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a) \(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-9}{4}\right):\frac{3}{7}\)
= \(\left(-\frac{3}{4}+\frac{2}{5}\right)\cdot\frac{7}{3}+\left(\frac{3}{5}+\frac{-9}{4}\right)\cdot\frac{7}{3}\)
= \(\left(-\frac{15}{20}+\frac{8}{20}\right)\cdot\frac{7}{3}+\left(\frac{12}{20}-\frac{45}{20}\right)\cdot\frac{7}{3}\)
= \(-\frac{7}{20}\cdot\frac{7}{3}-\frac{33}{20}\cdot\frac{7}{3}\)
=\(\frac{7}{3}\cdot\left(-\frac{7}{20}-\frac{33}{20}\right)\)
=\(\frac{7}{3}\cdot\left(-2\right)\)
=\(-\frac{14}{3}\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
Bài 2:
a, S = 1/11 + 1/12 + .. +1/20 với 1/2
SỐ số hạng tổng S: [20 - 11]: 1 + 1 = 10 số
mà 1/11 > 1/20
1/12 > 1/20
.........................
1/20 = 1/20
=> 1/11 + 1/12 + ... + 1/20 > 1/20 . 10 => S > 1/2
b, B = 2015/2016 + 2016/2017 và C = 2015+2016/2016+2017
Dễ dàng ta thấy: C = 4031/4033 < 1
B = 2015/2016 + 2016/2017
B = 2015/2016 + [1/2016 + 4062239/4066272]
B = [2015/2016 + 1/2016] + 4062239/4066272]
B = 1 +4062239/4066272
=> B > 1
Vậy B > C
c, [-1/5]^9 và [-1/25]^5
ta có: 255 = [52]5 = 52.5 = 510 > 59
=> [1/5]9 > [1/25]5
=> [-1/5]9 < [-1/25]5
d, 1/32+1/42+1/52+1/62 và 1/2
ta có: 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 = 1/9 + 1/16 + 1/25 + 1/36
mà: 1/9 < 1/8
1/16 < 1/8
1/25 < 1/8
1/36 < 1/8
=> 1/9+1/16+1/25+1/36 < 1/2
Vậy 1/32+1/42+1/52+1/62 < 1/2
Bài 1:
A = 3/4 . 8/9 . 15/16....2499/2500
A = [1.3/22][2.4/32]....[49.51/502]
A = [1.2.3.4.5...51 / 2.3.4....50][3.4.5...51 / 2.3.4...50]
A = 1/50 . 51/2
A = 51/100
B = 22/1.3 + 32/2.4 + ... + 502/49.51
B = 4/3.9/8....2500/2499
Nhận thấy B ngược A => B = 100/51 [cách tính tương tự tính A]
Bài 2:
a. S = 1/11+1/12+...+1/20 và 1/2
Số số hạng tổng S: [20 - 11]: 1 + 1 = 10 [ps]
ta có: 1/11 > 1/20
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm