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chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)
\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)
\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)
\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)
\(i.\sqrt{x+2+2\sqrt{x+1}}=\sqrt{x+1+2\sqrt{x+1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=\sqrt{x+1}+1\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
a)\(\frac{3+\sqrt{3}}{1+\sqrt{3}}\)=\(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{1+\sqrt{3}}\)=\(\sqrt{3}\)
b)\(\frac{2\sqrt{3}-6}{\sqrt{8}-\sqrt{2}}\)
\(\frac{y-2\sqrt{y}}{\sqrt{y}-2}\)=\(\frac{\sqrt{y}\left(\sqrt{y}-2\right)}{\sqrt{y}-2}\)=\(\sqrt{y}\)
d) \(\frac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}{\sqrt{x}-1}\)=\(\sqrt{x}\)+3
e)\(\frac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)=\(\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)=\(\sqrt{y}\)-1
g)\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x+3}}\)
chúc bạn học tốt
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
a: \(=\sqrt{10}\left(\sqrt{5}-1\right)+\sqrt{15}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}\left(\sqrt{5}-1\right)\left(\sqrt{2}+\sqrt{3}\right)\)
b: \(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-1\right)\)
c: \(=\sqrt{\left(x-y\right)\left(x+y\right)}+\sqrt{\left(x+y\right)^2}\)
\(=\sqrt{x+y}\cdot\left(\sqrt{x-y}+\sqrt{x+y}\right)\)
d: \(=\left(\sqrt{x-2}+1\right)^2\)
g: \(=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)