\(\left(2+\sqrt{2}\right)\sqrt{3-\sqrt{8}}\)
1)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
2)\(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
1) \(\left(3+\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3-\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}+1\right)^2}.\left(\sqrt{5}-1\right)\\ =2.\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\\ =2\left(5-1\right)=8=VP\)
=> ĐPCM
2) Bình phương 2 về, đpcm <=> \(2\sqrt{2}-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\) = \(2\left(\sqrt{2}-1\right)\)
VT = \(2\sqrt{2}-2\sqrt{2-1}=2\left(\sqrt{2}-1\right)\) = VP=> ĐPCM
1) Ta có: \(\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6-2\sqrt{5}}}{2}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(6-2\sqrt{5}\right)}{2}\)
=8
\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)
\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)
\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)
Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)
\(\Rightarrow\)PT vô nghiệm
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
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.
\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)
Gấp lắm . Giúp mình cảm ơn ạ
Bài 1
\(2\sqrt{\left(1+\sqrt{3}\right)^{ }3}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(\left(1+\sqrt{3}-\sqrt{5}\right).\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(\left(\sqrt[]{\dfrac{8}{3}}-\sqrt{5}\right)x\sqrt{6}\)
\(\left(5+4\sqrt{2}\right).\left(3+2\sqrt{1}+\sqrt{2}\right).\left(3-2\sqrt{1}+2\right)\)
\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
rút gọn
a) \(\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}\)
b) \(\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
c) \(\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)\)
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9.\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{4x^2+12x+9}=5\)
\(\sqrt{5x-6}-3=0\)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
rút gọn
a) \(\left(-7\sqrt{7}\right)\left(-2\sqrt{8}\right)\)
b) \(-\sqrt{33}.3\sqrt{3}\)
c) \(\left(3\sqrt{5}\right).\left(-10\sqrt{3}\right)\)
d) \(\dfrac{1}{2}\sqrt{5}.\left(-6\sqrt{2}\right)\)
e) \(\dfrac{2}{3}\sqrt{7}.\left(-\dfrac{9}{16}\sqrt{3}\right)\)
f) \(15\sqrt{6}:5\sqrt{3}\)
g) \(-25\sqrt{12}:\left(-5\sqrt{6}\right)\)
h) \(36\sqrt{8}:12\sqrt{2}\)
i) \(4\sqrt{27}:\left(-2\sqrt{3}\right)\)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
tính
\(\sqrt{5}.\left(\sqrt{3}+\sqrt{2}\right)\)
\(\left(\sqrt{3}-\sqrt{5}\right)^2\)
\(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{7}+\sqrt{2}\right)\)
\(\left(\sqrt{3}+\sqrt{5}\right)^2\)
\(\left(\sqrt{8+3\sqrt{7}}\right)+\left(\sqrt{8-3\sqrt{7}}\right)\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)