1, tìm x
a,\(\sqrt{2x}\)=12
b,\(\sqrt{9x^2-6x}\)+1=10
c,\(\sqrt{5}\times x^2-\sqrt{125}\)=0
1, tìm x
a,\(\sqrt{2x}\)=12
b,\(\sqrt{9x^2-6x}\)+1=10
c,\(\sqrt{5}\times x^2-\sqrt{125}\)=0
a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)
\(2x=144\)
\(x=72\)
b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)
\(\sqrt{9x^2-6x}=9\)
\(9x^2-6x=81\)
\(\left(3x-1\right)^2=82\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)
c) \(x^2\sqrt{5}-\sqrt{125}=0\)
\(x^2\sqrt{5}=5\sqrt{5}\)
\(x^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
bài 1: rut gọn
a, \(\sqrt{5\left\{1-a\right\}^2}\) với a>1
b,\(\sqrt{\dfrac{9\left[a^2+2a+1\right]}{144}}\)
c,\(\dfrac{2}{x-5}\times\sqrt{\dfrac{x^2\times10x+25}{64}}\)
d \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) với x≥0 và x≠1
a: \(\sqrt{5\left(1-a\right)^2}\)
\(=\sqrt{5\left(a-1\right)^2}\)
\(=\sqrt{5}\cdot\sqrt{\left(a-1\right)^2}\)
\(=\sqrt{5}\left|a-1\right|\)
\(=\sqrt{5}\left(a-1\right)\)(do a>1 nên a-1>0)
b: \(\sqrt{\dfrac{9\left|a^2+2a+1\right|}{144}}\)
\(=\sqrt{\dfrac{9}{144}\cdot\left|a^2+2a+1\right|}\)
\(=\sqrt{\dfrac{1}{16}\cdot\left|\left(a+1\right)^2\right|}\)
\(=\sqrt{\dfrac{1}{16}}\cdot\sqrt{\left|\left(a+1\right)^2\right|}\)
\(=\dfrac{1}{4}\cdot\left(a+1\right)^2\)
c:
ĐKXĐ: x<>5
Sửa đề:\(\dfrac{2}{x-5}\cdot\sqrt{\dfrac{x^2-10x+25}{64}}\)
\(=\dfrac{2}{x-5}\cdot\sqrt{\dfrac{\left(x-5\right)^2}{64}}\)
\(=\dfrac{2}{x-5}\cdot\dfrac{\sqrt{\left(x-5\right)^2}}{\sqrt{64}}\)
\(=\dfrac{2}{x-5}\cdot\dfrac{\left|x-5\right|}{8}\)
\(=\pm\dfrac{1}{4}\)
d: \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x}-\sqrt{x}\cdot1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\sqrt{x}\)
\(a,\sqrt{3x}.\sqrt{\dfrac{12}{x}}=\sqrt{\dfrac{3x.12}{x}}=\sqrt{36}=\sqrt{6^2}=6\\ b,\dfrac{\sqrt{7y^3}}{\sqrt{63y}}=\sqrt{\dfrac{7}{63}.\dfrac{y^3}{y}}=\sqrt{\dfrac{1}{9}y^2}=\sqrt{\left(\dfrac{1}{3}y\right)^2}=\dfrac{1}{3}y\)
\(c,\sqrt{14a^3}.\sqrt{\dfrac{25b^6}{126a}}=\sqrt{\dfrac{14.25.a^3.b^6}{126a}}\\ =\sqrt{\dfrac{350}{126}a^2b^6}=\sqrt{\dfrac{25}{9}a^2b^6}=\sqrt{\left(\dfrac{5}{3}ab^3\right)^2}=\left|\dfrac{5}{3}ab^3\right|=-\dfrac{5}{3}ab^3\\ d,\left(x-y\right)\sqrt{\dfrac{y^2}{x^2+y^2-2xy}}=\left(x-y\right)\sqrt{\dfrac{y^2}{\left(x-y\right)^2}}=\left(x-y\right).\sqrt{\left(\dfrac{y}{x-y}\right)^2}\\ =\left(x-y\right).\left|\dfrac{y}{x-y}\right|=-\left(x-y\right).\dfrac{y}{x-y}=-y\)
a) \(\sqrt[]{\dfrac{20^2-16^2}{16}}\)
\(=\sqrt[]{\dfrac{\left(20-16\right)\left(20+16\right)}{16}}\)
\(=\sqrt[]{\dfrac{4.36}{16}}\)
\(=\sqrt[]{\dfrac{36}{4}}=\sqrt[]{9}=3\)
b) \(\sqrt[]{\left(-2\right)^2.9.\dfrac{289}{49}}\)
\(=\sqrt[]{2^2.3^2.\dfrac{17^2}{7^2}}\)
\(=2.3.\dfrac{17}{7}=\dfrac{102}{7}\)
a: \(\Leftrightarrow2\cdot2\sqrt{x-3}+3\sqrt{x-3}-4\sqrt{x-3}=6\)
=>\(3\sqrt{x-3}=6\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7
b; \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\dfrac{3}{2}\sqrt{x-1}=6\)
=>\(\sqrt{x-1}=6\)
=>x-1=36
=>x=37
c: \(\Leftrightarrow2\sqrt{x^2-5}\cdot\dfrac{2}{3}+\dfrac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=2\)
=>\(-\sqrt{x^2-5}=2\)(vô lý)
Vậy: \(x\in\varnothing\)
a.
