\(\left(\sqrt{\left(\sqrt{20}\right)-\sqrt{19}}\right)^x+\left(\sqrt{\left(\sqrt{20}\right)+\sqrt{19}}\right)^x=2\)
Rút gọn biểu thức:
1) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
2) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
3) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
4) \(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
5) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
6) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)
1. Tính ( rút gọn)
a)\(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
b)\(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
c)\(\sqrt{8+2\sqrt{15}}+\sqrt{\left(\sqrt{2-\sqrt{5}}\right)^2}\)
d)\(\sqrt{12+6\sqrt{3}}.\left(3+\sqrt{3}\right)\)
e) \(\left(2-\sqrt{5}\right).\sqrt{9+4\sqrt{5}}\)
a: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)
\(=5-\sqrt{19}-\sqrt{19}+4\)
\(=9-2\sqrt{19}\)
b: Ta có: \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}\)
\(=3-2\sqrt{2}-3+2\sqrt{2}\)
=0
c.
Căn bậc 2 không xác định do $2-\sqrt{5}< 0$
d.
\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)
e.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
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RÚT GỌN BIỂU THỨC:
19) \(A = \left(\dfrac{1}{\sqrt{x}} - \dfrac{1}{\sqrt{x} - 1}\right) : \left(\dfrac{\sqrt{x} + 2}{\sqrt{x} - 1} - \dfrac{\sqrt{x} + 1}{\sqrt{x} - 2}\right)\)
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\) (ĐK: \(x>0;x\ne2;x\ne1\))
\(A=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(A=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{-3}\)
\(A=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
\(A=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)\(=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\dfrac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)
\(=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
#Urushi☕
Rút gọn
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right).\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
tìm x\(\in\)Z để \(-\frac{20\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}⋮20\)
Rút gọn biểu thức:
1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\)
2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)
9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\)
10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\)
11) \(\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}\)
12) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
13) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
14) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)
15) \(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{120}\)
16) \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)
17) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
18) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
19) \(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
20) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
a. \(\sqrt{\left(2x+3\right)^2}=x+1\)
b. \(\sqrt{\left(2x-1\right)^2}=x+1\)
c. \(\sqrt{x+3}=5\)
d. \(\sqrt{x+2}=\sqrt{7}\)
e. \(5\sqrt{x}=20\)
f. \(\sqrt{x+4}=7\)
g. \(\sqrt{\left(2x+1\right)^2}=3\)
a, \(\sqrt{\left(2x+3\right)^2}=x+1\)
\(\Leftrightarrow\left|2x+3\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x+3=x+1\\2x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x\ge-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
Vậy phương trình vô nghiệm.
TH2: \(\left\{{}\begin{matrix}-2x-3=x+1\\2x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\x< -\dfrac{3}{2}\end{matrix}\right.\Rightarrow\) vô nghiệm.
b,
a, \(\sqrt{\left(2x-1\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
TH1: \(\left\{{}\begin{matrix}2x-1=x+1\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=2\)
TH2: \(\left\{{}\begin{matrix}-2x+1=x+1\\2x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=0\)
4.
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)-\left(\dfrac{1}{x+\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
a. Rút gọn A.
b. Tính x khi \(A=\dfrac{1}{2}\)
5. CMR
\(\left(\dfrac{\sqrt{30}}{\sqrt{3}}-\dfrac{\sqrt{20}}{\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right).\sqrt{4+\sqrt{15}}=2\)
nhanh lên nha