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a/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
Ta được:
\(1+\frac{a^2-1}{3}=a\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x\left(1-x\right)}=0\\2\sqrt{x-x^2}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(1-x\right)=0\\-4x^2+4x-9=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\sqrt{x+5}=a\ge0\Rightarrow a^2-x=5\)
\(x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+a+x=0\)
\(\Leftrightarrow\left(a+x\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-x\left(x\le0\right)\\\sqrt{x+5}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2\left(x\le0\right)\\x+5=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
c/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
a/ ĐKXĐ: \(x^2+3x+2\ge0\)
\(\Leftrightarrow3-2\sqrt{x^2+3x+2}=1-2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\sqrt{x^2+3x+2}=\sqrt{x^2-x+1}+1\)
\(\Leftrightarrow x^2+3x+2=x^2-x+1+1+2\sqrt{x^2-x+1}\)
\(\Leftrightarrow2x=\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2-x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x^2+x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{-1+\sqrt{13}}{6}\\x=\frac{-1-\sqrt{13}}{6}\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(3x^2-7x+2\ge0\)
\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\) (1)
\(\Rightarrow3x^2-5x+7=9+3x^2-7x+2-6\sqrt{3x^2-7x+2}\)
\(\Rightarrow2-x=3\sqrt{3x^2-7x+2}\) (\(x\le2\))
\(\Rightarrow\left(2-x\right)^2=9\left(3x^2-7x+2\right)\)
\(\Rightarrow x^2-4x+4=27x^2-63x+18\)
\(\Rightarrow26x^2-59x+14=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)
Do bước biến đổi thứ 2 ko phải phép tương đương nên cần thay 2 nghiệm vào (1) để kiểm tra lại, bạn tự thay nhé
a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)
đặt\(x^2+x+1=t\left(t>0\right)\)
\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)
bình phương 2 vế pt trở thành:
\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)
\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m
vậy pt vô nghiệm
a/ ĐKXĐ: ...
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)
\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)
\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))
\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)
\(\Leftrightarrow11a^2+6a-25=0\)
Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó
b/
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)
\(\Leftrightarrow\sqrt{a^2+3a}=2\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
b: ĐKXĐ: x>=-1
\(\sqrt{x+1}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)
c: \(\sqrt{x-1}=1-x\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)
Do đó: x=1 là nghiệm của phương trình
d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)
\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)
\(\Leftrightarrow x^2+x-2=0\)
=>(x+2)(x-1)=0
=>x=-2(nhận) hoặc x=1(loại)