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1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
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\(0,81:\frac{x}{2}=\frac{16}{x^4}:\left(-0.9\right)\)
=> \(\frac{81}{100}:\frac{x}{2}=\frac{16}{x^4}:\frac{-9}{10}\)
=> \(\frac{81}{50x}=\frac{160}{-9.x^4}\)
=> \(81.\left(-9\right)x^4=50x.160\)
=> \(-729.x^4=8000.x\)
=> \(x^4:x=8000:\left(-729\right)\)
=> \(x^3=\frac{-8000}{729}\)
=> \(x^3=\frac{-20^3}{9^3}\)
=> \(x^3=\frac{-20}{9}^3\)
=> \(x=\frac{-20}{9}\)
\(\frac{x-5}{3}=\frac{6}{5}\)
\(x-5=\frac{6}{5}.3\)
\(x-5=\frac{18}{5}\)
\(x=\frac{18}{5}+5\)
\(\Rightarrow x=\frac{43}{5}\)
Vậy ...
\(\frac{1}{3}=\frac{2-x}{4}\)
\(\frac{4}{3}=2-x\)
\(x=2-\frac{4}{3}\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy ...
a) \(\frac{22}{7}\div\left(11-\chi\right)=\frac{7}{5}-\frac{2}{3}\)
\(\frac{22}{7}\div\left(11-\chi\right)=\frac{11}{15}\)
\(\left(11-\chi\right)=\frac{22}{7}\div\frac{11}{15}\)
\(\left(11-\chi\right)=\frac{30}{7}\)
\(\chi=11-\frac{30}{7}\)
\(\chi=\frac{47}{7}\)
b) (x+1)+(x+2)+(x+3)+...+(x+100)=5550
Từ 1 đến 100 có 100 số hạng => Có 100 x
(x + x + x + .... + x) + (1 + 2 + 3 + .. + 100) = 5550
Áp dụng tính chất cộng dãy số cách đều, ta có
(100.x) + 5050 = 5550
100.x = 5550 - 5050
100.x = 500
x = 500 : 100
x = 5
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)
\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)
\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)
\(\left(2x-1\right)=-\frac{4}{63}\)
2x= -4/63 + 1
2x = 59/63
x = 59/63 : 2
x = 59/126
1/3:(2.x-1)=-5-1/4
1/3:(2.x-1)=-21/4
2.x-1=1/3:-21/4
2.x-1=-4/63
2.x=-4/63+1
2.x=\(3\frac{59}{63}\)
x=\(3\frac{59}{63}\):2
x=\(1\frac{61}{63}\)
(1/1×2 + 1/2×3 + ... + 1/9×10) × x < 2/1×3 + 2/3×5 + ... + 2/9×11
(1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10) × x < 1 - 1/3 + 1/3 - 1/5 + ... + 1/9 - 1/11
(1 - 1/10) × x < 1 - 1/11
9/10 × x < 10/11
x < 10/11 : 9/10
x < 10/11 × 10/9
x < 100/99
Mà x là số tự nhiên => x = 0 hoặc 1
Ta có :
\(\frac{-16}{32}=\frac{-16:16}{32:16}=\frac{-1}{2}\)
+)\(\frac{-1}{2}=\frac{x}{-10}\)
=> (-10) x (-1) = X x 2
=> 10 = X x 2
=> X = 10 : 2
=> X = 5
+) \(\frac{-1}{2}=\frac{-7}{y}\)
=> (-1) x Y = (-7) x 2
=> -Y = -14
=> Y = 14
+)\(\frac{-1}{2}=\frac{z}{24}\)
=> (-1) x 24 = Z x 2
=> -24 = Z x 2
=> Z = -24 : 2
=> Z = -12
Kết luận : X = 5
Y = 14
Z = 12
Có:16(x-3)=24x=>16x-48=24x=>x=-6
16(x-3)=24x
<=> 16x-48= 24x
<=>-8x=48
<=>x= -6