Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)
b) Ta có: \(x^2+2x+1\)
\(=\left(x+1\right)^2\)
mà \(\left(x+1\right)^2\ge0\forall x\)
nên \(x^2+2x+1\ge0\forall x\)
Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x
c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)
\(\Leftrightarrow x\left(x-4\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)
Bài 3:
a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)
\(=\left|3-\sqrt{10}\right|\)
\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))
b) Ta có: \(\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|\sqrt{5}-2\right|\)
\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))
c) Ta có: \(3x-\sqrt{x^2-2x+1}\)
\(=3x-\sqrt{\left(x-1\right)^2}\)
\(=3x-\left|x-1\right|\)
\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)
Bài 1.
1. \(\sqrt{-3x+6}\) có nghĩa khi \(-3x+6\ge0\Leftrightarrow-3x\ge-6\Rightarrow x\le2\)
2.
\( a){\left( {\sqrt 7 - \sqrt 5 } \right)^2} + 2\sqrt {35} = 7 - 2\sqrt {35} + 5 + 2\sqrt {35} = 12\\ b)3\sqrt 8 - \sqrt {50} - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = 6\sqrt 2 - 5\sqrt 2 - \sqrt 2 + 1 = 1 \)
Bài 2.
\( M = \dfrac{{\sqrt a + 3}}{{\sqrt a - 2}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 2}} + \dfrac{{4\sqrt a - 4}}{{4 - a}}\\ M = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a + 3} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right) - \left( {4\sqrt a - 4} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{{4\sqrt a + 8}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{{4\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}\\ M = \dfrac{4}{{\sqrt a - 2}} \)
Bài 3.
1.
\( a)\sqrt {{{313}^2} - {{312}^2}} + \sqrt {{{17}^2} - {8^2}} = \sqrt {\left( {313 - 312} \right)\left( {313 + 312} \right)} + \sqrt {\left( {17 - 8} \right)\left( {17 + 8} \right)} \\ = \sqrt {625} + \sqrt {9.25} = 25 + 3.5 = 25 + 15 = 40\\ b)\dfrac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }} = \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{1 + \sqrt 2 }} = \sqrt 2 \)
2. \(\left\{{}\begin{matrix}2x+y=3\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y=6\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=7\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(1;1\right)\)
3.
\(
\sqrt {9\left( {x - 1} \right)} = 21\\
\Leftrightarrow 3\sqrt {x - 1} = 21\\
\Leftrightarrow \sqrt {x - 1} = 7\\
\Leftrightarrow x - 1 = 49\\
\Leftrightarrow x = 50
\)
Thử lại $x=50$ là nghiệm
Bài 1
1)
Đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi đó A=\(\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
2) Đề là \(5-2\sqrt{6}\)sẽ hợp lý hơn nha bn
Đkxđ\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-\sqrt{2}\ne0\end{matrix}\right.\)
Ta có \(5-2\sqrt{6}=\left(1-\sqrt{6}\right)^2\)
Khi đó
B= \(\frac{1-\sqrt{6}}{1-\sqrt{6}-\sqrt{2}}\)
1)
đk: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Rgọn
A=\(\frac{x+12}{x-4}+\frac{1}{\sqrt{x}+2}-\frac{4}{\sqrt{x}-2}\)
= \(\frac{x+12+\sqrt{x}-2-\left(4\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
2)
B=\(\frac{3\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{10\sqrt{x}}{x-4}\) đk \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
= \(\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
= \(\frac{3x-5\sqrt{x}-2-\left(x+3\sqrt{x}+2\right)+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{3x-5\sqrt{x}-2-x-3\sqrt{x}-2+10\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2x+2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(2x+2\sqrt{x}\right)-\left(4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{x}\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
=\(\frac{\left(\sqrt{x}+2\right)2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2\)
Chúc bn học tốt
Nhớ tích cho mk nhé