rut gon A=\(\dfrac{10^4\cdot81-16\cdot15^2}{4^4\cdot675}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DM
15 tháng 6 2017
Lần sau ghi dấu ra xíu nhé :v
a) Đặt \(\sqrt{x}=a\Rightarrow B=\left(\dfrac{a}{a+4}+\dfrac{4}{a-4}\right):\dfrac{a^2+16}{a+2}\)
Quy đồng,rút gọn : \(B=\dfrac{a+2}{a^2-16}\Rightarrow B=\dfrac{\sqrt{x}+2}{x-16}\)
b) \(B\left(A-1\right)=\dfrac{\sqrt{x}+2}{x-16}\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)=\dfrac{2}{x-16}\)
x - 16 là ước của 2 => \(x\in\left\{14;15;17;18\right\}\)
mới làm quen toán 9 ;v có gì k rõ ae chỉ bảo nhé :))
26 tháng 5 2021
\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)
LH
0
LH
1
4 tháng 12 2015
\(M=1.\left(a+b\right)\left(a^2+b^2\right).......\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)....\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)....\)
\(=\left(a^4-b^4\right)\left(a^4+b^4\right)......\)
\(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\)
\(=a^{32}-b^{32}\)
Y
21 tháng 5 2023
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+2}\right):\dfrac{x+4}{\sqrt{x}+2}\left(dkxd:x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{x+4}\right)\)
\(=\dfrac{x+2\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+4}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
Vậy \(B=\dfrac{1}{\sqrt{x}-2}\)
LN
1
PT
21 tháng 9 2018
Ta có:
\(A=\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)
\(A=\left(x^2-4x-2x+8\right)-\left(x^2-x-3x+4\right)\)
\(A=\left(x^2-6x+8\right)-\left(x^2-4x+4\right)\)
\(A=x^2-6x+8-x^2+4x-4\)
\(A=-2x+4\)
Thay \(x=1\dfrac{3}{4}=\dfrac{7}{4}\) vào A ta được:
\(A=-2.\dfrac{7}{4}+4\)
\(A=-\dfrac{7}{2}+4\)
\(A=\dfrac{1}{2}\)
\(A=\dfrac{2^4\cdot5^4\cdot3^4-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)
\(=\dfrac{2^4\cdot3^2\cdot5^2\left(3^2\cdot5^2-1\right)}{2^8\cdot3^3\cdot5^2}=\dfrac{1}{16}\cdot\dfrac{1}{3}\cdot\dfrac{15^2-1}{1}\)
\(=\dfrac{224}{48}=\dfrac{14}{3}\)