\(1+\frac{2}{n\left(n+3\right)}=\frac{n^2+3n+2}{n\left(n+3\right)}=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\)
\(\Rightarrow A=\frac{2.3}{1.4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}...\frac{2020.2021}{2019.2022}\)
\(\Rightarrow A=\frac{2.3.4...2020}{1.2.3...2019}.\frac{3.4.5...2021}{4.5.6...2022}=\frac{2020}{1}.\frac{3}{2022}=\frac{1010}{337}\)
Nhìn bài toán xong còn bạn nào có thể làm cho mình ko
1. x=\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-8\sqrt{2}}}}}-\sqrt{3}\)
2.Chứng minh: a + b + c = 2019 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2019\) thì 1 trong 3 số phải có 1 số bằng 2019
3. Giải
a, \(\left|x-2\right|\cdot\left(x-1\right)\cdot\left(x+1\right)\cdot\left(x+2\right)=4\)
b, \(\frac{15x}{x^2-3x+4}=\frac{12}{x+4}+\frac{4}{x-1}+1\)
Giúp tớ với
a)\(\frac{\left(x-4\right)^2}{7}-\frac{\left(x+4\right)^2}{5}\ge\frac{2\left(x^2-3\right)}{35}\)
b)\(\frac{\left(2x-1\right)^2}{2}-\frac{\left(2x+1\right)^2}{4}< \left(x-3\right)^2\)
c)\(|-5x+1|=6-3x\)
d)\(6\cdot|x+3|=-8x\)
e)\(7\cdot|x+1|=6-x\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{-3}{4}\right)^2\cdot\left(-1\right)^{2019}}{\left(\dfrac{2}{5}\right)^2\cdot\left(\dfrac{-5}{12}\right)^3}\)
Tính hộ mk với
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
cm gtbt:P=\(\dfrac{\left(x^2+a\right)\cdot\left(1+a\right)+a^2\cdot x^2+1}{\left(x^2-a\right)\cdot\left(1-a\right)+a^2\cdot x^2+1}\) không phụ thuộc vào x
Tính GTCBT
\(A=\dfrac{4x^4+1}{4\left(x+1\right)^2}\cdot\dfrac{4\left(x+2\right)^2+1}{4\left(x+3\right)^4+1}\cdot\cdot\cdot\dfrac{4\left(x+10\right)^4+1}{4\left(x+11\right)^4+1}\)
Tại x =19,092014
CMR:
a,\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
b,\(\left(x+y+z\right)^2\ge3\cdot\left(xy+yz+xz\right)\)
Đặt ẩn phụ:
a, \(\left(x^2+3x+1\right)\cdot\left(x^2+3x-3\right)-5\)
b, \(\left(x^2+2x\right)^2-2x^2-4x-3\)
c, \(\left(x+1\right)\cdot\left(x+3\right)\cdot\left(x+5\right)\cdot\left(x+7\right)+15\)
d, \(x^2-2xy+y^2-7x+7y+12\)
Tính:\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2019^2}\right)\)
