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Bài 1:
Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)
Số giá trị nguyên thỏa mãn điều kiện là:
\(\left(2+4\right)+1=7\)
\(a,\dfrac{3}{\sqrt{12x-1}}\) xác định \(\Leftrightarrow12x-1>0\Leftrightarrow12x>1\Leftrightarrow x>\dfrac{1}{12}\)
\(b,\sqrt{\left(3x+2\right)\left(x-1\right)}\) xác định \(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}3x+2\ge0\\x-1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}3x+2\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\le1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-\dfrac{2}{3}\\x\ge1\end{matrix}\right.\)
\(c,\sqrt{3x-2}.\sqrt{x-1}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3x-2\ge0\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{2}{3}\\x\ge1\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)
\(d,\sqrt{\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}}\) xác định \(\Leftrightarrow-x+5>0\Leftrightarrow x< 5\)
Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)
c) Đk: \(-3\le x< 5\)
d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:
\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)
\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.
Câu 1:
a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)
\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)
+) Với \(x\ge-2\):
\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)
+) Với \(x< -2\):
\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)
b) \(B=\sqrt{m^2-6m+9-2m}\)
\(B=\sqrt{m^2-8m+9}\)
Bạn xem lại đề nhé :)
c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(C=1+\sqrt{x-1}\)
d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Xét \(x\ge8\):
\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Xét \(4< x< 8\):
\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Vậy....
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
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
b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)
mà 80>75
nên \(4\sqrt{5}>5\sqrt{3}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
1. a)\(2\&\sqrt{5}\)
\(2=\sqrt{4}\)
=> \(2< \sqrt{5}\)
b)\(5\&\sqrt{23}\)
\(5=\sqrt{25}\)
=> \(5>\sqrt{23}\)
c) \(\sqrt{23}+\sqrt{13}\&\sqrt{83}\)
\(\left(\sqrt{23}+\sqrt{13}\right)^2=36+2\sqrt{229}\)
\(\left(\sqrt{83}\right)^2=83\)
\(\Rightarrow36+2\sqrt{299}< 83\)
=> \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2. a) \(\sqrt{x}=5;x\ge0\)
=> x = 25
b) \(3\sqrt{x}=6;x\ge0\)
=> x = 4
c) trùng
d) \(3-\sqrt{3+1}=1\)
\(3-\sqrt{3+1}=3-2=1\)
1)
a)\(2=\sqrt{4}< \sqrt{5}\)
b) \(5=\sqrt{25}>\sqrt{23}\)
c) \(\sqrt{83}>\sqrt{81}=9\)
\(\left\{{}\begin{matrix}\sqrt{23}< \sqrt{25}=5\\\sqrt{13}< \sqrt{16}=4\end{matrix}\right.\)
\(\sqrt{23}+\sqrt{13}< 4+5=9\)
Vậy \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2) Ta có:
\(\sqrt{x}=5\Rightarrow x=25\)
\(3\sqrt{x}=6\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(3-\sqrt{3+1}=1\)
Nên:
\(3-2=1\)(luôn đúng)