vận dụng bđt để giải Pt sau
\(\sqrt{2x-1}+\sqrt{19-2x}=\frac{6}{-x^2+10x-24}\)\(\left|x+1\right|+\left|x+2\right|+...+\left|x+2005\right|=2006x\)x2=2x8+\(\frac{3}{8}\)\(x+\sqrt{3+\sqrt{x}}=3\)\(8x^2+\sqrt{\frac{1}{x}}=\frac{5}{2}\)a)Giải các phương trình sau bằng phương pháp đặt ẩn phụ:
1) \(x^2-3x-3=\frac{3\left(\sqrt[3]{x^3-4x^2+4}-1\right)}{1-x}\) ;2)\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
b) Giải các phương trình sau(không giới hạn phương pháp):
1)\(2\left(1-x\right)\sqrt{x^2+2x-1}=x^2-2x-1\) ; 2)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
3)\(\frac{3x^2+3x-1}{3x+1}=\sqrt{x^2+2x-1}\) ; 4) \(\frac{2x^3+3x^2+11x-8}{3x^2+4x+1}=\sqrt{\frac{10x-8}{x+1}}\)
5)\(13x-17+4\sqrt{x+1}=6\sqrt{x-2}\left(1+2\sqrt{x+1}\right)\);
6)\(x^2+8x+2\left(x+1\right)\sqrt{x+6}=6\sqrt{x+1}\left(\sqrt{x+6}+1\right)+9\)
7)\(x^2+9x+2+4\left(x+1\right)\sqrt{x+4}=\frac{5}{2}\sqrt{x+1}\left(2+\sqrt{x+4}\right)\)
8)\(8x^2-26x-2+5\sqrt{2x^4+5x^3+2x^2+7}\)
À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)
Giải pt : a) \(8x^2-13x+7=\left(1+\frac{1}{x}\right)\sqrt[3]{3x^2-2}\)
b) \(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
c) \(2\sqrt{x+1}+6\sqrt{9-x^2}+6\sqrt{\left(x+1\right)\left(9-x^2\right)}=38+10x-2x^2-x^3\)
Vũ Minh Tuấn, Băng Băng 2k6, Hoàng Tử Hà, đề bài khó wá, Lê Gia Bảo, Aki Tsuki, Nguyễn Việt Lâm,
Lê Thị Thục Hiền, Nguyễn Trúc Giang, Học 24h, @tth_new, @Akai Haruma
Help me! Cần gấp
thanks!
Giải pt:
\(\sqrt{x^2+10x+21}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
\(4\left(x+1\right)^2=\left(2x+10\right)\left(1-\sqrt{3+2x}\right)^2\)
\(\frac{1}{1-\sqrt{1-x}}-\frac{1}{1+\sqrt{1-x}}=\frac{\sqrt{3}}{x}\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\sqrt{x-2}+\sqrt{4-x}=x^2-6x+11\)
a) ĐKXĐ: x\(\ge\)-3
PT\(\Leftrightarrow\sqrt{\left(x+7\right)\left(x+3\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)
Đặt \(\left(\sqrt{x+3},\sqrt{x+7}\right)=\left(a,b\right)\) \(\left(a,b\ge0\right)\)
PT\(\Leftrightarrow ab=3a+2b-6\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Leftrightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)(TM ĐK)
TH 1: a=2\(\Leftrightarrow\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)(tm)
TH 2: b=3\(\Leftrightarrow\sqrt{x+7}=3\Leftrightarrow x+7=9\Leftrightarrow x=2\)(tm)
Vậy tập nghiệm phương trình S={1; 2}
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 \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
giải pt
a) \(x+\sqrt{x+8}\left(1-\sqrt{x+8}\right)=\sqrt{x}+\sqrt{x+3}-8\)
b) \(2\left(2-x\right)=\sqrt{2x-4}\left(\sqrt{5-x}-\sqrt{3x-3}\right)\)
c) \(\sqrt[3]{24+x}.\sqrt{12-x}-6\sqrt{12-x}=x-12\)
d) \(\frac{x-1}{2\sqrt{3-2x}-3}=\frac{x-1}{3-2\sqrt[3]{5+3x}}\)
a/ ĐKXĐ: ...
\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)
\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)
\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))
\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)
\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)
\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\le12\)
\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)
\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)
\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)
\(\Leftrightarrow a^3+a^2-12a=0\)
\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)
d/ ĐKXĐ: \(x\le\frac{3}{2}\) ; \(x\ne\frac{3}{8};x\ne-\frac{13}{24}\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2\sqrt{3-2x}-3}-\frac{1}{3-2\sqrt[3]{5+3x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{1}{2\sqrt{3-2x}-3}=\frac{1}{3-2\sqrt[3]{5+3x}}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{3-2x}-3=3-2\sqrt[3]{5+3x}\)
\(\Leftrightarrow\sqrt[3]{5+3x}+\sqrt{3-2x}=3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{5+3x}=a\\\sqrt{3-2x}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a+b=3\\2a^3+3b^2=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=3-a\\2a^3+3b^2=19\end{matrix}\right.\)
\(\Leftrightarrow2a^3+3\left(3-a\right)^2=19\)
\(\Leftrightarrow2a^3+3a^2-18a+8=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-4\\a=\frac{1}{2}\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{5+3x}=-4\\\sqrt[3]{5+3x}=\frac{1}{2}\\\sqrt[3]{5+3x}=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5+3x=-64\\5+3x=\frac{1}{8}\\5+3x=8\end{matrix}\right.\)
Áp dụng nội suy niu tơn để giải pt sau
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
giải pt :
a, \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
b, \(2\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0\)
c, \(\left(x+3\right)\sqrt{-x^2-8x+48}=x-24\)
d, \(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)\left(2-x\right)}=3\)
e, \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
Giải các phương trình sau: (hệ phương trình)
1.\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
2.\(\sqrt{3-4x}+\sqrt{4x+1}=-16x^2-8x+1\)
3. \(\sqrt{x^2-2x+5}+\sqrt{x+1}=2\)
4. \(\left(-4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\)
5. \(\sqrt{-4x-1}+\sqrt{4x^2+8x+3}=-4x^2-4x\)
6. \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\sqrt{\frac{x+1}{x-3}}=-3\)
7. \(\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)
Ai làm được 4 bài hoặc nhiều hơn mik sẽ tick nha :)