\(\sqrt{x-1+x\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}=2\)
Những câu hỏi liên quan
LARGE 7)Gfrac{sqrt{x}+1}{x-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2sqrt{x}+2}{x-1}Gfrac{sqrt{x}+1}{sqrt{x}-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2(sqrt{x}+1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{sqrt{x}+1}{x-1}+frac{sqrt{x}-1}{sqrt{x}+1}-frac{2}{sqrt{x}-1}Gfrac{(sqrt{x}+1)^2+(sqrt{x}-1)^2-2(sqrt{x}+1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2x-2sqrt{x}}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2sqrt{x}(sqrt{x}-1)}{(sqrt{x}-1)(sqrt{x}+1)}Gfrac{2sqrt{x}}{sqrt{x}+1}8)H(frac{1}{sqrt{x}-1}+frac{sqrt{x}}{x-1}):(frac{sqrt{x}}{sqrt{x}-1}-...
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\[\LARGE 7)G=\frac{\sqrt{x}+1}{x-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{x-1}\\G=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{\sqrt{x}+1}{x-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}\\G=\frac{(\sqrt{x}+1)^2+(\sqrt{x}-1)^2-2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2x-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\G=\frac{2\sqrt{x}}{\sqrt{x}+1}\\8)H=(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}):(\frac{\sqrt{x}}{\sqrt{x}-1}-1)\\H=(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}):(\frac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}-1})\\H=\frac{\sqrt{x}+1+\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\\H=\frac{2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{1}{\sqrt{x}-1}\\H=\frac{2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}.(\sqrt{x}-1)\\H=\frac{2\sqrt{x}}{\sqrt{x}+1}\]
frac{x+1}{2left(x-1right)}+frac{2}{2left(sqrt{x}+1right)}+frac{sqrt{x}}{sqrt{x}left(sqrt{x}-1right)}.frac{left(x+1right).sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}-1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2sqrt{x}left(sqrt{x}+1right)}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}.frac{xsqrt{x}+sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x-2sqrt{x}}{2sqrt{x}left(sqrt{x}-1right)left(sqrt{x}+1right)}+frac{2x+2...
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\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
3 tháng 7 2023 lúc 15:15
Giải pt:1) x - 2sqrt{x - 1} 162) sqrt{1:-:x} - sqrt{x - 3} 03) sqrt{x - 4} + 2 74) x - sqrt{x - 2sqrt{x:-:1}} 05) sqrt{x - 2} - sqrt{x^2 - 2x} 06) sqrt{3:-:2sqrt{2}} - sqrt{x^2 + 2xsqrt{2}+2} 0
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Giải pt:
1) x - 2\(\sqrt{x - 1}\) = 16
2) \(\sqrt{1\:-\:x}\) - \(\sqrt{x - 3}\) = 0
3) \(\sqrt{x - 4}\) + 2 = 7
4) x - \(\sqrt{x - 2\sqrt{x\:-\:1}}\) = 0
5) \(\sqrt{x - 2}\) - \(\sqrt{x^2 - 2x}\) = 0
6) \(\sqrt{3\:-\:2\sqrt{2}}\) - \(\sqrt{x^2 + 2x\sqrt{2}+2}\) = 0
1
ĐK: \(x\ge1\)
Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow x=t^2+1\)
Khi đó:
\(x-2\sqrt{x-1}=16\)
\(\Leftrightarrow t^2-2t+1=16\\ \Leftrightarrow\left(t-1\right)^2=4^2\\ \Leftrightarrow t-1=4\\ \Leftrightarrow t=4+1=5\left(tm\right)\)
\(\Leftrightarrow\sqrt{x-1}=5\)
\(\Leftrightarrow x-1=5^2=25\\ \Leftrightarrow x=25+1=26\left(tm\right)\)
Vậy PT có nghiệm duy nhất x = 26.
2 ĐK: \(3\le x\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{1-x}=0\\\sqrt{x-3}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Từ điều kiện và bài giải ta kết luận PT vô nghiệm.
3 ĐK: \(x\ge4\)
\(\Leftrightarrow\sqrt{x-4}=7-2=5\\ \Leftrightarrow x-4=5^2=25\\ \Leftrightarrow x=25+4=29\left(tm\right)\)
Vậy PT có nghiệm duy nhất x = 29.
