Với x\(\ne-1\) \(\left(\dfrac{x^2+2x+2}{x+1}\right)^{2018}=a_0+a_1x+a_2x^2+...+a_kx^{2018}+\dfrac{b_1}{x+1}+\dfrac{b_2}{\left(x+1\right)^2}+...+\dfrac{b_{2018}}{\left(x+1\right)^{2018}}.\). Tính: S=\(\sum\limits^{2018}_{k=1}bx\)
cho khai triển \(\left(\dfrac{x^2+2x+2}{x+1}\right)^{2020}=a_0+a_1x+a_2x^2+...+a_{2020}x^{2020}+\dfrac{b_1}{x+1}+\dfrac{b_2}{\left(x+1\right)^2}+...+\dfrac{b_{2020}}{\left(x+1\right)^{2020}}\) tính tổng \(S=b_1+b_2+...+b_{2020}\)
cho khai triển \(\left(2018x^2+x+2018\right)^{2018}=a_0+a_1x+a_2x^2+...+a_{4036}x^{4036}\)
tính \(T=a_0-a_2+a_4-...-a_{4032}+a_{4036}\)
Thực hiện phép tính:
\(\left(1-\dfrac{1}{x+1}\right)\left(1-\dfrac{1}{x+2}\right)\left(1-\dfrac{1}{x+3}\right)...\left(1-\dfrac{1}{x+2018}\right)\)
\(\left(1-\frac{1}{x+1}\right)\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right)...\left(1-\frac{1}{x+2018}\right)\)
\(=\frac{x}{x+1}.\frac{x+1}{x+2}.\frac{x+2}{x+3}....\frac{x+2017}{x+2018}\)
\(=\frac{x}{x+2018}\)
1. Giải \(a,\sqrt{4}-\sqrt{9x}+\sqrt{25x}=8\) \(b,\sqrt{\dfrac{1}{4x}}+\sqrt{\dfrac{1}{9x}}-\sqrt{\dfrac{1}{36x}}=\dfrac{2}{3}\)
2. \(A=\dfrac{1}{\sqrt{1\cdot2018}}+\dfrac{1}{\sqrt{2\cdot2017}}+...+\dfrac{1}{\sqrt{k\left(2018-k+1\right)}}+...+\dfrac{1}{\sqrt{2018\cdot1}}\)
So sánh A với \(2\cdot\dfrac{2018}{2019}\)
3.Cho abc=201. Tính\(\dfrac{201a}{ab+201+a+201}+\dfrac{b}{cb+b+201}+\dfrac{c}{ac+c+1}\)
4.\(B=\left(\dfrac{1-x^3}{1-x}+x\right)\cdot\left(\dfrac{1+x^3}{1+x}-x\right)\) a, Rút gọn B b, tìm x để B=64
5. Tìm x: \(\left|x-2\right|-2\left|x+1\right|=3-2\left(1-2x\right)\)
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
Biết \(\left(1-3x+3x^2\right)^{2018}\left(1+3x-3x^2\right)^{2019}=a_0+a_1x+a_2x^2+...+a_{8074}x^{8074}.\). Tính giá trị biểu thức: \(A=a_0+a_1+a_2+...+a_{8074}\)
giải phương trình
\(\dfrac{\left(2017-x\right)^2+\left(2017-x\right)\left(x-2018\right)+\left(x-2018\right)^2}{\left(2017-x\right)^2-\left(2017-x\right)\left(x-2018\right)+\left(x-2018\right)^2}=\dfrac{19}{49}\)
Cho hỏi caćh làm ạ!!
Rút gọn:
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.......\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
Dạng này mình làm suốt rồi, bạn cứ yên tâm.
Biết \(\left(x+2\right)^{2019}+\left(x+2\right)^{2018}+...+\left(x+2\right)^{2010}=a_0+a_1x+a_2x^2+...+a_{2019}x^{2019}\). Tính giá trị của biểu thức \(B=a_0-a_1+a_2-...-a_{2019}\)
Tìm GTNN
a) \(y=\sqrt{x^3+2\left(1+\sqrt{x^3+1}\right)}+\sqrt{x^3+2\left(1-\sqrt{x^3+1}\right)}\)
b) \(f\left(x\right)=\dfrac{x}{2}+\dfrac{2}{x-1}\) với x>1
c) \(y=\dfrac{x-2017}{\sqrt{x-2018}}\)
a. ĐKXĐ: \(x\ge-1\)
\(y=\sqrt{x^3+1+2\sqrt{x^3+1}+1}+\sqrt{x^3+1-2\sqrt{x^3+1}+1}\)
\(=\sqrt{\left(\sqrt{x^3+1}+1\right)^2}+\sqrt{\left(\sqrt{x^3+1}-1\right)^2}\)
\(=\left|\sqrt{x^3+1}+1\right|+\left|1-\sqrt{x^3+1}\right|\ge\left|\sqrt{x^3+1}+1+1-\sqrt{x^3+1}\right|=2\)
b.
\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
c.
\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)