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Câu 2:
a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)
\(=3x-\left|x-5\right|\)
\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)
b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)
a: ĐKXĐ: \(x>0\)
b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
Bài 1
Mình làm mẫu một số câu thôi nhé
\(9,\sqrt{5}=\left(\sqrt{5}\right)^2=5\\ \sqrt{6}=\left(\sqrt{6}\right)^2=6\)
Vì \(5< 6\)
\(\Rightarrow\sqrt{5}< \sqrt{6}\)
\(10,2\sqrt{5}=\left(2\sqrt{5}\right)^2=20\\ \sqrt{7}=\left(\sqrt{7}\right)^2=7\)
Vì \(20>7\)
\(\Rightarrow2\sqrt{5}>\sqrt{7}\)
\(11,5\sqrt{2}=\left(5\sqrt{2}\right)^2=50\\ 2\sqrt{3}=\left(2\sqrt{3}\right)^2=12\)
Vì \(50>12\Rightarrow5\sqrt{2}>2\sqrt{3}\)
\(12,2\sqrt{6}=\left(2\sqrt{6}\right)^2=24\\ 5=5^2=25\)
Vì \(25>24\Rightarrow5>2\sqrt{6}\)
\(13,\sqrt{7}=\left(\sqrt{7}\right)^2=7\\ 2=2^2=4\)
Vì \(7>4\Rightarrow\sqrt{7}>2\)
\(14,3=3^2=9\\ \sqrt{5}=\left(\sqrt{5}\right)^2=5\)
Vì \(9>5\Rightarrow3>\sqrt{5}\)
\(15,3\sqrt{6}=\left(3\sqrt{6}\right)^2=54\)
Vì \(54>1\Rightarrow3\sqrt{6}>1\)
\(16,2\sqrt{2}=\left(2\sqrt{2}\right)^2=8\\ 3=3^2=9\)
Vì \(8< 9\Rightarrow2\sqrt{2}< 3\)
Phương pháp làm dạng bài này là bình phương hai vế rồi so sánh
Bài 2
Gợi ý : Biểu thức dưới dấu căn \(\ge\) 0
Lưu ý : Nếu biểu thức dưới dấu căn ở dưới mẫu thì \(>0\)
\(21,ĐK:4x^2-12x+9>0\\ \Rightarrow\left(2x-3\right)^2>0\\ \Leftrightarrow x\ne\dfrac{3}{2}\)
\(22,ĐK:x^2-8x+15\ge0\\ \Rightarrow\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\)
\(23,ĐK:\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)
\(24,ĐK:\left\{{}\begin{matrix}\dfrac{2+x}{5-x}\ge0\\5-x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2+x\ge0\\5-x\ge0\\x\ne5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ge-2\\x\le5\\x\ne5\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ge-2\\x< 5\end{matrix}\right.\left(t/m\right)\)
Hoặc
\(\left\{{}\begin{matrix}2+x\le0\\5-x\le0\\5-x\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\le-2\\x\ge5\\x\ne5\end{matrix}\right.\left(loại\right)\)
a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}-1}{3-\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{2\sqrt{a}-9-a+9+2a-5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-3\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}-1}{\sqrt{a}-3}\)
b: A là số nguyên
=>\(\sqrt{a}-3+2⋮\sqrt{a}-3\)
=>\(\sqrt{a}-3\in\left\{1;-1;2;-2\right\}\)
=>a thuộc {16;25;1}
\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)
\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)
a) A có nghĩa khi:
\(\left(x+1\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\x-3\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\ge3\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\le3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-1\end{matrix}\right.\)
b) Ta có:
\(B=\sqrt{x+1}\cdot\sqrt{x-3}=\sqrt{\left(x+1\right)\left(x-3\right)}\)
Nên: A=B nên tập nghiệm xác định như nhau
c) \(A=B\) khi:
\(\sqrt{\left(x+1\right)\left(x-3\right)}=\sqrt{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow1=1\) (luôn đúng)
\(\Rightarrow x\in R\)
a: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1-2⋮\sqrt{x}+1\)
=>căn x+1 thuộc {1;2}
=>căn x thuộc {0;1}
mà x<>1
nên x=0
a: ĐKXĐ: \(\left[{}\begin{matrix}x\le-\dfrac{3}{2}\\x>3\end{matrix}\right.\)
b: ĐKXĐ: x>3
c: Ta có: A=B
\(\Leftrightarrow\sqrt{\dfrac{2x+3}{x-3}}=\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
\(\Leftrightarrow0x=0\)(luôn đúng với mọi x>3)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{x-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}-1}\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3x-5}{x-1}\ge0\\x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{3}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{3}\\x< 1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x< 1\end{matrix}\right.\)
Vậy ...
b, Ta có : \(A=\sqrt{\dfrac{3x-5}{x-1}}=3\)
\(\Leftrightarrow3x-5=9x-9\)
\(\Leftrightarrow x=\dfrac{2}{3}\left(TM\right)\)
Vậy ...