ĐK: \(x\ge3\)
PT trở thành:
\(2\sqrt{4\left(x-3\right)}+3\sqrt{x-3}-\sqrt{16\left(x-3\right)}=6\\ \Leftrightarrow4\sqrt{x-3}+3\sqrt{x-3}-4\sqrt{x-3}=6\\ \Leftrightarrow\left(4+3-4\right)\sqrt{x-3}=6\\ \Leftrightarrow\sqrt{x-3}=2\\ \Leftrightarrow x-3=2^2=4\\ \Leftrightarrow x=7\left(tm\right)\)
a, \(\sqrt[]{x^2-10x+25}\) = x-2
b, \(\sqrt[]{x+2}\) + \(\sqrt[]{9x+18}\) + \(\sqrt[]{4x+8}\) = 2
\(a,\sqrt{x^2-10x+25}=x-2\\ ĐK:x-2\ge0\Leftrightarrow x\ge2\\ \sqrt{x^2-10x+25}=x-2\\ \Leftrightarrow x^2-10x+25=x^2-4x+4\\ \Leftrightarrow x^2-x^2-10x+4x=4-25\\ \Leftrightarrow-6x=-21\\ \Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\\ Vậy.S=\left\{\dfrac{7}{2}\right\}\\ b,\sqrt{x+2}+\sqrt{9x+8}+\sqrt{4x+8}=2\\ \Leftrightarrow\sqrt{x+2}+3\sqrt{x+2}+2\sqrt{x+2}=2\\ \Leftrightarrow6\sqrt{x+2}=2\\ \Leftrightarrow\left[{}\begin{matrix}x+2\ge0\\\sqrt{x+2}=\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\ge-2\\x+2=\dfrac{1}{9}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\ge-2\\x=\dfrac{1}{9}-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\ge-2\\x=-\dfrac{17}{9}\left(tm\right)\end{matrix}\right.\\ Vậy.S=\left\{-\dfrac{17}{9}\right\}\)
giúp em với ạ
d) \(\dfrac{3x}{7y}\cdot\sqrt{\dfrac{49y^2}{9x^2}}\)
\(=\dfrac{3x}{7y}\cdot\sqrt{\dfrac{\left(7y\right)^2}{\left(3x\right)^2}}\)
\(=\dfrac{3x}{7y}\cdot\dfrac{7y}{3x}\)
\(=\dfrac{21xy}{21xy}\)
\(=1\)
e) \(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}\)
\(=\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}+\dfrac{2\left(a+1-a-1\right)}{a\left(a+1\right)}}\)
\(=\sqrt{1^2+\left(\dfrac{1}{a}\right)^2+\left(\dfrac{1}{a+1}\right)^2+2\cdot1\cdot\dfrac{1}{a}-2\cdot1\cdot\dfrac{1}{a+1}-2\cdot\dfrac{1}{a}\cdot\dfrac{1}{a+1}}\)
\(=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}\)
\(=1+\dfrac{1}{a}-\dfrac{1}{a+1}\)
a, \(\dfrac{\sqrt[]{7-2\sqrt[]{6}}}{\sqrt[]{6}-1}\)
b, 2.|x+y|.\(\sqrt[]{\dfrac{1}{x^2+2xy+y^2}}\) (x+y>0)
c, \(\dfrac{\left(x-5\right)^4}{\left(4-x\right)^2}\)-\(\dfrac{x^2-25}{x-4}\)(x<4)