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Kiểm tra đề lại nhé b. Sao rút gọn rồi mà còn phức tạp thế
a/ Để P có nghĩa thì
\(\hept{\begin{cases}a>0\\\sqrt{a}-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
b/ \(P=\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}:\frac{\sqrt{a}+2}{\sqrt{a}-1}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\)
\(=\frac{1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\)
\(=\frac{2\sqrt{a}+a-a+2\sqrt{a}-1-a\sqrt{a}-4a-4\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
\(=-\frac{1+4a+a\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)
a, Ta có: \(A=\left(\frac{x+4}{3x+6}-\frac{1}{x^2+4x+4}\right).\left(1+\frac{x-1}{x+5}\right)\)
\(=\left(\frac{x+4}{3\left(x+2\right)}-\frac{1}{\left(x+2\right)^2}\right).\left(\frac{x+5}{x+5}+\frac{x-1}{x+5}\right)\)
\(=\left(\frac{\left(x+4\right)\left(x+2\right)}{3\left(x+2\right)^2}-\frac{1.3}{3\left(x+2\right)^2}\right).\frac{x+5+x-1}{x+5}\)
\(=\frac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}.\frac{2x+4}{x+5}\)
\(=\frac{x^2+2x+4x+8-3}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{x^2+6x+5}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{\left(x+1\right)\left(x+5\right)}{3\left(x+2\right)^2}.\frac{2\left(x+2\right)}{x+5}\)
\(=\frac{\left(x+1\right)\left(x+5\right).2\left(x+2\right)}{3\left(x+2\right)^2\left(x+5\right)}\) \(=\frac{2\left(x+1\right)}{3\left(x+2\right)}\)
b, Với \(x\ne-2,x\ne-5\) ta có:
\(A=\frac{2\left(x+1\right)}{3\left(x+2\right)}=\frac{2}{3}.\frac{x+1}{x+2}=\frac{2}{3}.\frac{\left(x+2\right)-1}{x+2}=\frac{2}{3}.1.\frac{-1}{x+2}=\frac{2}{3}.\frac{-1}{x+2}\)
Để A có giá trị là một số nguyên \(\Leftrightarrow\frac{2}{3}.\frac{-1}{x+2}\) có giá trị là một số nguyên \(\Leftrightarrow\frac{-1}{x+2}\in Z\) (vì \(\frac{2}{3}\in Z\))\(\Leftrightarrow-1⋮\left(x+2\right)\)
\(\Leftrightarrow x+2\inƯ\left(-1\right)=\left\{1;-1\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3\right\}\)
Đối chiếu ĐKXĐ \(\Rightarrow x\in\left\{-1;-3\right\}\)
Vậy để A có giá trị là một số nguyên thì \(x\in\left\{-1;-3\right\}\)
\(
1)Q = \left( {\dfrac{1}{{y - \sqrt y }} + \dfrac{1}{{\sqrt y - 1}}} \right):\left( {\dfrac{{\sqrt y + 1}}{{y - 2\sqrt y + 1}}} \right)\\
Q = \left( {\dfrac{1}{{\sqrt y \left( {\sqrt y - 1} \right)}} + \dfrac{1}{{\sqrt y - 1}}} \right).\dfrac{{y - 2\sqrt y + 1}}{{\sqrt y + 1}}\\
Q = \dfrac{{1 + \sqrt y }}{{\sqrt y \left( {\sqrt y - 1} \right)}}.\dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y + 1}}\\
Q = \dfrac{{\sqrt y - 1}}{{\sqrt y }}
\)
b) Thay \(y=3-2\sqrt{2}\) vào biểu thức ta được:
\(\dfrac{{\sqrt {3 - 2\sqrt 2 } - 1}}{{\sqrt {3 - 2\sqrt 2 } }} = \dfrac{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} - 1}}{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} }} = \dfrac{{ \sqrt 2 - 1-1}}{{\sqrt 2 -1}} \\= \dfrac{{\sqrt 2-2 }}{{ \sqrt 2 -1}} = \dfrac{{(\sqrt 2 -2)\left( { \sqrt 2+1 } \right)}}{{\left( { \sqrt 2-1 } \right)\left( {\sqrt 2+1 } \right)}} = - \sqrt 2 \)
