P=\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)duoc bieu dien duoi dang tong cua 3 can thuc bac hai nhu sau P=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)
khi do a+b+c=?
Những câu hỏi liên quan
\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\) được biểu diễn dưới dạng tổng 3 căn thức bậc 2 như sau: P=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\). khi đó a+b+c=.......
\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)
\(=\sqrt{2+5+7+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{5.7}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{5}+\sqrt{7}\right)^2}=\sqrt{2}+\sqrt{5}+\sqrt{7}\)
\(\Rightarrow a+b+c=2+5+7=14\)
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Can bac hai cua 64 co the viet duoi dang nhu sau: \(\sqrt{64}\)= 6+\(\sqrt{4}\)
Hoi co ton tai hay khong cac so co 2 chu so co the viet can bac hai cua chung duoi dang nhu tren va la mot so nguyen? Hay chi ra toan bo cac so do.
Cho \(P=\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)
Được biểu diễn dưới dạng: \(P=\sqrt{a}+\sqrt{b}+\sqrt{c}\)
Tính a+b+c
Ta có
\(P=\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)
\(\Leftrightarrow P=\sqrt{\left(\sqrt{5}+\sqrt{2}+\sqrt{7}\right)^2}\)
\(\Leftrightarrow P=\sqrt{5}+\sqrt{2}+\sqrt{7}\)
Mà \(P=\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{5}+\sqrt{2}+\sqrt{7}\)
Suy ra \(a+b+c=5+2+7=14\)
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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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tính :
A=\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
B=\(\sqrt{14+\sqrt{40}+\sqrt{56}+\sqrt{140}}\)
\(A=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
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B=\(\sqrt{14+2\sqrt{10}+2\sqrt{14}+2\sqrt{35}}\)
B=\(\sqrt{7+5+2+2\sqrt{2.5}+2\sqrt{2.7}+2\sqrt{7.5}}\)
B=\(\left(\sqrt{\sqrt{7}+\sqrt{5}+\sqrt{2}}\right)^2\)
B=\(\sqrt{7}+\sqrt{5}+\sqrt{2}\)
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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: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap
Giup minh nhe!!
1. Rut gon bieu thuc:
a) Asqrt{11-6sqrt{2}}+3+sqrt{2}
b) Bsqrt{12+2sqrt{11}}+sqrt{12-2sqrt{11}}
2. Rut gon bieu thuc:
a) Asqrt{x-2+2sqrt{x-3}}+sqrt{x+6+6sqrt{x-3}},voix3
3. Tim GTNN cua bieu thuc:
a) Asqrt{4x^2-12x+9}+sqrt{x^2-10x+25}+sqrt{9x^2-6x+1}+sqrt{16x^2-72x+81}
b) Bdfrac{1}{2}sqrt{x^2}+sqrt{x^2-2x+1}
Đọc tiếp
Giup minh nhe!!
1. Rut gon bieu thuc:
a) \(A=\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
b) \(B=\sqrt{12+2\sqrt{11}}+\sqrt{12-2\sqrt{11}}\)
2. Rut gon bieu thuc:
a) \(A=\sqrt{x-2+2\sqrt{x-3}}+\sqrt{x+6+6\sqrt{x-3}},voix>=3\)
3. Tim GTNN cua bieu thuc:
a) \(A=\sqrt{4x^2-12x+9}+\sqrt{x^2-10x+25}+\sqrt{9x^2-6x+1}+\sqrt{16x^2-72x+81}\)
b) \(B=\dfrac{1}{2}\sqrt{x^2}+\sqrt{x^2-2x+1}\)
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
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