(x ^ 4 - x ^ 2 + 1)/(x ^ 2 * (1 - x ^ 2)) + 5/(2x * sqrt(1 - x ^ 2)) + 2 = 0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Số tự nhiên lớn nhất có bốn chữ số mà trong đó không có hai chữ số nào giống nhau là: 9876
a, Với \(x\ge0;x\ne1\):
\(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=\sqrt{x}\left(1-\sqrt{x}\right)=\sqrt{x}-x\)
b, Thay \(x=7-4\sqrt{3}\) vào P, ta được:
\(P=\sqrt{7-4\sqrt{3}}-\left(7-4\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.2+2^2}+4\sqrt{3}-7\)
\(=\sqrt{\left(\sqrt{3}-2\right)^2}+4\sqrt{3}-7\)
\(=\left|\sqrt{3}-2\right|+4\sqrt{3}-7\)
\(=2-\sqrt{3}+4\sqrt{3}-7\) (vì \(\sqrt{3}< 2\))
\(=-5+3\sqrt{3}\)
$Toru$
a) \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\left(x\ge0,x\ne1\right)\\ =\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(x-1\right)^2}{2}\\ =\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{2}\\ \)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\\ =\left[x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)\right].\dfrac{\sqrt{x}-1}{2}\\ \)
\(=-2\sqrt{x}.\dfrac{\sqrt{x}-1}{2}\\ =-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) \(x=7-4\sqrt{3}\left(TMDK\right)\)
\(\sqrt{x}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
Thay vào biểu thức P, ta được:
\(P=-\left(7-4\sqrt{3}\right)+2-\sqrt{3}=-5+3\sqrt{3}\)
Ta có: \(E=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)
\(3E=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(3E-E=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+..+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\)
\(2E=1+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6E=3+1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(4E=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow E=\dfrac{3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}}{4}=\dfrac{3}{4}-\dfrac{\dfrac{100}{3^{99}}+\dfrac{100}{3^{100}}}{4}< \dfrac{3}{4}\) (đpcm)
Lời giải:
$51:32:72=\frac{51}{32\times 72}=\frac{17\times 3}{32\times 3\times 24}=\frac{17}{32\times 24}=\frac{17}{768}$
44444444444444444444444455555555555555555555555555544444444444444444444444445555555555555555555555554444445555555555555555555554444444444455555555555555555554444444444444444444444555555555555555555555444444444444444455555555555555555555544444444444445555555544444445555554444445555545554545454545454545545454545454545454545454545454545454545444444455555544444545444444444444444444444444444444444444444444444455555555555555555555555554444444444444444444444444444444444444444444444444444444555555555555555555555555555555555
Lúc đầu anh trai nhiều hơn em trai số quả bóng là:
13 + 13 = 26 (quả)
Đ/s: 26 quả bóng
a: \(\dfrac{8}{9}=1-\dfrac{1}{9}\)
\(\dfrac{108}{109}=1-\dfrac{1}{109}\)
Vì 9<109 nên \(\dfrac{1}{9}>\dfrac{1}{109}\)
=>\(-\dfrac{1}{9}< -\dfrac{1}{109}\)
=>\(-\dfrac{1}{9}+1< -\dfrac{1}{109}+1\)
=>\(\dfrac{8}{9}< \dfrac{108}{109}\)
b: \(\dfrac{97}{100}=0,97;\dfrac{98}{99}=0,\left(98\right)\)
mà 0,97<0,(98)
nên \(\dfrac{97}{100}< \dfrac{98}{99}\)
c: \(\dfrac{19}{18}=1+\dfrac{1}{18}\)
\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)
Vì 18<2020 nên \(\dfrac{1}{18}>\dfrac{1}{2020}\)
=>\(1+\dfrac{1}{18}>1+\dfrac{1}{2020}\)
=>\(\dfrac{19}{18}>\dfrac{2021}{2020}\)
d: \(\dfrac{131}{171}=\dfrac{130+1}{170+1}>\dfrac{130}{170}=\dfrac{13}{17}\)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề và hỗ trợ bạn nhanh hơn nhé.