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a: \(A=\dfrac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: \(A-5=\dfrac{2x-4\sqrt{x}+2}{\sqrt{x}}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>=0\)
=>A>=5
a: \(B=\dfrac{1}{\sqrt{x}+1}\)
\(B-1=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}>=0\)
=>B>=1
b: \(P=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(P\cdot\sqrt{x}+2x-\sqrt{x}=3x-2\sqrt{x-4}+3\)
=>\(x+\sqrt{x}+1+2x-\sqrt{x}=3x+3-2\sqrt{x-4}\)
=>\(-2\sqrt{x-4}+3=1\)
=>x-4=1
=>x=5
Phần a,b,c bạn có thể tham khảo bài bên dưới.
Phần d.
ĐKXĐ: $x\geq 0; x\neq 4$
$A>5\Leftrightarrow \frac{x+9}{2\sqrt{x}}>5$ ($x> 0$)
$\Leftrightarrow x+9> 10\sqrt{x}$
$\Leftrightarrow x-10\sqrt{x}+9>0$
$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-9)>0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}-1>0\\ \sqrt{x}-9>0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}-1<0\\ \sqrt{x}-9<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>1\\ x>81\end{matrix}\right.\\ \left\{\begin{matrix} 0\leq x< 1\\ 0\leq x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x>81\\ 0\leq x< 1\end{matrix}\right.\)
Kết hợp với đkxđ suy ra $x>81$ hoặc $0< x< 1$
a
Với: x \(\ge0,x\) \(\ne4\) có:
\(A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right):\left(\dfrac{\left(\sqrt{x}+2\right)^2}{x-4}-\dfrac{\left(\sqrt{x}-2\right)^2}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4}{x-4}-\dfrac{x-4\sqrt{x}+4}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{2\sqrt{x}}{x-4}\right)\)
\(=\dfrac{\left(x+9\right)\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}=\dfrac{x+9}{2\sqrt{x}}\)
b
Giải \(x^2-5x+4=0\)
Nhẩm nghiệm: a + b + c = 0 (1 - 5 + 4 = 0)
\(\Rightarrow x_1=1;x_2=\dfrac{c}{a}=\dfrac{4}{1}=4\)
Thay x = 1 vào A:
\(A=\dfrac{1+9}{2\sqrt{1}}=\dfrac{10}{2}=5\)
Thay x = 4 vào A:
\(A=\dfrac{4+9}{2.\sqrt{4}}=\dfrac{13}{2.2}=\dfrac{13}{4}\)
c
ĐK: x > 0
\(A=0\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}=0\)
=> \(x+9=0\Rightarrow x=-9\) (không thỏa mãn)
Vậy không xác định được giá trị x
d
ĐK: x > 0
\(A>5\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}>5\)
\(\Leftrightarrow x+9>5.2\sqrt{x}\Leftrightarrow x+9>10\sqrt{x}\)
\(\Leftrightarrow\left(x+9\right)^2>\left(10\sqrt{x}\right)^2=100x\)
<=> \(x^2+18x+81-100x>0\)
<=> \(x^2-82x+81>0\)
<=> \(x^2-81x-x+81>0\)
<=> \(x\left(x-81\right)-\left(x-81\right)>0\)
<=> \(\left(x-1\right)\left(x-81\right)>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1>0\\x-81>0\end{matrix}\right.\\\left[{}\begin{matrix}x-1< 0\\x-81< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x>81\end{matrix}\right.\\\left[{}\begin{matrix}x< 1\\x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>81\\x< 81\end{matrix}\right.\)
Vậy để A > 5 thì x > 81 và 0 < x < 81
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
b) Xét hiệu:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)
\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)
\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)
Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\)
\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)
Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)
(giúp cậu nó nha)
\(a,A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dk:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(b,x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Thay \(x=4\) vào A :
\(A=\dfrac{\sqrt{4}-1}{\sqrt{4}+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)
câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm
1) So sánh:
N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)
M = \(\sqrt{18}-\sqrt{8}\)
\(=3\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
Ta có: \(1=\sqrt{1}\)
Mà 1 < 2
\(\Rightarrow\sqrt{1}< \sqrt{2}\)
Hay 1 \(< \sqrt{2}\)
Vậy N < M
a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)
c: 2x-3căn x-5=0
=>2x-5căn x+2căn x-5=0
=>2căn x-5=0
=>x=25/4
Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)
Ta có:
\(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) (ĐK: \(x\ne4;x\ge0\))
\(B=\dfrac{x}{\left(\sqrt{x}\right)^2-2^2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(\Rightarrow P=\dfrac{A}{B}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}}{\sqrt{x}-2}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\sqrt{x}}=\dfrac{x-4}{x}\) (ĐK: \(x\ne0\))
Theo đề ta có:
\(P\cdot x\le10\sqrt{x}-29-\sqrt{x}+25\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\dfrac{x-4}{x}\cdot x\le9\sqrt{x}-4\)
\(\Leftrightarrow x-4\le9\sqrt{x}-4\)
\(\Leftrightarrow x-9\sqrt{x}\le0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-9\right)\le0\)
Mà: \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}-9\le0\)
\(\Leftrightarrow\sqrt{x}\le9\)
\(\Leftrightarrow x\le81\)
Kết hợp với đk:
\(0\le x\le81\)
tồn tại \(\sqrt{A}\Rightarrow x>4\)
\(B=x^2+14x-5x\sqrt{x}-153\sqrt{x}+452\)
\(B=\left(\sqrt{x}-4\right)\left(x\sqrt{x}-x+10\sqrt{x}-113\right)\)
khi
\(0\le x< 16\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-4< 0\\x\sqrt{x}-x+10\sqrt{x}-113< 64+40-113-x=-9-x< 0\end{matrix}\right.\) B>0
hay B không có nghiệm khi x<16
Kết luận x>16 \(\Rightarrow\sqrt{x}-2>1\Rightarrow\dfrac{1}{\sqrt{x}-2}< 1\Rightarrow A< 1\Rightarrow A^4< A< \sqrt{A}\)