\sqrt{x-1}-2=1\\\sqrt{x-1}-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=0\left(loại\right)\end{matrix}\right.\)
Giải (2) cũng ra x=8
Hệ \(\Leftrightarrow x+1+3x-1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{3x-1}\right)=x-1\)
\(\Leftrightarrow3x+1+3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow3x+1=-3\sqrt[3]{\left(x+1\right)\left(3x-1\right)\left(x-1\right)}\)
\(\Leftrightarrow27x^3+9x+27x^2+1=-27\left(x^2-1\right)\left(3x-1\right)\)
\(\Leftrightarrow27x^3+9x+27x^2+1+81x^3-81x-27x^2+27=0\)
\(\Leftrightarrow108x^3-72x+28=0\)
\(\Leftrightarrow x^3-\dfrac{2}{3}x+\dfrac{7}{27}=0\)
- AD công thức các đa nô :
\(\Rightarrow x=\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}+\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}+\sqrt[3]{-\dfrac{-\dfrac{2}{3}}{2}-\sqrt{\dfrac{\left(-\dfrac{2}{3}\right)^2}{4}+\dfrac{\left(\dfrac{7}{27}\right)^3}{27}}}\)
\(\Rightarrow x\approx-0,96685\)
ĐKXĐ: x>=1
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{1}{2}\left(x+3\right)\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\dfrac{1}{2}\left(x+3\right)\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{1}{2}\left(x+3\right)\)
=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{1}{2}\left(x+3\right)\)
TH1: \(x>=2\)
PT sẽ tương đương với \(\sqrt{x-1}+1+\sqrt{x-1}-1=\dfrac{1}{2}\left(x+3\right)\)
=>\(2\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)
=>\(4\sqrt{x-1}=x+3\)
=>\(\sqrt{16x-16}=x+3\)
=>x>=-3 và (x+3)^2=16x-16
=>x>=-3 và x^2+6x+9-16x+16=0
=>x>=-3 và x^2-7x+25=0
=>Loại
TH2: 1<=x<2
PT sẽ là \(\sqrt{x-1}+1+1-\sqrt{x-1}=\dfrac{1}{2}\left(x+3\right)\)
=>1/2(x+3)=2
=>x+3=4
=>x=1(nhận)
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
Đặt căn x=a; căn 1-x=b
Theo đề, ta có: a+b=1+2/3ab
=>3a+3b=3+2ab
=>3a+3b-2ab=3
=>a(3-2b)+3b-4,5=-1,5
=>-a(2b-3)+3(b-1,5)=-1,5
=>-2a(b-1,5)+3(b-1,5)=-1,5
=>(-2a+3)(b-1,5)=-1,5
=>(2a-3)(b-1,5)=1,5
=>(2a-3)(2b-3)=3
=>(2a-3;2b-3) thuộc {(1;3); (3;1);(-1;-3); (-3;-1)}
=>(a,b) thuộc {(2;3); (3;2); (1;0); (0;1)}
TH1: a=2; b=3
=>căn x=2 và căn 1-x=3
=>x=4 và 1-x=9
=>Loại
TH2: a=3 và b=2
=>căn x=3 và căn 1-x=2
=>x=9 và 1-x=4(loại)
TH3: a=1 và b=0
=>x=1 và 1-x=0
=>x=1
TH4: a=0 và b=1
=>x=0 và 1-x=1
=>x=0
ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)
\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)
- TH1: Với \(x>1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)
Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)
BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)
\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)
TH2: với \(x< 1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)
Tương tự như trên, đặt \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)
\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)
Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)
1) \(\sqrt{x^2+1}=\sqrt{5}\)
\(\Leftrightarrow x^2+1=5\)
\(\Leftrightarrow x^2=5-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=3+1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=\dfrac{4}{2}\)
\(\Leftrightarrow x=2\left(tm\right)\)
3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))
\(\Leftrightarrow43-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1=43-x\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))
\(\Leftrightarrow\sqrt{4x-3}=x-2\)
\(\Leftrightarrow4x-3=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+4=4x-3\)
\(\Leftrightarrow x^2-8x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1^2\)
\(\Leftrightarrow x=1\left(tm\right)\)
1)
\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy PT có nghiệm `x=2` hoặc `x=-2`
2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
Vậy PT có nghiệm `x=2`
3)
\(ĐKXĐ:x\le43\)
PT trở thành:
\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm `x=-6` hoặc `x=7`
4)
ĐKXĐ: \(x\ge\dfrac{3}{4}\)
PT trở thành:
\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)
5)
ĐKXĐ: \(x\ge0\)
PT trở thành:
\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
Khi đó:
(1)\(\Leftrightarrow3t^2+8t+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)
Vậy PT vô nghiệm.