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DV
2
22 tháng 11 2017
Bài này có trong đề Violympic toán 9 vòng 7 năm học 2017 2018
Đề bài này bị sai, trong căn thứ nhất không có x2 mà x thôi. Mình đã sửa đề và dùng shift solve ( hoặc biến đổi) được kết quả đúng là 2
23 tháng 11 2017
\(\sqrt{x+3+2\sqrt{3x}}-\sqrt{x+3-2\sqrt{3x}}=2\sqrt{2}\)
<=> \(\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{3}\sqrt{x}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{x}\right)^2-2\sqrt{3}\sqrt{x}+\left(\sqrt{3}\right)^2}=2\sqrt{2}\)
<=>\(\sqrt{\left(\sqrt{x}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{x}-\sqrt{3}\right)^2}=2\sqrt{2}\)
<=>\(\left(\sqrt{x}+\sqrt{3}\right)+\left(\sqrt{x}-\sqrt{3}\right)=2\sqrt{2}\)
<=>\(2\sqrt{x}=2\sqrt{2}\)
<=>\(\sqrt{x}=\sqrt{2}\)
<=>\(x=2\)
21 tháng 7 2018
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)
\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)
\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).
ND
21 tháng 7 2018
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)
\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)
\(i.\sqrt{x+2+2\sqrt{x+1}}=\sqrt{x+1+2\sqrt{x+1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=\sqrt{x+1}+1\)
BT
10 tháng 8 2017
a/ \(P=\left(\frac{x-7\sqrt{x}+12}{x-4\sqrt{x}+3}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}.\)
\(P=\left(\frac{x-7\sqrt{x}+12}{\left(x-4\sqrt{x}+4\right)-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2\right)^2-1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-2-1\right)\left(\sqrt{x}-2+1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\left(\frac{x-7\sqrt{x}+12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{x-7\sqrt{x}+12+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(P=\frac{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}-1\right)}\) => \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}\)
b/ Để P>3/4 => \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}>\frac{3}{4}\)
+/ TH1: x>1 => \(4\left(\sqrt{x}+3\right)>3\left(\sqrt{x}-1\right)\)
<=> \(\sqrt{x}>-16\) => x>1
+/ TH2: 0<x<1 => \(4\left(\sqrt{x}+3\right)< 3\left(\sqrt{x}-1\right)\) => \(\sqrt{x}< -16\)=> Loại
ĐS: x>1
c/ P=2 <=> \(P=\frac{\sqrt{x}+3}{\sqrt{x}-1}=2\)
<=> \(\sqrt{x}+3=2\left(\sqrt{x}-1\right)\)
<=> \(\sqrt{x}=5=>x=25\)
AT
17 tháng 8 2018
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
NV
15 tháng 7 2016
ĐKXĐ: \(x\ge0\) và \(x\ne9\)
a/ \(\frac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\frac{2\sqrt{x}-6}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x\sqrt{x}-3-\left(2\sqrt{x}-6\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x\sqrt{x}+8\sqrt{x}-3x-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}\left(x+8\right)-3\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{x+8}{\sqrt{x}+1}\)
b/ Thay \(x=14-6\sqrt{5}\) vào P ta được:
\(P=\frac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\frac{22-6\sqrt{5}}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}\)
18 tháng 9 2020
B1:
\(C=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{3^2-\left(\sqrt{5}\right)^2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{3-\sqrt{5}}.\sqrt{2}+\sqrt{3+\sqrt{5}}.\sqrt{2}\right)\)
\(=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
30 tháng 6 2021
a)\(\frac{\sqrt{a-2\sqrt{ab}+b}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{\sqrt{\sqrt{a}-\sqrt{b}}}=\sqrt{a}-\sqrt{b}\) (vì a > b > 0)
b) \(\frac{\sqrt{x-3}}{\sqrt{\sqrt{x}+\sqrt{3}}}:\frac{\sqrt{\sqrt{x}-\sqrt{3}}}{\sqrt{3}}=\frac{\sqrt{3}.\sqrt{x-3}}{\sqrt{\left(\sqrt{x}+\sqrt{3}\right)\left(\sqrt{x}-\sqrt{3}\right)}}=\frac{\sqrt{3\left(x-3\right)}}{\sqrt{x-3}}=\sqrt{3}\)
c) \(2y^2\sqrt{\frac{x^4}{4y^2}}=2y^2\cdot\frac{x^2}{-2y}=-x^2y\) (vì y < 0)
d) \(\frac{y}{x}\cdot\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}\cdot\frac{x}{y^2}=\frac{1}{y}\)(vì x > 0)
e) \(5xy\cdot\sqrt{\frac{25x^2}{y^6}}=5xy\cdot\frac{-5x}{y^3}=\frac{-25x^2}{y^2}\) (Vì x < 0, y > 0)