Cho x=\(\sqrt{5}+\sqrt{3}\)
a) Tính \(x^{^{ }3}\)
b) CM; \(x^4-16x^2+4=0\)
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18 tháng 8 2023
a: Khi x=25 thì \(A=\dfrac{7}{5+8}=\dfrac{7}{13}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{x-9}\)
\(=\dfrac{x+5\sqrt{x}-24}{x-9}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{x-9}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
c: P=A*B
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\cdot\dfrac{7}{\sqrt{x}+8}=\dfrac{7}{\sqrt{x}+3}\)
P là số nguyên
=>căn x+3 thuộc Ư(7)
=>căn x+3=7
=>x=16
10 tháng 9 2018
a) ta có : \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3}+\sqrt{\left(\sqrt{2}+2\right)^3}=1-\sqrt{2}+\sqrt{2}+2=3\)
b) ở đây : https://hoc24.vn/hoi-dap/question/650070.html
DN
11 tháng 9 2018
a, Ta có: \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3+\sqrt[3]{\left(\sqrt{2}+2\right)^3}}\)
\(=1-\sqrt{2}+\sqrt{2}+2=1+2=3\)
Vậy ...
\(b,x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
Đặt \(\sqrt{2+\sqrt{3}=t}\) , ta có:
\(x=\sqrt{6-3.t}-\sqrt{2+t}\)
\(\Rightarrow x^2=2+t+3.\left(2-t\right)-2\sqrt{3}\left(2+t\right)\left(2-t\right)\)
\(=8-2t-2\sqrt{3\left(4-t^2\right)}\)
\(=8-2t-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)
\(=8-2t-\sqrt{6}.\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)\)
\(=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
\(=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\)
\(\Rightarrow\left(x^2-8\right)^2=32\)
\(\Rightarrow x^4-16x^2+64=32\)
\(\Rightarrow x^4-16x^2+64-32=0\)
\(\Rightarrow x^4-16x^2+32=0\) (đpcm)
Chúc bạn hok tốt!!!
10 tháng 8 2023
2/
a) Ta có:
\(3\sqrt{2}=\sqrt{3^2\cdot2}=\sqrt{9\cdot2}=\sqrt{18}\)
\(2\sqrt{3}=\sqrt{2^2\cdot3}=\sqrt{4\cdot3}=\sqrt{12}\)
Mà: \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Rightarrow2\sqrt{3}< 3\sqrt{2}\)
b) Ta có:
\(4\sqrt[3]{5}=\sqrt[3]{4^3\cdot5}=\sqrt[3]{320}\)
\(5\sqrt[3]{4}=\sqrt[3]{5^3\cdot4}=\sqrt[3]{500}\)
Mà: \(320< 500\Rightarrow\sqrt[3]{320}< \sqrt[3]{500}\Rightarrow4\sqrt[3]{5}< 5\sqrt[3]{4}\)
10 tháng 8 2023
3/
a)ĐKXĐ: \(x\ne1;x\ge0\)
b) \(A=\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\)
\(A=\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\)
\(A=1^2-\left(\sqrt{x}\right)^2\)
\(A=1-x\)
11 tháng 8 2017
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
20 tháng 8 2021
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
31 tháng 8 2021
a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)
b: Ta có: P=A:B
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
13 tháng 5 2022
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
\(x^3=\left(\sqrt{5}+\sqrt{3}\right)^3=\sqrt{5^3}+3.5.\sqrt{3}+3.\sqrt{5}.3+\sqrt{3^3}\)
\(=14\sqrt{5}+18\sqrt{3}\)
\(x^2=\left(\sqrt{5}+\sqrt{3}\right)^2=8+2\sqrt{15}\)
\(\frac{1}{x^2}=\frac{1}{8+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8^2-4.15}=\frac{4-\sqrt{15}}{2}\)
\(\Leftrightarrow\frac{4}{x^2}=8-2\sqrt{15}\)
\(\Leftrightarrow x^2+\frac{4}{x^2}=16\)
\(\Leftrightarrow x^4-16x^2+4=0\)
Ta có đpcm.