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Ta có: \(\frac{-4}{8}=\frac{-1}{2}=\frac{x}{-10}\)
\(\Rightarrow x=\frac{\left(-10\right).\left(-1\right)}{2}=5\)
Thay x = 5 được \(\frac{5}{-10}=\frac{-1}{2}=\frac{-7}{y}\)
\(\Rightarrow y=\frac{\left(-7\right).2}{-1}=14\)
Thay y = 14 được \(\frac{-7}{14}=\frac{-1}{2}=\frac{z}{-2}\)
\(\Rightarrow z=\frac{\left(-2\right).\left(-1\right)}{2}=1\)
Vậy x = 5 ; y = 14 và z = 1
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 :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
a. \(\frac{x}{9}< \frac{7}{x}\)=> \(x.x< 9.7\)
=> \(x^2< 63\)
\(\frac{7}{x}< \frac{x}{6}\)=> \(7.6< x.x\)
=> \(42< x^2\)
Vậy \(42< x^2< 63\)
=> \(x^2=49\)
=> \(x=7\)
b. \(\frac{3}{y}< \frac{y}{7}\)=> \(7.3< y.y\)
=> \(21< y^2\)
\(\frac{y}{7}< \frac{4}{y}\)=> \(y.y< 4.7\)
=> \(y^2< 28\)
Vậy \(21< y^2< 28\)
=> \(y^2=25\)
=> \(y=5\)
\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\frac{8}{99}\)
\(A=\frac{4}{33}\)
b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)
\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)
c] Ta đặt \(\left(8n+5,6n+4\right)=d\)
\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)
Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản
\(\frac{x}{2}-\frac{3}{2}=\frac{10}{y}-\frac{x}{y}\)
\(\frac{x-3}{2}=\frac{10-x}{y}\)
\(\Leftrightarrow\left(x-3\right)\cdot y=2\cdot\left(10-x\right)\)
\(xy-3y=20-2x\)
\(xy+2x-3y-6=14\)
\(x\left(y+2\right)-3\left(y+2\right)=14\)
\(\left(y+2\right)\left(x-3\right)=14\)
\(y+2\) | -1 | 1 | -2 | 2 | -7 | 7 | -14 | 14 |
\(x-3\) | -14 | 14 | -7 | 7 | -2 | 2 | -1 | 1 |
\(y\) | -3 | -1 | -4 | 0 | -9 | 5 | -16 | 12 |
\(x\) | -11 | 17 | -4 | 10 | 1 | 5 | 2 | 4 |
Vậy các cặp (x;y) là: (-11;-3) (17;-1) (-4;-4) (10;0) (1;-9) (5;5) (2;-16) (4;12)
Quy đồng ta thấy 28<x^2<40 => x^2=36 hay x=6
Ta có \(\frac{7}{x}< \frac{x}{4}< \frac{10}{x}\Rightarrow\frac{28}{4x}< \frac{x^2}{4x}< \frac{40}{4x}\Rightarrow28< x^2< 40\Rightarrow5^2\le x^2\le6^2\)\(\Rightarrow x^2=5^2;6^2\)
+) nếu x^2=5^2 thì x=5;-5
+) nếu x^2=6^2 thì x=6;-6
vậy x=5;-5;6;-6