Tính:

b) \(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)

\(=\left(-2\right)^3.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

\(=\left(-8\right).\frac{1}{2}:\frac{13}{12}\)

\(=\left(-4\right):\frac{13}{12}\)

\(=-\frac{48}{13}.\)

Tìm x

a) \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

⇒ \(\frac{13}{3}:\frac{x}{4}=20\)

⇒ \(\frac{x}{4}=\frac{13}{3}:20\)

⇒ \(\frac{x}{4}=\frac{13}{60}\)

⇒ \(x.60=4.13\)

⇒ \(x.60=52\)

⇒ \(x=52:60\)

⇒ \(x=\frac{13}{15}\)

Vậy \(x=\frac{13}{15}.\)

b) \(\left(2^3:4\right).2^{x+1}=64\)

⇒ \(\left(8:4\right).2^{x+1}=64\)

⇒ \(2.2^{x+1}=64\)

⇒ \(2^{x+1}=64:2\)

⇒ \(2^{x+1}=32\)

⇒ \(2^{x+1}=2^5\)

⇒ \(x+1=5\)

⇒ \(x=5-1\)

⇒ \(x=4\)

Vậy \(x=4.\)

Chúc bạn học tốt!

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

Bài 3:

a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)

2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)

2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)

3A = \(1-\frac{1}{2^6}\)

=> 3A < 1 

=> A < \(\frac{1}{3}\)(đpcm)

b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)

4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)       (1)

Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)

4B = \(3-\frac{1}{3^{99}}\)

=> 4B < 3

=> B < \(\frac{3}{4}\)   (2)

Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)

bài 1:

5n+7 chia hết cho 3n+2

=> [3(5n+7) - 5(3n + 2)] chia hết cho 3n+2

=> (15n + 21 - 15n - 10) chia hết cho 3n+2

=> 11 chia hết cho 3n + 2

=> 3n + 2 thuộc Ư(11) = {1;-1;11;-11}

Ta có bảng:

Vậy n = {-1;3}

Bài 2:

1, chữ số tận cùng

a, Xét 7¹⁹⁹⁹

Ta có: 7¹⁹⁹⁹ = 7¹⁹⁹⁶.7³ = (7⁴)⁴⁹⁹.343 = (...1)⁴⁹⁹.343 = (....1).343 = ....3 (1)

Vậy số 57¹⁹⁹⁹ có tận cùng là 3

b, Xét 3¹⁹⁹⁹

Ta có: 3¹⁹⁹⁹ = 3¹⁹⁹⁶.3³ = (3⁴)⁴⁹⁹.27 = (...1)⁴⁹⁹.27 = (...1) . 27 = ....7  (2)

Vậy số 93¹⁹⁹⁹ có chữ số tận cùng là 7

2, 

Từ (1) và (2) suy ra A = 999993¹⁹⁹⁹ + 555557¹⁹⁹⁹ = ...7 + ...3 = ....0

Vì A có chữ số tận cùng là 0 nên A chia hết cho 5. 

\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)

\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)

\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)

\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)

\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)

\(=3,75.\left(7,2+2,8\right)\)

\(=3,75.10=37,5\)

\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)

\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)

\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)

\(=\frac{-3}{7}+-\frac{4}{7}=-1\)

\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)

\(=9-\frac{1}{8}.8+0,2\)

\(=9-1+0,2=8+0,2=8,2\)

\(a-c\left(tựlm\right)\)

\(b.\left(x-1\right)^5=-32\)

\(\Rightarrow\left(x-1\right)^5=\left(-2\right)^5\)

\(\Rightarrow x-1=-2\)

\(x=-2+1=-1\)

\(d.\left(2^3:4\right).2^{x+1}=64\)

\(2.2^{x+1}=64\)

\(\Rightarrow2^{1+x+1}=64=2^6\)

\(\Rightarrow2+x=6\Rightarrow x=6-2=4\)

Gọi ba cạnh của tam giác lần lượt là a, b, c

Ta có a/3 = b/4 = c/5 và a + b + c = 13,2 (cm)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

a/3 = b/4 = c/5 = ( a + b + c ) / (3 + 4 + 5) = 13,2 / 12 = 11/10 (cm)

Suy ra:

a/3 = 11/10 => a = 11/10 x 3 = 3,3 (cm)

b/4 = 11/10 => b = 11/10 x 4 =4,4 (cm)

c/5 = 11/10 => c = 11/10 x 5 =5,5 (cm)

Vậy độ dài ba cạnh của tam giác là 3,3 ; 4,4 ; 5,5 (cm)

Bài 1 : Ta có:

\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)

= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)

= \(\frac{7}{9}\)

Bài 2 :

 \(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)

=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)

=> 50x = 10

=> x = 10 : 50

=> x = 1/5

Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3

                                         <=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy 

a) Ta có: \(\left(x-2\right)^3+\frac{8}{27}=0\)

\(\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}\)

\(\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Leftrightarrow x-2=\frac{-2}{3}\)

hay \(x=\frac{-2}{3}+2=\frac{4}{3}\)

Vậy: \(x=\frac{4}{3}\)

b) Ta có: \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

\(\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20\)

\(\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}\)

hay \(x=\frac{13\cdot4}{60}=\frac{13}{15}\)

Vậy: \(x=\frac{13}{15}\)

c) Ta có: \(\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)

\(\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}\)

\(\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}\)

\(\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16\)

\(\Leftrightarrow3x=-160\)

hay \(x=\frac{-160}{3}\)

Vậy: \(x=\frac{-160}{3}\)

d) Ta có: \(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)

\(\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{8}{3};\frac{4}{3}\right\}\)

a/ (x - 2)³ + \(\frac{8}{27}\) = 0

=> (x - 2)³ = 0 - \(\frac{8}{27}\) = \(\frac{-8}{27}\)

=> x - 2 = \(-\frac{2}{3}\)

=> x = \(-\frac{2}{3}+2=\frac{4}{3}\)

b/ \(4\frac{1}{3}:\frac{x}{4}=6:0,3\)

=> \(4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20\)

=> \(\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}\)

=> \(x=\frac{13}{60}.4=\frac{13}{15}\)

c/ \(\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}\)

=> \(\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}\)

=> \(0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}\)

=> \(\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16\)

=> \(x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}\)