Tính Q = \(\left(\dfrac{6}{8}+1\right)\left(\dfrac{6}{18}+1\right)\left(\dfrac{6}{30}+1\right)...\left(\dfrac{6}{10700}+1\right)\)
BT2: Tính nhanh
7) \(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{71}-\left(-\dfrac{2}{7}\right)+\dfrac{4}{35}-\dfrac{7}{18}\)
8)\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
Tính giá trị của biểu thức sau:
B=\(\left(-1\dfrac{1}{6}\right):\left(-3\dfrac{1}{3}+2\dfrac{1}{4}\right)-\left(-\dfrac{3}{8}\right):\left(8-6\dfrac{3}{8}\right)\)
B = \(\left(-1\dfrac{1}{6}\right)\) : \(\left(\dfrac{-10}{3}+\dfrac{9}{4}\right)\) - \(\left(-\dfrac{3}{8}\right)\) : \(\left(8-\dfrac{51}{8}\right)\)
B = \(\dfrac{-7}{6}\) : \(\dfrac{-13}{12}\) - \(\left(-\dfrac{3}{8}\right)\) : \(\dfrac{13}{8}\)
B = \(\dfrac{14}{13}\) - \(\dfrac{-3}{13}\)
B = \(\dfrac{17}{13}\)
Tính :
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)....\left(29^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right).....\left(30^4+\dfrac{1}{4}\right)}\)
Ta có một số phân tích sau : \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)
Nhân mỗi biểu thức trong ngoặc ở cả tử thức với \(16\)\(=\)\(2^4\), ta được :
\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)
\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)
Kết hợp với phân tích nêu trên, khi đó :
\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)
\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)
ĐỀ 1:
1. Tính:
a. \(\dfrac{7}{23}\left[\left(\dfrac{-8}{6}\right)-\dfrac{45}{18}\right]\)
b. \(\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)\)
c. \(\dfrac{3}{5}.\left(\dfrac{-8}{3}\right)-\dfrac{3}{5}:\left(-6\right)\)
d. \(\dfrac{1}{2}\left(\dfrac{4}{3}+\dfrac{2}{5}\right)-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)\)
e. \(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
2. Tìm x, biết:
a. \(1\dfrac{2}{5}x+\dfrac{3}{7}=\dfrac{4}{5}\)
b. |x - 1,5| = 2
c. \(\dfrac{4}{5}-\left|x-\dfrac{1}{6}\right|=\dfrac{2}{3}\)
d. 3x . 2x = 216
e. \(\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)+\dfrac{-1}{2}=\dfrac{3}{4}\)
* Lm nhanh nha
Ths you <3
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Tính giá trị các biểu thức sau:
a, A = \(\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
b, B= \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) c, C = \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{210}\right)\)
\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
a, \(\dfrac{1}{24}-\left\{\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{ }8\right)\right\}\)
b,\(\left(\dfrac{5}{7}-\dfrac{7}{5}\right)-\left\{\dfrac{1}{2}\left(\dfrac{2}{7}-\dfrac{1}{10}\right)\right\}\)
c,\(3-\left(\dfrac{-6}{7}\right)^6+\left(\dfrac{1}{2}\right)^2:2\)
d,\(\left(5^{-5}\right)^{-1}.\left(\dfrac{1}{2}\right)^2.\dfrac{1}{10^5}\)
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
a, \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}.\sqrt{\dfrac{49}{4}}\right)\): \(\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]\): \(\dfrac{1704}{445}\)
b, \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)
c, \(\left(1-\dfrac{1}{2}\right)\)x\(\left(1-\dfrac{1}{3}\right)\)x.....x\(\left(1-\dfrac{1}{n+1}\right)\) (n ϵ N)
d, -66 x \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)\) + 124 x -37 + 63 x -124
e, \(\dfrac{7}{4}\) x \(\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)