Tính nhanh hộ mình nhé:
\(\left(1999x1998+1998x1997\right)x\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)
rút gọn C=\(\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\)
làm nhanh hộ mình nhé, mình đang gấp
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\) Tìm x
Giải nhanh hộ mình nhé
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x+\left(-1\right)-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=1-\frac{1}{3}-\frac{5}{6}\)
\(x.\left(-5-\frac{1}{2}-\frac{3}{2}\right)=\frac{-1}{6}\)
\(x.\left(-7\right)=\frac{-1}{6}\)
x=\(\frac{1}{42}\)
\(b,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+...+\left(x+\frac{1}{1024}\right)=1\)
\(a,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(c,\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{x.x+3}=\frac{34}{103}\)
\(d,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
HỘ MÌNH VỚI . GẤP . MAI MÌNH ĐI HỌC RỒI . THANKS TRƯỚC NÀ . MÌNH SẼ TICK CHO BẠN NÀO TRẢ LỜI ĐÚNG NHẤT VÀ NHANH NHẤT NHÉ
a) số số x là 4 nên ta có:
(x.4)+1/2+1/4+1/8+1/16=1 mà 1/2+1/4+1/8+1/16=15/16 nên x=1-15/16=1/16:4=1/64
Tính M = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{20^2}-1\right)\)
Ai trả lời đúng và nhanh nhất mình sẽ tích nhé !
Mình chưa học lớp 7
Mình mới học lớp 5 thôi
Xin lỗi nha
Tìm x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
giải nhanh hộ mình với, mai mình nộp rồi
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}:3\)
\(\Rightarrow x=\frac{1}{18}\)
Tính \(A=\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{2}{9}\right).....\left(1-\frac{2}{2011}\right)\)
- giải hộ mình ạ :< Nhanh + đúng = Tick -.-
\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)\cdot\cdot\cdot\left(1-\frac{2}{2011}\right)\)
\(A=\left(\frac{5-2}{5}\right)\left(\frac{7-2}{7}\right)\left(\frac{9-2}{9}\right)\cdot\cdot\cdot\left(\frac{2011-2}{2011}\right)\)
\(A=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot\cdot\cdot\frac{2009}{2011}\)(các thừa số trên tử giống dưới mẫu mình lượt bỏ đi nhé!)
\(A=\frac{3}{2011}\)
\(A=\left(1-\frac{2}{5}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{2}{9}\right)...\left(1-\frac{2}{2011}\right)\)
\(=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}...\frac{2009}{2011}\)
\(=\frac{3}{2011}\)
\(A=\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{2}{9}\right).....\left(1-\frac{2}{2011}\right)\)
\(A=\frac{3}{5}.\frac{5}{7}.\frac{7}{9}........\frac{2009}{2011}\)
\(A=\frac{3}{2011}\)
tính bằng cách thuận tiện
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x\left(1-\frac{1}{6}\right)\))
giúp mik nhé ai nhanh mik k
(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6)
= 1/2 x 2/3 x 3/4 x 4/5 x 5/6
= 1/6 (tử và mẫu triệt tiêu cho nhau)
Tĩm x:
1,\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+5}\)
2,\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}+......+x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Giải chi tiết hộ mình nha mình tik cho
a ) Tính \(A=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
b ) Tìm x biết \(\left|\left|3x-3\right|+2x+\left(-1\right)^{2016}\right|=3x+2017^0\)
Nhanh lên nhé các bạn , các bạn học giỏi vào giúp mình ạ !!