\(\left(x-1\right)^{x+2}=1\)
b)\(\left(x+\frac{1}{2}\right)^4=1\)
\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{2-2\sqrt{x}+2+2\sqrt{x}}{\left(2+2\sqrt{x}\right)\left(2-2\sqrt{x}\right)}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{4}{4-4x}-\frac{x^2+1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{1+x-x^2-1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)=\frac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}.\frac{x+1}{x}=1\)
Nếu bạn bảo kiểm tra thì lời giải đúng rồi nhé!
a) Chứng minh: \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x\left(x+1\right)}\)
b). Tính nhẩm tổng sau: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
Bài 4.1: Tìm x, biết
a) \(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\)
b) \(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
c) \(\left|2\frac{1}{5}-x\right|+\left|x-\frac{1}{5}\right|+8\frac{1}{5}=1,2\)
d) \(2\left|x+3\frac{1}{2}\right|+\left|x\right|-3\frac{1}{2}=\left|2\frac{1}{5}-x\right|\)
Giải phương trình \(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)}+\frac{1}{\left(x^2+1\right)}\)
AYUASGSHXHFSGDB HAGGAHAJF
a) chứng minh: \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x\left(x+1\right)}\)
b) tính nhẩm tổng sau:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
a,\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b,Áp dụng câu a:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
a) Chứng minh: \(\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x\left(x+1\right)}\)
b) Đố: Đố bạn tính nhẩm được tổng sau:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
a)
\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
b) S =\(\frac{1}{x}-\frac{1}{x+5}+\frac{1}{x+5}=\frac{1}{x}\)
Tính:
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}+\frac{1}{\left(x+3\right).\left(x+4\right)}+\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{x+5}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
Giải phương trình :
a) \(\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)
b) \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)^2\left(x+\frac{1}{x}\right)^2=\left(x-4\right)^2\)
\(\frac{x^2}{\left(x+2\right)}=3x^2-6x-3,x\ne-2\)
\(\Rightarrow x^2=\left(3x^2-6x-3\right)\left(x+2\right)^2\)
\(\Rightarrow x^2-\left(3x^2-6x-3\right)\left(x+2\right)^2=0\)
\(\Rightarrow x^2-\left(3x^4+12x^3+12x^2-6x^3-24x^2-24x-3x^2-12x-12\right)=0\)
\(\Rightarrow x^2-\left(3x^4+6x^3-15x^2-36x-12\right)=0\)
\(\Rightarrow16x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-2x^2+18x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-x^2\left(3x^2+6x+2\right)+\left(3x^2+6x+2\right)=0\)
\(\Rightarrow-\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(3x^2+6x=2\right)=0\\x^2-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{-3+\sqrt{3}}{3}\\\frac{-3-\sqrt{3}}{3},x\ne-2\\x=-\sqrt{6}\\x=\sqrt{6}\end{matrix}\right.\)
Rút gọn \(B=\left(x^4-x+\frac{x-3}{x^3+1}\times\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right)\times\frac{4x^2+6x+1}{\left(x+3\right)\left(4-x\right)}\)
Tính nhanh tổng sau:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\)
quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6
=1\x-1\x+6
=6\x(x+6)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)