chứng minh rằng
\(\left(3-2\sqrt{6}-\sqrt{33}\right)\times\left(\sqrt{6}+\sqrt{4}+4\right)=24\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
NT
Những câu hỏi liên quan
Chứng minh đẳng thức:
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(4\sqrt{\frac{1}{2}}+12\right)=-14\sqrt{2}\)
rg: \(\sqrt{\left(\sqrt{7}-4\right)}^2\) = 3
chứng minh:
\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\dfrac{1}{10}}+10\right)=3.3\sqrt{10}\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}\left(5\sqrt{\dfrac{1}{2}}+12\right)=-14.5\sqrt{2}\)
a: \(=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{2}+2\sqrt{5}\right)\cdot\sqrt{5}\cdot\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=\left(-3\sqrt{10}+10\right)\left(\dfrac{3}{10}\sqrt{10}+10\right)\)
\(=-9-30\sqrt{10}+3\sqrt{10}+100=91-27\sqrt{10}\)
b: \(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}\cdot\left(\dfrac{5}{2}\sqrt{2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\left(5\sqrt{3}+12\sqrt{6}\right)\)
\(=-60-144\sqrt{2}+30\sqrt{2}+144\)
\(=84-114\sqrt{2}\)
Đúng 0
Bình luận (0)
TC
19 tháng 8 2021 lúc 10:04
Tính
A= \(\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
Đúng 1
Bình luận (0)
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right)\times\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right)\times\left(-\sqrt{6}\right)\)
Giải:
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)
\(=\sqrt{2}\)
Vậy ...
Đúng 0
Bình luận (0)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Rút gọn :
dfrac{sqrt{x+sqrt{4left(x-1right)}}-sqrt{x-sqrt{4left(x-1right)}}}{sqrt{x^2-4left(x-1right)}}.left(sqrt{x-1}-dfrac{1}{sqrt{x-1}}right)
b)left(sqrt{2}+1right)left(sqrt{3}+1right)left(sqrt{6}+1right)left(5-2sqrt{2}-sqrt{3}right)
c)left(sqrt{5}+1right)left(sqrt{7}+1right)left(sqrt{35}+1right)left(34-4sqrt{7}-6sqrt{5}right)
d) left(sqrt{7}+1right)left(2sqrt{2}-1right)left(2sqrt{14}-1right)left(55+12sqrt{2}-7sqrt{7}right)
e)left(3sqrt{2}+1right)left(2sqrt{3}+1right)left(6sqrt{6}+1...
Đọc tiếp
Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Chứng minh các đẳng thức sau:
e) \(\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}-4.\sqrt{\dfrac{3}{2}}\right).\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}+4.\sqrt{\dfrac{3}{2}}\right)=-\sqrt{2}\)
`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`
`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`
`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`
`=((9sqrt6)/6+(4sqrt6)/6)^2-24`
`=((13sqrt6)/6)^2-24`
`=13^2/6-24`
`=25/6`
Đúng 0
Bình luận (0)
chứng minh rằng
\(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\times\left(15+2\sqrt{6}\right)=201\)
\(VT=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right).\left(15+2\sqrt{6}\right)=201\)
\(=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right).\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)\)
\(=15^2-\left(2\sqrt{6}\right)^2=201=VP\) (đpcm)
Đúng 0
Bình luận (0)
Chứng minh: (\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)=-14,5\sqrt{2}\)
\(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)\)
\(=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)\)
\(=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}\)
\(=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139\)
Ta có: \(-14,5\sqrt{2}\simeq-20,506\)
\(VT\ne VP\)
Đẳng thức không xảy ra
Đúng 0
Bình luận (0)