Tính : \(\sqrt{17-12\sqrt{3}}\)
ai giúp e vs !!HELP ME
Giúp vs, làm câu nào cx đc, làm hết thì tốt
a) \(\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
b) \(\sqrt{27-10\sqrt{2}}+\sqrt{18-8\sqrt{2}}\)
c) \(\sqrt{3-\sqrt{5}}.\sqrt{8}\)
d) \(\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{2}}.\sqrt{8}\)
e) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
g) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-\left(\sqrt{2}+3\right)\)
h) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
i) \(\left(\sqrt[3]{25}-\sqrt[3]{10}+\sqrt[3]{4}\right).\left(\sqrt[3]{5}+\sqrt[3]{2}\right)\)
k) \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
L) \(A=\sqrt[3]{10+14\sqrt{2}}+\sqrt[3]{10-14\sqrt{2}}\)
k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)
\(=\sqrt{3}-1\)
Bài 1: Tính
A=\(\frac{\sqrt{5+\sqrt{17}}-\sqrt{5-\sqrt{17}}-\sqrt{10-4\sqrt{2}}+4}{\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}+2-\sqrt{2}}\)
Bài 2: Tìm x,y nguyên thỏa mãn:
y2 = 1 + \(\sqrt{9-x^2-4x}\)
Help me!!! Giúp với... Mình gần thi học sinh giỏi rồi.... Có ai wan tâm ko? Huhuhu...
C/m A, B ∈ Z, với:
A= \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B= \(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Help me
*\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}+1=2\)
\(\Rightarrow A\in Z\)
* \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-2\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) \(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\) \(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\) \(=\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{9-8}\)
\(=2\)
\(\Rightarrow B\in Z\)
Bài 1: Tính :
\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}}\)
Help me!!! Tối học nhà thầy r ạ!!! Tình hình nguy cấp lắm r... !!! Giúp với !!!!
\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}}=\sqrt{\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}}\)
\(=\sqrt{\sqrt{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}}=\sqrt{\sqrt{5-\sqrt{6-2\sqrt{5}}}}=\sqrt{\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}}\)
\(=\sqrt{\sqrt{5-\left(\sqrt{5}-1\right)}}=\sqrt{\sqrt{6-\sqrt{5}}}\)
1) Rút gọn
h)\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)
e)\(\frac{\sqrt{28}-2\sqrt{12}-2\sqrt{18}}{3\sqrt{7}-2\sqrt{27}-\sqrt{102}}\)
f)\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
mọi ng giúp mình vs đang cần gấp tks !!!đc câu nào giúp câu đấy cx đk
\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)
Cho x=\(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\),y=\(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)
Tính P=\(x^3+y^3-3\left(x+y\right)+1979\)
mong mọi người giúp thanks you
\(x^3=3+2\sqrt{2}+3-2\sqrt{2}+3\cdot\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\)
\(y^3=17+12\sqrt{2}+17-12\sqrt{2}+3\sqrt[3]{\left(17-12\sqrt{2}\right)\left(17+12\sqrt{2}\right)}\left(\sqrt[3]{17-12\sqrt{2}}+\sqrt[3]{17+12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3x\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\)
Thay vào P, ta được
\(P=x^3+y^3-3x-3y+1979\\ P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979\\ P=6+34+1979=2019\)
\(x^3=6+3\sqrt[3]{\left(3+2\sqrt[]{2}\right)\left(3-2\sqrt[]{2}\right)}\left(\sqrt[3]{3+2\sqrt[]{2}}+\sqrt[3]{3-2\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=6+3x\)
\(\Rightarrow x^3-3x=6\)
Tương tự:
\(y^3=34+3\sqrt[3]{\left(17+12\sqrt[]{2}\right)\left(17-12\sqrt[]{2}\right)}\left(\sqrt[3]{17+12\sqrt[]{2}}+\sqrt[3]{17-12\sqrt[]{2}}\right)\)
\(\Rightarrow y^3=34+3y\)
\(\Rightarrow y^3-3y=34\)
Do đó:
\(P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979=6+34+1979=...\)
Rút gọn biểu thức
\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}-3\sqrt{5\sqrt{48}}}\)
\(2\sqrt{5\sqrt{3}}-2\sqrt{8\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
Help me plsssssss
Help me plssssssss
a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}\)
b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)
Rút gọn biểu thức: \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
CÁC BN GIÚP MK VS,,,
Tính:
1) \(\sqrt{14-2\sqrt{33}}\)
2) \(\sqrt{12-2\sqrt{35}}\)
3) \(\sqrt{16-2\sqrt{55}}\)
4) \(\sqrt{14-6\sqrt{5}}\)
5) \(\sqrt{17-12\sqrt{2}}\)
6) \(\sqrt{27-12\sqrt{5}}\)
7) \(\sqrt{4+\sqrt{15}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)