giai phuong trinh
\(\frac{1}{\sqrt{x}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}=1\)
giai phuong trinh : \(\frac{1}{1+x}+\frac{1}{1+\sqrt{x}}=\frac{2+\sqrt{x}}{2x}\)
GIAI PHUONG TRINH:
\(2\left(\sqrt{\frac{x-1}{4}-3}\right)=2\sqrt{\frac{4x-4}{9}}-\frac{1}{3}\)
giai phuong trinh \(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
giai phuong trinh sau:
\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}}=2\)
m.m giup minh cau nay aj. thank m.n
Giai phuong trinh : \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{\sqrt{x}+2}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
Xin lỗi bạn nhiều nhiều lắm mình không biết làm bài này vì mình chưa học
không cần đâu bạn à bài này mình giải được rồi
2.giai phuong trinh sau:
a.\(\sqrt{\frac{2x-3}{x-1}}=2\)
b.\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
Đkiện: x <1 hoặc x \(\ge\frac{3}{2}\)
\(\sqrt{\frac{2x-3}{x-1}}=2\) (1)
(1) => \(\frac{2x-3}{x-1}=4\)
=> 2x - 3 = 4x - 4
<=> 2x - 4x = -4 + 3
<=> -2x = -1
<=> x = \(\frac{1}{2}\)( TMĐK)
Vậy x = \(\frac{1}{2}\)
b, Đkiện: x \(\ge\frac{3}{2}\)
(1) => \(\sqrt{2x-3}=2\sqrt{x-1}\)
=>2x - 3 = 4(x - 1)
<=> 2x -3 = 4x -4
<=> -2x = -1
<=> x = \(\frac{1}{2}\)(ko TMĐK)
Vậy pt vô nghiệm
b. \(x>0;x\ne1\)
\(\Rightarrow\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\Rightarrow2x-3=4x-4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
Giair phuong trinh
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (ĐK: \(x\ge\frac{-1}{2}\) )
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)
Vậy .....
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)(1)
Vì VT > 0 nên VP >0
\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow x\ge-\frac{1}{2}\)
Khi đó \(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)
Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))
A=(\(\frac{1}{\sqrt{x}-1}\)+\(\frac{1}{\sqrt{x}+1}\))2.\(\frac{x^2-1}{2}\)-\(\sqrt{1-x^2}\)
1)tim dk cua x de bieu thuc A co nghia
2)rut gon bieu thuc A
3) giai phuong trinh theo x khi A=-2
giải phuong trinh
\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)
\(ĐKXĐ:-\frac{1}{4}\le x\le2\)
\(PT\Leftrightarrow x+\sqrt{x+\frac{1}{4}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}}=2\)
\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)
\(\Leftrightarrow\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2=2\)
\(\Leftrightarrow\sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\Leftrightarrow x+\frac{1}{4}=\left(\sqrt{2}-\frac{1}{2}\right)^2=\frac{9-4\sqrt{2}}{4}\)
\(\Rightarrow x=\frac{9-4\sqrt{2}-1}{4}=\frac{8-4\sqrt{2}}{4}=2-\sqrt{2}\) (TMĐKXĐ)
Vậy PT trên có nghiệm là \(x=2-\sqrt{2}\)