\(\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}\)
K
Khách
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1 tháng 10 2016
\(A=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)=2\)
\(B=\sqrt{18+8\sqrt{2}}+\sqrt{18-8\sqrt{2}}=\sqrt{\left(\sqrt{2}+4\right)^2}+\sqrt{\left(4-\sqrt{2}\right)^2}=4+\sqrt{2}+4-\sqrt{2}=8\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+\frac{2\sqrt{2}}{\sqrt{2}}.\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
S
2 tháng 9 2017
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{2}}{2}\)
___________
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
__________
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
14 tháng 8 2023
a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)
GH
21 tháng 6 2023
\(E=2\sqrt{3}+3\sqrt{3^3}-\sqrt{100.3}\\ =2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\\ =\left(2+9-10\right)\sqrt{3}=\sqrt{3}\)
\(F=\sqrt{3^2.2}+4\sqrt{18}=\sqrt{18}+4\sqrt{18}=\left(1+4\right)\sqrt{18}=5\sqrt{18}\)
\(G=2\sqrt{3}-4\sqrt{3^3}+5\sqrt{4^2.3}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=\left(2-12+20\right)\sqrt{3}=10\sqrt{3}\)
\(H=\left(3\sqrt{25.2}-5\sqrt{9.2}+3\sqrt{2^3}\right)\sqrt{2}\\ =\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}\\ =6\sqrt{2}.\sqrt{2}=6\)
17 tháng 12 2023
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)
30 tháng 9 2021
a)
\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{2}+1=\sqrt{3}+1\)
b)
\(\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{16}+\sqrt{2}\right)^2}=\sqrt{9}+\sqrt{2}-\sqrt{16}-\sqrt{2}=3-4=-1\)
c)
\(=\sqrt{2\left(2-\sqrt{3}\right)}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
18 tháng 9 2021
k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)
\(=\sqrt{3}-1\)
AH
Akai Haruma
Giáo viên
7 tháng 1 2019
Lời giải:
\(x=\sqrt{4+\sqrt{8}}.\sqrt{(2+\sqrt{2+\sqrt{2}})(2-\sqrt{2+\sqrt{2}})}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2^2-(2+\sqrt{2})}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2(2+\sqrt{2})}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{(2+\sqrt{2})(2-\sqrt{2})}\)
\(=\sqrt{2}.\sqrt{2^2-2}=\sqrt{2}.\sqrt{2}=2\)
\(y=\frac{3.2\sqrt{2}-2.2\sqrt{3}+2\sqrt{5}}{3.3\sqrt{2}-2.3\sqrt{3}+3\sqrt{5}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}{3(3\sqrt{2}-2\sqrt{3}+\sqrt{5})}=\frac{2}{3}\)
12 tháng 9 2021
Ta có: \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
Ta có: \(a=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
=2
Thay a=2 và \(b=\dfrac{2}{3}\) vào M, ta được:
\(M=\dfrac{1+2\cdot\dfrac{2}{3}}{2+\dfrac{2}{3}}-\dfrac{1-2\cdot\dfrac{2}{3}}{2-\dfrac{2}{3}}\)
\(=\dfrac{7}{8}+\dfrac{1}{4}\)
\(=\dfrac{7}{8}+\dfrac{2}{8}=\dfrac{9}{8}\)
NT
29 tháng 8 2021
4. \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}=3\sqrt{2}=8\sqrt{2}\)
5. \(\dfrac{\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}}{2\sqrt{3}}=\dfrac{2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}}{2\sqrt{3}}=\dfrac{-16\sqrt{3}}{2\sqrt{3}}=-8\sqrt{3}\)
29 tháng 8 2021
4: Ta có: \(A=3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}\)
\(=8\sqrt{2}\)
5: Ta có: \(\left(\sqrt{12}-\sqrt{48}-\sqrt{108}-\sqrt{192}\right):2\sqrt{3}\)
\(=\left(2\sqrt{3}-4\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):2\sqrt{3}\)
\(=-16\sqrt{3}:2\sqrt{3}=-8\)
\(\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}\)
\(=\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-8\sqrt{2}+2}}\)
\(=\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{4^2-2.4.\sqrt{2}+\sqrt{2}^2}}\)
\(=\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{\left(4-\sqrt{2}\right)^2}}\)
\(=\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}\)
\(=\sqrt{2\sqrt{3}+4}\)
\(=\sqrt{1+2\sqrt{3}+3}\)
\(=\sqrt{1+2\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=1+\sqrt{3}\)