a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
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Những câu hỏi liên quan
7 tháng 3 2016
Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9
\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)
b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)
<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)
<=>x=-43/30
Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại
7 tháng 3 2016
a,A=1863/623:2/7
A=1863/178
b,
ta có:
1863/178.x+5/6=-3/4
1863/178.x=-19/12
=>x=-1691/11178
31 tháng 12 2016
Bài 1:
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}=\frac{5}{7}\)
Bài 2:
a) \(\frac{x}{7}+\left(\frac{-3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(\Rightarrow\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)
b) \(\left(x-1\right)^{x+6}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+4}=0\)
\(\Rightarrow\left(x-1\right)^{x+4}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(x-1\right)^{x+1}=0\) hoặc \(\left(x-1\right)^2-1=0\)
+) \(\left(x-1\right)^{x+1}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(x-1\right)^2-1=0\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\left(x-1\right)=\pm1\)
+ \(x-1=1\Rightarrow x=2\)
+ \(x-1=-1\Rightarrow x=0\)
Vậy \(x\in\left\{0;2;1\right\}\)
HN
31 tháng 12 2016
1)
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)
\(=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}\)
\(=\frac{5}{7}\)
2) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
11 tháng 5 2022
Bài 3:
\(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
15 tháng 3 2019
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
PT
28 tháng 9 2017
a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
PT
28 tháng 9 2017
c) \(\left(x+2\right)-\left(x+3\right)>0\)
\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\ge\dfrac{-7}{3}\)
TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\le5\)
TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
28 tháng 9 2017
a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)
\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)
\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)
Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
Vậy x = -5
b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)
\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)
\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)
\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)
\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
\(\Rightarrow x+102=0\Rightarrow x=-102\)
Vậy x = -102
c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3
= x - x + 2 - 3
= -1
mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0
d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)
\(\Rightarrow x\ge\frac{-7}{3}\)
Vậy \(x\ge\frac{-7}{3}\)
24 tháng 10 2017
a, 7/5 : x + 3/2 = 16/3
7/5 : x = 16/3 - 3/2
7/5 : x = 23/6
x = 7/5 : 23/6
x = 42/115
b, x : 1/5 + 1/7 = 3/5 . 18/21
x : 1/5 + 1/7 = 18/35
x : 1/5 = 18/35 - 1/7
x : 1/5 = 13/35
x = 13/35 . 1/5
x = 13/175
c, x - 1 và 1/3 : 2 = 5/7
x - 4/3 : 2 = 5/7
x - 4/3 = 5/7 . 2
x - 4/3 = 10/7
x = 10/7 + 4/3
x = 58/21
d, x + 2 và 3/5 . 1/6 = 35/36
x + 13/5 . 1/6 = 35/36
x + 13/5 = 35/36 : 1/6
x + 13/5 = 35/6
x = 35/6 - 13/5
x = 97/30
e, ( x + 3/2 ) : 2 = 7/10 + 1/5
( x + 3/2 ) : 2 = 9/10
x + 3/2 = 9/10 . 2
x + 3/2 = 9/5
x = 9/5 - 3/2
x = 3/10
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
Bạn có thể viết thay dòng "Từ (1) và (2)" thành "Từ các điều kiện trên" bạn nhé !(bạn ko cần phải sửa, đây chỉ là gợi ý)