b) \(x^2+8x-3\) = \(2\sqrt{x\left(8+x\right)}\)
c)\(\sqrt{\dfrac{x^2+x+1}{x}}+\sqrt{\dfrac{x}{x^2+x+1}}=\dfrac{7}{4}\)
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a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
a) ĐK: \(x^2+7x+7\ge0\)
Đặt \(a=\sqrt{x^2+7x+7}\) \(\left(a\ge0\right)\)
PT \(\Rightarrow3a^2-3+2a=2\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x^2+7x+7=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) (Thỏa mãn)
Vậy ...
b) ĐK: \(x^2-6x+6\ge0\)
Đặt \(a=\sqrt{x^2-6x+6}\) \(\left(a\ge0\right)\)
PT \(\Rightarrow a^2+3=4a\) \(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\) (Thỏa mãn)
+) Với \(a=3\) \(\Rightarrow x^2-6x+6=9\) \(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{3}\\x=3-2\sqrt{3}\end{matrix}\right.\) (Thỏa mãn)
+) Với \(a=1\) \(\Rightarrow x^2-6x+6=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) (Thỏa mãn)
Vậy ...
c)C1: Áp dụng bđt AM-GM \(\Rightarrow VT\ge2>\dfrac{7}{4}\)
=> Dấu = ko xảy ra hay pt vô nghiệm
C2: Đk:\(x>0\)
Đặt \(a=\sqrt{\dfrac{x^2+x+1}{x}}\left(a>0\right)\) \(\Rightarrow\dfrac{1}{a}=\sqrt{\dfrac{x}{x^2+x+1}}\)
Pttt: \(a+\dfrac{1}{a}=\dfrac{7}{4}\Leftrightarrow4a^2-7a+4=0\)
\(\Delta =-15<0 \) => Pt vô nghiệm
Vậy...
d) Đk: \(x\le-8;x\ge0\)
Đặt \(t=\sqrt{x\left(8+x\right)}\left(t\ge0\right)\)
Pttt: \(t^2-3=2t\Leftrightarrow t^2-2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x\left(8+x\right)}=3\Leftrightarrow x^2+8x-9=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\) (tm)
Vậy...
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\dfrac{-4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)
=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)
=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)
=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)
Để BPT luôn đúng thì m<-0,3
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
Giúp em với ạ! Em cảm ơn
\(b,\)Đặt \(\sqrt{x^2+8x}=a\left(a\ge0\right)\)
Khi đó phương trình trở thành:
\(a^2-3=2a\\ \Leftrightarrow a^2-2a-3=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-1\\a=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+8x}=-1\\\sqrt{x^2+8x}=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+8x=1\\x^2+8x=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+8x-1=0\\x^2+8x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4+\sqrt{17}\\x=-4-\sqrt{17}\\x=1\\x=-9\end{matrix}\right.\)