cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
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cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
A= \(a^{2017}\left(a^2-8a+11\right)+b^{2017}\left(b^2-8b+11\right)=\)\(a^{2017}\left(a^2-8a+16-5\right)+b^{2017}\left(b^2-8b+16-5\right)=\)\(a^{2017}\left(\left(a-4\right)^2-\sqrt{5^2}\right)+b^{2017}\left(\left(b-4\right)^2-\sqrt{5^2}\right)\)=\(a^{2017}\left(a-4-\sqrt{5}\right)\left(a-4+\sqrt{5}\right)+b^{2017}\left(b-4-\sqrt{5}\right)\left(b-4+\sqrt{5}\right)\)= 0+0= 0
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1, cho a = \(4+\sqrt{5}\),b=\(4-\sqrt{5}\)
Tính A=\(\left(a^{2018}-8a^{2017}+11a^{2016}\right)+\left(b^{2018}-8b^{2017}+11b^{2016}\right)\)
2, cho \(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+48}=18\)
Tính A=\(\sqrt{4x^2-24x+256}-2\sqrt{x^2-6x+36}\)
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
cho bieu thuc A = \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri cua bieu thuc A tai x=7+4√3
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
cho bieu thuc A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a.Tim x de bieu thuc A co nghia ? rut gon A ?
b. Tinh gia tri cua bieu thuc A tai x=7=4√3
1)Tính:
a)\(\sqrt{13a}.\sqrt{\frac{52}{a}}\left(a< 0\right)\)
b)\(\left(2+\sqrt{5}\right).\left(2-\sqrt{5}\right)\)
c)\(\sqrt{b^4\left(a-b\right)^2}.\frac{1}{a-b}\left(a< 0\right)\)
d)\(\left(\sqrt{2019}-\sqrt{2018}\right).\left(\sqrt{2018}+\sqrt{2019}\right)\)
Giúp mk vs mấy bn, mk đang cần gấp
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
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Cho \(A=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2018}\)
Chứng minh A không là số nguyên.
Cho \(xy+\sqrt{\left(1+x\right)^2\left(1+y\right)^2}=\sqrt{2017}\)
Tinh gia tri cua bieu thuc: \(P=x+\sqrt{1+y^2}+y\sqrt{1+x^2}\)
Giup mk!!!
