giải phương trình:
\(\sqrt{x^2-1}=x+1+\sqrt{x+1}\)
Giải phương trình:
\(\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)
ĐKXĐ: \(x\ge-1\)
\(\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-6\sqrt{x+1}+9}=2\sqrt{x+1-2\sqrt{x+1}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-3\right)^2}=2\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\)
Ta có:
\(\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|\ge\left|\sqrt{x+1}+1+\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\)
Dấu "=" xảy ra khi và chỉ khi:
\(\sqrt{x+1}-3\ge0\Rightarrow x\ge8\)
Vậy nghiệm của pt là \(x\ge8\)
Giải phương trình:
\(\sqrt{x+2\sqrt{x}+1}-\sqrt{x-2\sqrt{x}+1}=2\)
\(\sqrt{x+2\sqrt{x}+1}-\sqrt{x-2\sqrt{x}+1}=2\left(x\ge0\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow\sqrt{x}+1-\left|\sqrt{x}-1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1-\left(\sqrt{x}-1\right)=2,\forall\sqrt{x}-1\ge0\\\sqrt{x}+1-\left(1-\sqrt{x}\right)=2,\forall\sqrt{x}-1< 0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0\sqrt{x}=0,\forall x\ge1\\\sqrt{x}=1,\forall x< 1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in R,x\ge1\\x=1,x< 1\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x\in R,x\ge1\)
Giải phương trình (bằng phương pháp ẩn phụ): \(\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=2\).
ĐKXĐ: \(x\ge1\)
Do \(\sqrt{x-\sqrt{x^2-1}}.\sqrt{x+\sqrt{x^2-1}}=\sqrt{x^2-x^2+1}=1\)
Đặt \(\sqrt{x-\sqrt{x^2-1}}=t\Rightarrow\sqrt{x+\sqrt{x^2-1}}=\dfrac{1}{t}\)
Phương trình trở thành:
\(t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\Rightarrow t=1\)
\(\Rightarrow\sqrt{x-\sqrt{x^2-1}}=1\Leftrightarrow x-\sqrt{x^2-1}=1\)
\(\Leftrightarrow x-1=\sqrt{x^2-1}\)
\(\Rightarrow x^2-2x+1=x^2-1\)
\(\Rightarrow x=1\) (thỏa mãn)
Giải phương trình
\(\sqrt{x+ 2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{1}{2}(x+3)\)
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)
Giải phương trình P=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}+2}{x-1}\)
ĐKXĐ \(x\ge1\)
\(P=\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}+\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{2\sqrt{x}+2}{x-1}\)
\(P=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-2\sqrt{x}-2}{x-1}\)
\(P=\dfrac{2x-2\sqrt{x}}{x-1}\)
\(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
Giải phương trình ???
Giải phương trình \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}\) ta được nghiệm là
`\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\sqrt{x-1}(x>=1)`
`<=>\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\sqrt{x-1}`
`<=>\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=2\sqrt{x-1}`
`<=>|\sqrt{x-1}+1|+|\sqrt{x-1}-1|=2\sqrt{x-1}`
`<=>\sqrt{x-1}+1+|\sqrt{x-1}-1|=2\sqrt{x-1}`
`<=>|\sqrt{x-1}-1|=\sqrt{x-1}-1`
`<=>\sqrt{x-1}-1>=0``
`<=>sqrt{x-1}>=1`
`<=>x-1>=1`
`<=>x>=2`
Vậy `S={x|x>=2}`
giải phương trình:\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{x-1}-1=0\)
\(\Leftrightarrow x-1-\sqrt{x-1}-1=0\) (1)
Đặt \(\sqrt{x-1}\) = t (t \(\ge0\))
pttt : t2 - t - 1 =0
\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\\t=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)
=> \(\sqrt{x-1}=\dfrac{1+\sqrt{5}}{2}\)
\(\Leftrightarrow x-1=\dfrac{3+\sqrt{5}}{2}\)
\(\Leftrightarrow x=\dfrac{5+\sqrt{5}}{2}\) (tm)
p/s: thử lại hộ mình nhaa
giải phương trình \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)
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
\(\sqrt{x-1}+\sqrt{x^2-1}=x\sqrt{x}\)
Giải phương trình
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{\left(x-1\right)\left(x+1\right)}=x\sqrt{x}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}+1\right)=x\sqrt{x}\)
\(\Leftrightarrow\dfrac{\sqrt{x-1}.x}{\sqrt{x+1}-1}=x\sqrt{x}\)
\(\Leftrightarrow\dfrac{\sqrt{x-1}}{\sqrt{x+1}-1}=\sqrt{x}\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{x^2+x}-\sqrt{x}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x}=\sqrt{x^2+x}\)
\(\Leftrightarrow2x-1+2\sqrt{x^2-x}=x^2+x\)
\(\Leftrightarrow x^2-x-2\sqrt{x^2-x}+1=0\)
\(\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\)
\(\Leftrightarrow x^2-x-1=0\)
Giải phương trình (bằng phương pháp đánh giá): \(\sqrt{1-x^2}+\sqrt[4]{x^2+x-1}+\sqrt[6]{1-x}=1\).