\(2)B = \dfrac{{\sqrt y - 1}}{{{y^2} - y}}:\left( {\dfrac{1}{{\sqrt y }} - \dfrac{1}{{\sqrt y + 1}}} \right)\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {y - 1} \right)}}:\dfrac{{\sqrt y + 1 - \sqrt y }}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {\sqrt y - 1} \right)\left( {\sqrt y + 1} \right)}}:\dfrac{1}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{1}{{y\left( {\sqrt y + 1} \right)}}.\sqrt y \left( {\sqrt y + 1} \right)\\ B = \dfrac{{\sqrt y }}{y} \)
b) Thay \(y=3+2\sqrt{2}\) vào biểu thức ta được:
\(B = \dfrac{{\sqrt {3 + 2\sqrt 2 } }}{{3 + 2\sqrt 2 }} = \dfrac{{\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} }}{{3 + 2\sqrt 2 }} = \dfrac{{\left( {1 + \sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}} = 3 - 2\sqrt 2 + 3\sqrt 2 - 4 = - 1 + \sqrt 2 \)
Nhiều quá @@
a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)
\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)
b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
c: P=A/B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
\(P-2=\dfrac{2\sqrt{x}}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{-2}{\sqrt{x}+1}< 0\)
=>P<2
\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)
Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
\(1.\sqrt{x-2}+2=x\) ( x ≥ 2 )
⇔ \(\sqrt{x-2}=x-2\)
⇔ \(x-2=x^2-4x+4\)
⇔ \(x^2-5x+6=0\)
⇔ \(\left(x-3\right)\left(x-2\right)=0\)
⇔ \(x=3\left(TM\right)orx=2\left(TM\right)\)
\(2a.\sqrt{x+2\sqrt{x}+1}=3\)
⇔ \(x+2\sqrt{x}+1=9\)
⇔ \(x-2\sqrt{x}+4\sqrt{x}-8=0\)
⇔ \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)=0\)
⇔ \(x=4\)
\(b.\sqrt{x-2\sqrt{x}+1}=\sqrt{x-1}\) ( x ≥ 1 )
⇔ \(x-2\sqrt{x}+1=x-1\)
⇔ \(2\sqrt{x}=2\)
⇔ \(x=1\left(TM\right)\)
\(3.a.A=\dfrac{a+b-\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}-1}{a-b}=\dfrac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{\sqrt{a}-\sqrt{b}-1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}+1}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{1}{a-b}\)
( a # b ; a ; b ≥ 0 )
\(b.a-b=1\)
Thay vào A : \(A=\dfrac{1}{1}=1\)
Tương tự bài cuối nhé , dài ~
1. Có \(\sqrt{x-2}\) + 2 = x
<=> \(\sqrt{x-2}\) = x-2
ĐKXĐ: x - 2 ≥0 <=> x ≥ 2
<=> (\(\sqrt{x-2}\))2 = (x-2)2
<=> x- 2 = x2 - 4x + 4
<=> x2 - 5x + 6 = 0
<=> (x2 - 2.x.\(\dfrac{5}{2}\) + \(\dfrac{25}{4}\)) + 6 - \(\dfrac{25}{4}\) = 0
<=> (x - \(\dfrac{5}{2}\))2 - (\(\dfrac{1}{2}\))2 = 0
<=> (x - \(\dfrac{5}{2}\) - \(\dfrac{1}{2}\))(x - \(\dfrac{5}{2}\) + \(\dfrac{1}{2}\)) = 0
<=> (x - 3)(x - 2) = 0
Vậy x = 3 (TM) hoặc x = 2 (TM)
2a) \(\sqrt{x+2\sqrt{x}+1}\) = 3 ĐKXĐ: \(\sqrt{x}\)+1 ≥0 <=> \(\sqrt{x}\) ≥ - 1
<=> x + 2\(\sqrt{x}\) +1 = 9 (bình phương cả 2 vế)
<=> (\(\sqrt{x}\) +1)2 = 9
<=> \(\sqrt{x}\) +1 = 3
<=> \(\sqrt{x}\) = 2
<=> x = 4 (TM)
b) \(\sqrt{x+2\sqrt{x}+1}\) = \(\sqrt{x-1}\)
ĐKXĐ: x - 1 ≥ 0 <=> x ≥ 1
=> x + 2\(\sqrt{x}\) + 1 = x - 1 (bình phương cả 2 vế)
<=> 2\(\sqrt{x}\) +2 = 0
<=> 2(\(\sqrt{x}\) +1) = 0
<=> \(\sqrt{x}\) +1 = 0
<=> \(\sqrt{x}\) = -1 (KTM)
Vậy phương trình vô nghiệm