Bài 3 giải phương trình
a) \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x\)
b) \(\frac{\sqrt{x}+5}{\sqrt{x}-4}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
Những câu hỏi liên quan
Bài Toán :
Giải phương trình sau :
\(\frac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4.\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{5\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
Bài toán :
Giải phương trình :
\(\frac{3.\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-5\right)}+\frac{5\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...
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b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy
Giải phương trình
\(\frac{3\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)}{\left(1-\sqrt{3}\right)\left(1-\sqrt{5}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{5}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{5}\right)}+\frac{5\left(x-1\right)\left(x+\sqrt{3}\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}-\sqrt{3}\right)}=3x-2\)
giải các phương trình
a \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)
b \(\sqrt{3x^2-4x}=2x-3\)
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
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4 tháng 7 2021 lúc 20:43
* giải phương trình
a. \(\sqrt{\left(x-3\right)^2}=2\)
b.\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7(thỏa ĐK)
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a) \(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy:.....
b) ĐKXĐ: x ≥ -2
\(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)
\(\Leftrightarrow2.\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
<=> x + 2 = 9
<=> x = 7
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Tham khảo ạ:
a) Ta có: √(x−3)2=2(x−3)2=2
⇔|x−3|=2⇔|x−3|=2
⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1
b) ĐKXĐ: x≥−2x≥−2
Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6
⇔2√x+2=6⇔2x+2=6
⇔x+2=9⇔x+2=9
hay x=7(thỏa ĐK)
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Bài 1:a, Asqrt{2-sqrt{3}}+sqrt{2+sqrt{3}}b, B left(frac{1}{sqrt{5}-sqrt{2}}-frac{1}{sqrt{5}+sqrt{2}}+1right).frac{1}{left(sqrt{2}+1right)^2}Bài 2: Giải pta,frac{5sqrt{x}-2}{8sqrt{x}+2,5}frac{2}{7}b, sqrt{x^2-6x+9}sqrt{4+2sqrt{3}}Bài 3:Aleft(frac{x+2}{xsqrt{x}+1}-frac{1}{sqrt{x}+1}right).frac{4sqrt{x}}{3}
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Bài 1:
a, A=\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b, B= \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right).\frac{1}{\left(\sqrt{2}+1\right)^2}\)
Bài 2: Giải pt
a,\(\frac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\frac{2}{7}\)
b, \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
Bài 3:
A=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)
6 tháng 7 2021 lúc 20:37
* giải phương trình
a. \(\sqrt{\left(x+1\right)^2}=3\)
b. \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)
a) Ta có: \(\sqrt{\left(x+1\right)^2}=3\)
\(\Leftrightarrow\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)
\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)
\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)
\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)
\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)
\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)
\(\Leftrightarrow30\sqrt{x-3}=7x+18\)
\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)
\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)
\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)
Vậy: Phương trình vô nghiệm
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a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b)Đk:\(x\ge-1\)
Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)
Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{x+1}=5\)
\(\Leftrightarrow x=24\left(tm\right)\)
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a. \(\sqrt{\left(x+1\right)^2}\) \(=3\)
⇔ \(\left|x+1\right|=3\)
⇔ \(\left|x\right|=2\)
⇒ \(x=2\) và \(x=-2\)
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Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
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Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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