\(\sqrt{ }\)x+3 + \(\sqrt{ }\)6-x - \(\sqrt{ }\)x+3×\(\sqrt{ }\)6-x = 3
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: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
3.
ĐKXĐ: \(x\ge-1\)
\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
1,\(\sqrt{x-5}+\sqrt{x+4}=3\)
2,\(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+6-6\sqrt{x}}=1\)
3,\(\sqrt{x+3}-\sqrt{x-4}=1\)
4,\(\sqrt{15-x}+\sqrt{3-x}=6\)
5,\(\sqrt{10-x}+\sqrt{x+3}=5\)
6,\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
7,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
8,\(\sqrt{x^2-5x+6}+\sqrt{x-2-3\sqrt{x-3}}=3\)
9,\(2x^2-x+4=2\sqrt{2x+3}\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
6.
\(DK:x\ge2\)
\(\Leftrightarrow\left(\sqrt{2x-1}-\sqrt{x+1}\right)+\sqrt{x-2}=0\)
\(\Leftrightarrow\frac{x-2}{\sqrt{2x-1}+\sqrt{x+1}}+\sqrt{x-2}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\frac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1\right)=0\)
Vi \(\frac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1>0\)
\(\Rightarrow x=2\left(n\right)\)
Vay nghiem cua PT la \(x=2\)
\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
1) \(\sqrt{x^2-4x+5}+3=4x-x^2\)
2) \(4\sqrt{x^2-6+6}=x^2-6x +9\)
3) \(\sqrt{x^2-3x^3}+\sqrt{x^2-3x+6}=3\)
4) \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
rút gọn
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) ( x>6)
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
\(\sqrt{28-6\sqrt{3}}\)
\(\sqrt{6-\sqrt{20}}\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\cdot\left(x+2\right)}}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
A=(\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\))/(\(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\))
\(\sqrt{28-6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=3\sqrt{3}-1\)
\(\sqrt{6-\sqrt{20}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-1\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\left(x+2\right)}}\)
\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x+1}\right)^2}\)
\(=\sqrt{x+2}+\sqrt{x+1}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{\left(x-1\right)\left(x+3\right)}+\left(x+3\right)}\)
\(=\sqrt{\left(\sqrt{x+3}-\sqrt{x-1}\right)^2}\)
\(=\left|\sqrt{x+3}-\sqrt{x-1}\right|\)
\(\sqrt{21-6\sqrt{6}}+\sqrt{21+6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\)
\(=6\sqrt{2}\)
\(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(\left[\dfrac{\left(\sqrt{x}+1\right)-\left(3-\sqrt{x}\right)}{\sqrt{x}+1}\right]\)
\(=\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]\times\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}\times2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
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 phương trình: \(x=\sqrt{x-2}.\sqrt{x-3}+\sqrt{x-3}.\sqrt{x-6}+\sqrt{x-6}.\sqrt{x-2}\)