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a)\(\left(\frac{2}{3}+\frac{2}{5}\right)x=\frac{1}{5}-2\frac{1}{2}\)
\(\frac{16}{15}x=\frac{1}{5}-1\)
\(\frac{16}{15}x=-\frac{4}{5}\)
\(x=-\frac{4}{5}\div\frac{16}{15}\)
\(x=-\frac{3}{4}\)
b)\(\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}\)
\(\frac{4}{7}x=\frac{1}{5}+\frac{2}{3}\)
\(\frac{4}{7}x=\frac{13}{15}\)
\(x=\frac{13}{15}\div\frac{4}{7}\)
\(x=\frac{91}{60}\)
\(\left(\frac{2}{3}+\frac{1}{5}\right)\)CHỨ HK PHẢI LÀ \(\left(\frac{2}{3}+\frac{2}{5}\right)\)ĐÂU Ạ
CHO MK XIN LỖI VÌ GHI SAI ĐẦU BÀI
Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)
Ta có: \(\frac{1}{2^2}< \frac{8}{9}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3}-\frac{1}{4}\)
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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{8}{9}-\frac{1}{9}=\frac{7}{9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{8}{9}\)(1)
Đảo ngược công thức trên lại,ta lại có: \(\frac{1}{a+1}+\frac{1}{a}=\frac{1}{\left(a+1\right)a}< \frac{1}{a.a}=\frac{1}{a^2}\)
SAu đó bạn làm tương tự như trên sẽ được . Giờ mình bận rồi=)))
Đây là toán nhé =))
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1) và (2) => \(\frac{2}{5}< S< \frac{8}{9}\)
Ta có:\(A=\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)
Để\(A\inℤ\Leftrightarrow\frac{3}{n+2}\inℤ\)
\(\Leftrightarrow n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow n=\left\{-1;-3;1;-5\right\}\)
DKXD \(n\ne-2\)
A=\(\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)
=> A nguyen<=> \(\frac{3}{n+2}\) nguyen ma n lai la so nguyen theo gt
=> \(\left(n+2\right)\in U\left(3\right)\)
=> n+2 = 1
n+2=3
n+2=-1
n+2=-3
=> n=-1
n=1
n=-3
n=-5
tatca deu TMDKXD
Vay \(n\in\left(-5;-3;-1,1\right)\)
Chuc ban hoc tot
a, \(x-\frac{5}{6}=\frac{-2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
b, \(\frac{-7}{5}+x=\frac{-4}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
c, \(x-\frac{2}{5}=-\frac{1}{6}-\frac{3}{-4}\)
\(\Leftrightarrow x-\frac{2}{5}=-\frac{1}{6}+\frac{3}{4}\)
\(\Leftrightarrow x-\frac{2}{5}=\frac{7}{12}\Leftrightarrow x=\frac{59}{60}\)
\(\frac{3}{8}+\left(17-5\right)^4=\frac{3}{8}+12^4=\frac{3}{8}+20736=\frac{165891}{8}\)
\(\frac{3}{8}+\left(17-5\right)^4\)
\(=\frac{3}{8}+\left(2\right)^4\)
\(=\frac{3}{8}+16\)
\(=\text{0.375}+16\)
\(=16.375\)
KHÔNG VÀO NHÓM NHƯNG CHƠI FF
KB ĐI: 1874233559 ( TUI RANK HUYỀN THOẠI )
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)>\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\right)\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\)và \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{9}{10}\)nên \(D< 1\)
Bạn ghi đề sai rồi nhé.
a/ \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow3\left(x-1\right)=72\)
\(\Leftrightarrow x-1=24\)
\(\Leftrightarrow x=25\)
Vậy ..
b/ \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x^2=6^2=\left(-6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy ..
c/ \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x\left(x+1\right)=72\)
\(\Leftrightarrow x\left(x+1\right)=8.9\)
\(\Leftrightarrow x=8\)
Vậy ..
\(\left(\frac{1}{2}-\frac{3}{4}\right):\frac{1}{2}+\left(\frac{1}{2}\right)^2\)
\(=\left(\frac{2}{4}-\frac{3}{4}\right):\frac{1}{2}+\frac{1}{4}\)
\(=\frac{-1}{4}:\frac{1}{2}+\frac{1}{4}\)
\(=\frac{-1}{4}.\frac{2}{1}+\frac{1}{4}\)
\(=\frac{-1}{2}+\frac{1}{4}\)
\(=\frac{-2}{4}+\frac{1}{4}\)
\(=\frac{-1}{4}\)
\(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\)
\(\Rightarrow A>\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{50\cdot51}\)
\(\Rightarrow A>\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow A>\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)
Mà \(\frac{16}{51}>\frac{1}{4}\Rightarrow A>\frac{16}{51}>\frac{1}{4}\Rightarrow A>\frac{1}{4}\)