Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\sqrt{2a}.\sqrt{18a}=\sqrt{2a}.3\sqrt{2a}=3.2a=6a\)
\(b.\sqrt{3a.27ab^2}=\sqrt{9a^2b^2.9}=9\text{ |}ab\text{ |}\)
\(c.2y^2.\sqrt{\dfrac{x^4}{4y^2}}=2y^2.\dfrac{x^2}{-2y}=-x^2y\)
\(d.\dfrac{y}{x}.\sqrt{\dfrac{x^2}{y^4}}=\dfrac{y}{x}.\dfrac{x}{y^2}=\dfrac{1}{y}\)
\(e.\sqrt{\dfrac{9a^2}{16}}=\dfrac{3\text{ |}a\text{ |}}{4}\)
\(f.\sqrt{10.16a^2}=-4a\sqrt{10}\)
\(g.\sqrt{a^4\left(3-a\right)^2}=a^2\left(a-3\right)\)
\(h.\sqrt{\dfrac{2a^2b^4}{98}}\sqrt{\dfrac{a^2b^4}{49}}=\dfrac{b^2\text{ |}a\text{ |}}{7}\)
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\) với \(x>0;x\ne1\)
\(\Rightarrow A=\dfrac{x}{\sqrt{x-1}}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
= \(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
= \(\sqrt{x}-1\)
b) Với \(x>0;x\ne1\)
A=\(\sqrt{x}-1\)
Ta có : \(x=3+2\sqrt{2}\) ( Thỏa mãn ĐKXĐ )
Thay \(x=3+2\sqrt{2}\) vào biểu thức A ta có :
A=\(\sqrt{3+2\sqrt{2}}-1\)= \(\sqrt{2}+1-1\)=\(\sqrt{2}\)
\(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
a ) Rút gọn :
\(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
\(\Rightarrow A=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
\(\Rightarrow A=\sqrt{x}-1\)
b ) \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)
Thay x vào A, ta có :
\(\sqrt{\left(\sqrt{2}+1\right)^2}-1=\sqrt{2}+1-1=\sqrt{2}\)
Vậy ...............
a: \(\left|x_1-x_2\right|=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\left(\dfrac{1}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{1}{4}+4}\)
\(=\sqrt{\dfrac{17}{4}}\)
=>\(\left[{}\begin{matrix}x_1-x_2=\dfrac{\sqrt{17}}{2}\\x_1-x_2=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\)
c,d:Vì pt có hai nghiệm trái dấu
nên chắc chắn hai biểu thức này sẽ không tính được vì sẽ có một căn bậc hai mà biểu thức trong căn âm
a: \(P=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}}{x^2-1}+\dfrac{4\sqrt{x}}{x-1}\right)-\dfrac{5x}{x^2-1}\)
\(=\sqrt{x}\left(\dfrac{\sqrt{x}+4\sqrt{x}\left(x+1\right)}{\left(x^2-1\right)}\right)-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x\left(x+1\right)}{x^2-1}-\dfrac{5x}{x^2-1}\)
\(=\dfrac{x+4x^2+4x-5x}{x^2-1}\)
\(=\dfrac{4x^2}{x^2-1}\)
Khi x=4 thì \(P=\dfrac{4\cdot16}{16-1}=\dfrac{64}{15}\)
b: Để P/Q=0 thì P=0
=>x=0
Pt hoành độ giao điểm:
\(\frac{1}{2}x^2=-x+m\Leftrightarrow x^2+2x-2m=0\)
\(\Delta'=1+2m>0\Rightarrow m>-\frac{1}{2}\)
Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=-2m\end{matrix}\right.\)
\(x_1x_2+y_1y_2=5\)
\(\Leftrightarrow x_1x_2+\frac{1}{4}x_1^2x_2^2=5\)
\(\Leftrightarrow\left(x_1x_2\right)^2+4x_1x_2-20=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1x_2=-2+2\sqrt{6}\\x_1x_2=-2-2\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2m=-2+2\sqrt{6}\\-2m=-2-2\sqrt{6}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\sqrt{6}-1\\m=\sqrt{6}+1\end{matrix}\right.\)
ta thấy pt luôn có no . Theo hệ thức Vi - ét ta có:
x1 + x2 = \(\dfrac{-b}{a}\) = 6
x1x2 = \(\dfrac{c}{a}\) = 1
a) Đặt A = x1\(\sqrt{x_1}\) + x2\(\sqrt{x_2}\) = \(\sqrt{x_1x_2}\)( \(\sqrt{x_1}\) + \(\sqrt{x_2}\) )
=> A2 = x1x2(x1 + 2\(\sqrt{x_1x_2}\) + x2)
=> A2 = 1(6 + 2) = 8
=> A = 2\(\sqrt{3}\)
b) bạn sai đề
Lời giải:
a. Tại $x_0=\sqrt{5}$ thì:
$y=f(x_0)=\frac{x_0}{2}-\sqrt{x_0^2-1}+2$
$=\frac{\sqrt{5}}{2}-\sqrt{5-1}+2=\frac{\sqrt{5}}{2}$
b. Tại $x=\frac{1}{4}$ thì $x^2-1=\frac{-15}{16}< 0$ nên căn thức $\sqrt{x^2-1}$ không xác định. Do đó không tính được.