4
ĐK: \(x\ge1\)
Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow x=t^2+1\)
Khi đó:
\(x-\sqrt{x-2\sqrt{x-1}}=0\\ \Leftrightarrow t^2+1-\sqrt{t^2-2t+1}=0\\ \Leftrightarrow t^2+1-\sqrt{\left(t-1\right)^2}=0\\ \Leftrightarrow t^2+1-\left|t-1\right|=0\left(1\right)\)
Trường hợp 1:
Với \(0\le t< 1\) thì:
\(\left(1\right)\Leftrightarrow t^2+1-\left(1-t\right)=0\\ \Leftrightarrow t^2+t=0\\ \Leftrightarrow t\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\Rightarrow\sqrt{x-1}=0\Rightarrow x=1\left(nhận\right)\\t=-1\left(loại\right)\end{matrix}\right.\)
Trường hợp 2:
Với \(t\ge1\) thì:
\(\left(1\right)\Leftrightarrow t^2+1-\left(t-1\right)=0\\ \Leftrightarrow t^2-t+2=0\)
\(\Delta=\left(-1\right)^2-4.2=-7< 0\)
=> Loại trường hợp 2.
Vậy PT có nghiệm duy nhất x = 1.
5
ĐK: \(x\ge2\)
Đặt \(\sqrt{x-2}=t\left(t\ge0\right)\Rightarrow x=t^2+2\)
Khi đó:
\(\sqrt{x-2}-\sqrt{x^2-2x}=0\\ \Leftrightarrow\sqrt{x-2}-\sqrt{x}.\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{t^2+2-2}-\sqrt{t^2+2}.\sqrt{t^2+2-2}=0\\ \Leftrightarrow\sqrt{t^2}-\sqrt{t^2+2}.\sqrt{t^2}=0\\ \Leftrightarrow t-\sqrt{t^2+2}.t=0\\ \Leftrightarrow t\left(1-\sqrt{t^2+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\Rightarrow\sqrt{x-2}=0\Rightarrow x=2\left(tm\right)\\\sqrt{t^2+2}=1\Rightarrow t^2+2=1\Rightarrow t^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất x = 2.
6 Không có ĐK vì đưa về tổng bình lên luôn \(\ge0\)
\(\Leftrightarrow\sqrt{\sqrt{2}^2-2.\sqrt{2}.\sqrt{1}+\sqrt{1}^2}-\sqrt{x^2+2x.\sqrt{2}+\sqrt{2}^2}=0\\ \Leftrightarrow\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}-\sqrt{\left(x+\sqrt{2}\right)^2}=0\\ \Leftrightarrow\left|\sqrt{2}-\sqrt{1}\right|-\left|x+\sqrt{2}\right|=0\\ \Leftrightarrow\sqrt{2}-1-\left|x+\sqrt{2}\right|=0\)
Trường hợp 1:
Với \(x\ge-\sqrt{2}\) thì:
\(\left(1\right)\Leftrightarrow\sqrt{2}-1-\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\sqrt{2}-1-x-\sqrt{2}=0\\ \Leftrightarrow-1-x=0\\ \Leftrightarrow x=-1\left(tm\right)\)
Với \(x< -\sqrt{2}\) thì:
\(\left(1\right)\Leftrightarrow\sqrt{2}-1--\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\sqrt{2}-1+x+\sqrt{2}=0\\ \Leftrightarrow2\sqrt{2}+1+x=0\\ \Leftrightarrow x=-1-2\sqrt{2}\left(tm\right)\)
Vậy phương trình có 2 nghiệm \(x=-1\) hoặc \(x=-1-2\sqrt{2}\)
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Bleft(frac{xsqrt{x}+x+sqrt{x}}{xsqrt{x}-1}-frac{sqrt{x}+3}{1-sqrt{x}}right).frac{x-1}{2x+sqrt{x}-1} ĐKXĐ: ...frac{left(xsqrt{x}+x+sqrt{x}right)left(1-sqrt{x}right)-left(sqrt{x}+3right)left(xsqrt{x}-1right)}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2x+2sqrt{x}-sqrt{x}-1}frac{xsqrt{x}+x+sqrt{x}-x^2-xsqrt{x}-x-x^2+sqrt{x}-3xsqrt{x}+3}{left(xsqrt{x}-1right)left(1-sqrt{x}right)}.frac{x-1}{2sqrt{x}left(sqrt{x}+1right)-left(sqrt{x}+1right)}frac{-3xsqrt{x}+2sqrt{x}-2x^2+3}{left(xsqrt{x}-1ri...
