Bài1: Tĩm x, biết: a) \sqrt{1,69}.(2.\sqrt{x}+\sqrt{81/121})=13/10
\(\sqrt{1,69}.\left(2\sqrt{x}+\sqrt{\dfrac{81}{121}}\right)=\dfrac{13}{10}\left(x>0\right)\)
GIÚP MIK VỚI
\(\sqrt{1.69}\left(2\sqrt{x}+\sqrt{\dfrac{81}{121}}\right)=\dfrac{13}{10}\)
\(\Leftrightarrow2\sqrt{x}+\dfrac{9}{11}=1\)
\(\Leftrightarrow2\sqrt{x}=\dfrac{2}{11}\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{11}\)
hay x=1/121
Tìm x biết
\(\sqrt{1,69}\left(2\sqrt{x}+\sqrt{\frac{81}{21}}\right)=\frac{13}{10}\)
Tìm x biết
\(a,\sqrt{1,69\cdot\left(2\cdot x+\sqrt{\frac{81}{121}}\right)=\frac{13}{10}}\)
\(b,2\cdot x^7=3^9\)
\(c,x^6=4\cdot x\)
Bài 1: Tìm x, biết:
a) \(\sqrt{1,69}.(2.\sqrt{x}+\sqrt{81/121})=13/10\)
b) \((2x^2-3).(3x^2-1/0,12).(x^2+1)=0\)
c) \(|2/5.\sqrt{x}-1/20|-3/4=1/5\)
d) \(x-5\sqrt{x}=0\)
e) \(2x^7=3x^9\)
f) \(x^6=4x^8\)
Giải hộ mik với nha! Thank you
\(d,x-5\sqrt{x}=0\)
\(ĐKXĐ:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}\)(Thỏa mãn ĐKXĐ)
Vậy...
Căn bậc hai của 1,69×( 2 căn x+ căn 81/121)=13/10
1. tính
a,\(\sqrt{\dfrac{1,44}{3,61}}\) ; b, \(\sqrt{\dfrac{0,25}{9}}\) ; c, \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}\)
d,\(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}\) ; e,\(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}\) ; g,
2. Tính
a, \(\dfrac{\sqrt{245}}{\sqrt{5}}\) ; b, \(\dfrac{\sqrt{3}}{\sqrt{75}}\) ; c, \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}\) ; d, \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}\)
3. Tính.
a, \(\sqrt{\dfrac{61^2-60^2}{81}}\) ; b, \(\sqrt{\dfrac{74^2-24^2}{121}}\)
4. Tìm số x không âm, biết:
a, 9 - 4 \(\sqrt{x}=1\) ; b, \(\sqrt{\dfrac{x}{5}}=4\) c, \(\sqrt{7x}< 9\)
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
Bài 3 :
Câu a : \(\sqrt{\dfrac{61^2-60^2}{81}}=\sqrt{\dfrac{\left(61-60\right)\left(61+60\right)}{81}}=\sqrt{\dfrac{121}{81}}=\dfrac{\sqrt{121}}{81}=\dfrac{11}{9}\)
Câu b : \(\sqrt{\dfrac{74^2-24^2}{121}}=\sqrt{\dfrac{\left(74+24\right)\left(74-24\right)}{121}}=\sqrt{\dfrac{98.50}{121}}=\dfrac{\sqrt{4900}}{\sqrt{121}}=\dfrac{70}{11}\)
Bài 4 :
Câu a : \(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Câu b : \(\sqrt{\dfrac{x}{5}}=4\Leftrightarrow\dfrac{x}{5}=16\Leftrightarrow x=80\)
Câu c : \(\sqrt{7x}< 9\Leftrightarrow7x< 81\Leftrightarrow x< \dfrac{81}{7}\)
Câu 1
1) Tính
a) \(\sqrt{25}+\sqrt{49}\) b) \(\sqrt{121}-\sqrt{81}\)
2) Với x > -2 thì \(\sqrt{2x+1}\) có nghĩa không
3) Rút gọn biểu thức sau :
a) \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\) c) \(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
1:
a: \(\sqrt{25}+\sqrt{49}=5+7=12\)
b: \(\sqrt{121}-\sqrt{81}=11-9=2\)
2: x>-2
=>2x>-4
=>2x+1>-3
=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa
3:
a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
c:
\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)
Cho biết : \(\sqrt{x^2-6x+13}-\sqrt{x^2-6x+10}\)=1
Tính : \(\sqrt{x^2-6x+13}+\sqrt{x^2-6x+10}\)=?
Tinh giá trị biểu thuc \(A=x^2+2016-2017\)
Biết \(x=\frac{\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}}{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right):\sqrt{\sqrt{13}+3}}\)