Rút gọn biểu thức chứa căn bậc hai - Hỏi đáp

b/Để P=-1
=>\(\frac{\sqrt{x}-1}{\sqrt{x}+1}=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\Leftrightarrow2\sqrt{x}=0\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
Vậy : để P =-1 thì x=0
c/Để P>2 thì \(\frac{\sqrt{x}-1}{\sqrt{x}+1}>2\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}-2>0\Leftrightarrow\frac{\sqrt{x}-1-2\sqrt{x}-2}{\sqrt{x}+1}>0\Leftrightarrow\frac{-\sqrt{x}-3}{\sqrt{x}+1}>0\)
Mà \(\sqrt{x}+1>0\)
\(\Leftrightarrow-\sqrt{x}-3>0 \Leftrightarrow-\sqrt{x}>3\Leftrightarrow\sqrt{x}< -3\)(vô lý)
Vậy không tồn tại x để P >2

b, Có P=-1

<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)=-1

<=> \(\sqrt{x}-1=-\sqrt{x}-1\) <=> \(2\sqrt{x}=0\) <=> x=0(tm đk của x)

Vậy để P=-1 <=>x=0

c, Để P>2

<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}>2\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-2>0\) <=> \(\frac{-\left(\sqrt{x}+3\right)}{\sqrt{x}+1}>0\) <=> \(\frac{\sqrt{x}+3}{\sqrt{x}+1}< 0\)(vô lý)

Vậy không có g/trị x thỏa mãn P>2

g, Có P=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}< 1\) (vì \(\frac{2}{\sqrt{x}+1}>0\))

Vậy P<1

h, Có \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

vì \(\sqrt{x}\ge0\) <=> \(\sqrt{x}+1\ge1\) <=> \(\frac{2}{\sqrt{x}+1}\le2\)

<=> \(1-\frac{2}{\sqrt{x}+1}\ge1-2=-1\) <=> \(P\ge-1\)

Dấu "=" xảy ra <=> x=0

Vậy minP=-1

i, Có \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

=\(\sqrt{3}+2+\left|2-\sqrt{3}\right|=\sqrt{3}+2+2-\sqrt{3}=4\)

=> x=4(tm đk của x)

Thay x=4 vào P đã rút gọn có:

P=\(\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)

Vậy P=\(\frac{1}{3}\) tại x=4

k, Có P<\(\frac{1}{2}\)

<=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}< \frac{1}{2}\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{1}{2}< 0\) <=> \(\frac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)

=> \(\left\{{}\begin{matrix}\sqrt{x}-3>0\\2\left(\sqrt{x}+1\right)>0\end{matrix}\right.\) (vì mẫu luôn >0) <=> \(\left\{{}\begin{matrix}\sqrt{x}>3\\\sqrt{x}+1>0\end{matrix}\right.\)_{(luôn đúng)} <=> \(x>9\)

Vậy để P<\(\frac{1}{2}\) thì x>9

a) ĐKXĐ: \(x\ge0,x\ne4\)

\(b)P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\\ P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {2 - \sqrt x } \right)}}\\ P = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - \left( {2 + 5\sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\ P = \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\ P = \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\ P = \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\ P = \dfrac{{3\sqrt x }}{{\sqrt x + 2}} \)

\(c)P=2\Rightarrow\)\(\dfrac{{3\sqrt x }}{{\sqrt x + 2}}=2 \)

\(\Leftrightarrow3\sqrt{x}=2\left(\sqrt{x}+2\right)\\ \Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\\ \Leftrightarrow3\sqrt{x}-2\sqrt{x}=4\\ \Leftrightarrow\sqrt{x}=4\\ \Leftrightarrow x=16\)

a)

\(N=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)

b)

\(N=x-\sqrt{x}+1=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy Min N = \(\frac{3}{4}\)khi x=\(\frac{1}{4}\)

Câu c) mk ko bt, sorry nha :<

\(M=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{2\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}>0\) \(\forall x>0\)

\(M-2=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{-2\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\frac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}< 0\) \(\forall x\ne1;x>0\)

\(\Rightarrow0< M< 2\Rightarrow\) để M nguyên thì \(M=1\)

\(\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow x-3\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\) \(\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)

a, *) ĐXKĐ: \(x>0\).

\(P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\\ =\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-1+1\\ =x+\sqrt{x}-2\sqrt{x}\\ =x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\)

*) Để P=2 thì:

\(x-\sqrt{x}=2\\ \Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow x-2\sqrt{x}+\sqrt{x}-2=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\\ \left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vn\right)\\\sqrt{x}=2\end{matrix}\right.\\ \Leftrightarrow x=4\left(t/m\right)\)

b, Với \(x>1\) thì \(\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=P>0\)

Suy ra \(\left|P\right|=P\), hay \(P-\left|P\right|=0\).

Chúc bạn học tốt nha.

\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{17-6\sqrt{8}}-\sqrt{24-8\sqrt{8}}\)

\(\sqrt{\left(3-\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}=\left|3-\sqrt{8}\right|-\left|4-\sqrt{8}\right|\)

\(=3-\sqrt{8}-4+\sqrt{8}=-1\)

\(A=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(A=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(A=\sqrt{xy}+\sqrt{xy}\)

\(A=2\sqrt{xy}\)

a/ĐK: \(a\ge0;a\ne1\)
Ta có: P\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\times\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}=\frac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)

= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]:\frac{x-6\sqrt{x}+9}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

= \(\left[\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\left(\sqrt{x}-3\right)^2}{\left(2-\sqrt{x}\right)\left(\sqrt{x}-3\right)}\)

= \(\left[\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:\frac{\sqrt{x}-3}{2-\sqrt{x}}\)

= \(\left[\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{nt}\right]:nt\)

\(=\left[\frac{8\sqrt{x}+4x}{nt}\right]:nt\)

\(=\left[\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]:nt\)

\(=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-3}\)

\(\dfrac{3\sqrt{x}+11}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+5}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{5}{\sqrt{x}+2}=3+\dfrac{5}{\sqrt{x}+2}\)

Để bt nguyên thì:

\(\dfrac{5}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(5\right)\)

\(\Leftrightarrow\sqrt{x}+2=\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow\sqrt{x}=\left\{3\right\}\Leftrightarrow x=9\)

vậy........

\(dk:x\ge\frac{-10}{3}\)

\(x^2+9x+20=2\sqrt{3x+10}\Leftrightarrow x^2+6x+10+\left(3x+10\right)-2\sqrt{3x+10}=0\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\Rightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}=1\end{matrix}\right.\Leftrightarrow x=-3\left(tmdk\right)\)

Điều kiện 3x + 10 ≥ 0 =>x ≥ -10 /3
Pt <=> (3x + 10)² + 7(3x + 10) + 10 = 18\(\sqrt{\left(3x+10\right)}\)
Đặt y = \(\sqrt{\left(3x+10\right)}\) ≥ 0 pt trở thành
y⁴ + 7y² - 18y + 10 = 0
<=> (y - 1)²(y² + 2y + 10) = 0
<=> (y-1)^2 [(y+1)^2 +9] =0
mà (y+1)^2 +9 > 0 =>y=1 => x= -3

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\Rightarrow\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{2+\sqrt{3}}{2+\sqrt{\left(1+\sqrt{3}\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{9-3}\)

\(=\frac{6}{6}=1\)

\(\Rightarrow A=\sqrt{2}\)