Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
\(P=B:A\)
\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(P=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{3}\Leftrightarrow3\sqrt{x}-6=\sqrt{x}+3\)
\(\Leftrightarrow2\sqrt{x}=9\Leftrightarrow\sqrt{x}=4,5\Leftrightarrow x=\dfrac{81}{4}\)
b. \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)
Ta có: \(-\dfrac{5}{\sqrt{x}+3}\ge-\dfrac{5}{\sqrt{0}+3}=-\dfrac{5}{3}\)
\(\Rightarrow1-\dfrac{5}{\sqrt{x}+3}\ge1-\dfrac{5}{3}=-\dfrac{2}{3}\)
Suy ra: \(P\ge-\dfrac{2}{3}\) khi \(x=0\)
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.5\sqrt{7}+7}+\sqrt{25-2.5\sqrt{7}+7}=5+\sqrt{7}+5-\sqrt{7}=10\)
\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{25+2.5.3\sqrt{2}+18}=5+3\sqrt{2}\) \(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{3-\sqrt{x}}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=\dfrac{1}{3+\sqrt{x}}\)
\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(f.\dfrac{x\sqrt{x}+64}{\sqrt{x}+4}=\dfrac{\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)}{\sqrt{x}+4}=x-4\sqrt{x}+16\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
Còn 2 con cuối làm tương tự nhé ( đăng dài quá ).
\(a.\sqrt{32+10\sqrt{7}}+\sqrt{32-10\sqrt{7}}=\sqrt{25+2.\sqrt{25}.\sqrt{7}+7}+\sqrt{25-2.\sqrt{25}.\sqrt{7}+7}=\sqrt{\left(5+\sqrt{7}\right)^2}+\sqrt{\left(5-\sqrt{7}\right)^2}=5+\sqrt{7}+5-\sqrt{7}=10\)\(b.\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.\sqrt{8}.1}+1}}=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}=\sqrt{13+30\sqrt{3+2\sqrt{2}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{\sqrt{25}+2.\sqrt{25}.\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+\sqrt{18}\)
\(c.\dfrac{3-\sqrt{x}}{9-x}=\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{9-x}{9-x}.\dfrac{1}{3+\sqrt{x}}=\dfrac{1}{3+\sqrt{x}}=\dfrac{3-\sqrt{x}}{9-x}\)\(d.\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)}=\sqrt{x}-2\)\(e.\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-1}=\sqrt{x}-2\)
\(g.\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(x\sqrt{x}-y\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{x^2+x\sqrt{xy}-y\sqrt{xy}-y^2}{x-y}=\dfrac{\sqrt{xy}\left(x-y\right)+\left(x-y\right)\left(x+y\right)}{x-y}=\dfrac{\left(x-y\right)\left(\sqrt{xy}+x+y\right)}{x-y}=x+y+\sqrt{xy}\)\(h.6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(x-3\right)^2}=6-2x-\left|x-3\right|=6-2x-3+x=3-x\)
\(i.\sqrt{x+2+2\sqrt{x+1}}=\sqrt{x+1+2\sqrt{x+1}+1}=\sqrt{\left(\sqrt{x+1}+1\right)^2}=\sqrt{x+1}+1\)
1) \(M=\dfrac{10}{\sqrt{x}+2};M_{\left(16\right)}=\dfrac{10}{\sqrt{16}+2}=\dfrac{10}{6}=\dfrac{5}{3}\)
2)\(N=\dfrac{2\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-18}{x-4}=2+\dfrac{4}{\sqrt{x}-2}+\dfrac{\sqrt{x}-18}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2+\dfrac{4\sqrt{x}+8+\sqrt{x}-18}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(N=2+\dfrac{5}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+9}{\sqrt{x}+2}\)
N khác 0 mọi x thuộc đk
\(\dfrac{M}{N}=M.\dfrac{1}{N}=\dfrac{10}{\sqrt{x}+2}.\dfrac{\sqrt{x}+2}{\left(2\sqrt{x}+9\right)}=\dfrac{10}{2\sqrt{x}+9}\)
\(\dfrac{M}{N}=\dfrac{12-\sqrt{x}}{13}=\dfrac{10}{2\sqrt{x}+9}\)
\(\Leftrightarrow\left(12-\sqrt{x}\right)\left(2\sqrt{x}+9\right)=130\)
\(15\sqrt{x}+12.9-2x=130\)
\(2x-15\sqrt{x}+22=0\)
\(\Delta_{\sqrt{x}}=15^2-4.2.22=137\)
\(\sqrt{x}=\dfrac{15+-\sqrt{137}}{4}\)
\(\left[{}\begin{matrix}x_1=\dfrac{181-15.\sqrt{137}}{8}\\x_2=\dfrac{181+15.\sqrt{137}}{8}\end{matrix}\right.\) tự kiểm tra số liểu (nhẩm tính có thể nhầm; thấy lẻ quá)
a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)
b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)
c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)
d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)
\(A=\dfrac{x+5}{\sqrt{x}-2}=\dfrac{10\sqrt{x}-20+\left(x-10\sqrt{x}+25\right)}{\sqrt{x}-2}\)
\(=10+\dfrac{\left(\sqrt{x}-5\right)^2}{\sqrt{x}-2}\ge10\)
Dấu "=" xảy ra <=> \(\sqrt{x}-5=0\Leftrightarrow x=25\left(tm\right)\)