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\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Cho hàm số \(y=\sqrt{x+\sqrt{x^2+1}}\). Tính đạo gàm của hàm số.
A. \(y'=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}}\)
B. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}\)
C. \(y'=\dfrac{\sqrt{x^2+1}}{2\sqrt{\sqrt{x+\sqrt{x^2+1}}}}\)
D. \(y'=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
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Bài: Rút gọn biểu thức sau1)(1-dfrac{sqrt{x}}{1-sqrt{x}}):(dfrac{sqrt{x}+3}{sqrt{x}-2}-dfrac{sqrt{x}+2}{3-sqrt{x}}+dfrac{sqrt{x}+2}{x-5sqrt{x}+6})2)(dfrac{1}{sqrt{x}+1}-dfrac{2sqrt{x}-2}{xsqrt{x}-sqrt{x}+x-1}):(dfrac{1}{sqrt{x}-1}-dfrac{2}{x-1})
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Bài: Rút gọn biểu thức sau
1)(1-\(\dfrac{\sqrt{x}}{1-\sqrt{x}}\)):(\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)-\(\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\)+\(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\))
2)(\(\dfrac{1}{\sqrt{x}+1}\)-\(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\)):(\(\dfrac{1}{\sqrt{x}-1}\)-\(\dfrac{2}{x-1}\))
1: \(=\left(1+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)
2: \(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
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Pleft(frac{sqrt{x}+2}{sqrt{x}+1}-frac{x-sqrt{x}-3}{x-sqrt{x}-2}right):left(frac{x-sqrt{x}}{x-sqrt{x}-2}+frac{2}{sqrt{x}-2}right)frac{left(sqrt{x}+2right)left(sqrt{x}-2right)-x+sqrt{x}+3}{left(sqrt{x}+1right)left(sqrt{x}-2right)}:frac{x-sqrt{x}+2left(sqrt{x}+1right)}{left(sqrt{x}+1right)left(sqrt{x}-2right)}frac{x-4-x+3+sqrt{x}}{left(sqrt{x}+1right)left(sqrt{x}-2right)}.frac{left(sqrt{x}+1right)left(sqrt{x}-2right)}{x-sqrt{x}+2sqrt{x}+2}frac{sqrt{x}-1}{x+sqrt{x}+2}
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\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???
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Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !
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Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?
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Xem thêm câu trả lời
a) 8sqrt{x+2} + sqrt{11-x} - 2sqrt{22+9x-x^2}+ 4 0b) sqrt{1+4x}+ 2sqrt{2-x}+2sqrt{left(1+4xright)left(2-xright)}3c) sqrt{8+sqrt{x}}+sqrt{5-sqrt{x}}5d) sqrt{x^4-1}-2 sqrt{x-1}- 2sqrt{x^3+x^2+x+1}
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a) \(8\sqrt{x+2}\) + \(\sqrt{11-x}\) - \(2\sqrt{22+9x-x^2}\)+ 4 =0
b) \(\sqrt{1+4x}\)+ \(2\sqrt{2-x}\)+\(2\sqrt{\left(1+4x\right)\left(2-x\right)}\)=3
c) \(\sqrt{8+\sqrt{x}}\)+\(\sqrt{5-\sqrt{x}}\)=5
d) \(\sqrt{x^4-1}\)-2 =\(\sqrt{x-1}\)- \(2\sqrt{x^3+x^2+x+1}\)
c) \(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}=5\)
\(\Leftrightarrow\left(\sqrt[]{8+\sqrt[]{x}}+\sqrt{5-\sqrt[]{x}}\right)^2=25\left(1\right)\left(đkxđ:0\le x\le25\right)\)
Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số dương \(\left(1;\sqrt[]{8+\sqrt[]{x}}\right);\left(1;\sqrt{5-\sqrt[]{x}}\right)\)
\(\left(1.\sqrt[]{8+\sqrt[]{x}}+1.\sqrt{5-\sqrt[]{x}}\right)^2\le\left(1^2+1^2\right)\left(8+\sqrt[]{x}+5-\sqrt[]{x}\right)=26\)
\(\left(1\right)\Leftrightarrow26=25\left(vô.lý\right)\)
Vậy phương trình đã cho vô nghiệm
b) \(\sqrt[]{1+4x}+2\sqrt[]{2-x}+2\sqrt[]{\left(1+4x\right)\left(2-x\right)}=3\) \(\left(đkxđ:-\dfrac{1}{4}\le x\le2\right)\)
\(\)\(\Leftrightarrow\sqrt[]{1+4x}+2\sqrt[]{2-x}=3-2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\)
\(\Leftrightarrow\left(\sqrt[]{1+4x}+2\sqrt[]{2-x}\right)^2=\left[3-2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\right]^2\left(1\right)\)
Áp dụng Bất đẳng thức Bunhiacopxki :
\(\left(1.\sqrt[]{1+4x}+2\sqrt[]{2-x}\right)^2\le\left(1^2+2^2\right)\left(1+4x+2-x\right)=5\left(3x+3\right)\)
Áp dụng Bất đẳng thức Cauchy :
\(2\sqrt[]{\left(1+4x\right)\left(2-x\right)}\le1+4x+2-x=3x+3\)
Dấu "=" xảy ra khi và chỉ khi
\(1+4x=2-x\)
\(\Leftrightarrow x=\dfrac{1}{5}\left(thỏa.đk\right)\)
\(pt\left(1\right)\Leftrightarrow5\left(4x+3\right)=4x+3\)
\(\Leftrightarrow4\left(4x+3\right)=0\)
\(\Leftrightarrow x=-\dfrac{3}{4}\left(k.thỏa.x=\dfrac{1}{5}.vô.lý\right)\)
Vậy phương trình đã cho vô nghiệm
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left(dfrac{x+2}{xsqrt{x}-1}+dfrac{sqrt{x}}{x+sqrt{x}+1}+dfrac{1}{1-sqrt{x}}right):dfrac{sqrt{x}-1}{2}
left(dfrac{x+2}{left(sqrt{x}-1right)left(x+sqrt{x}+1right)}+dfrac{sqrt{x}left(sqrt{x}-1right)}{left(sqrt{x}-1right)left(x+sqrt{x}+1right)}-dfrac{1}{sqrt{x}-1}right).dfrac{2}{sqrt{x}-1}
dfrac{x+2+x-sqrt{x}-x-sqrt{x}-1}{left(sqrt{x}-1right)left(x+sqrt{x}+1right)}.dfrac{2}{sqrt{x}-1}dfrac{sqrt{x}-1}{x+sqrt{x}+1}.dfrac{2}{sqrt{x}-1}dfrac{2}{x+sqrt{x}+1}
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\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
6 tháng 8 2021 lúc 19:56
Giải phương trình:1. sqrt{dfrac{42}{5-x}}+sqrt{dfrac{60}{7-x}}62. sqrt{x^2-3x+2}+sqrt{x+3}sqrt{x-2}+sqrt{x^2+2x-3}3. x^2+x+12sqrt{x+1}364. sqrt{x+2}-sqrt{x-6}25. sqrt[3]{x-1}-sqrt[3]{x-3}sqrt[3]{2}6. 5sqrt{1+x^3}2left(x^2+2right)6. left(sqrt{x+5}-sqrt{x+2}right)left(1+sqrt{x^2+7x+10}right)3
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Giải phương trình:
1. \(\sqrt{\dfrac{42}{5-x}}+\sqrt{\dfrac{60}{7-x}}=6\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)
3. \(x^2+x+12\sqrt{x+1}=36\)
4. \(\sqrt{x+2}-\sqrt{x-6}=2\)
5. \(\sqrt[3]{x-1}-\sqrt[3]{x-3}=\sqrt[3]{2}\)
6. \(5\sqrt{1+x^3}=2\left(x^2+2\right)\)
6. \(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
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b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
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3.
ĐKXĐ: \(x\ge-1\)
\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
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