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Lời giải:
a,Ta có: \(\frac{33}{a}=\frac{45}{-120}=\frac{-y}{8}=\frac{z}{160}=\frac{3}{-8}\)
Do: \(\frac{33}{a}=\frac{3}{-8}\Rightarrow-8.33=3.a\Leftrightarrow-264=3.a\Leftrightarrow a=-88\)
\(\frac{-y}{8}=\frac{3}{-8}\Rightarrow8.y=3.8\Leftrightarrow y=3\)
\(\frac{z}{160}=\frac{3}{-8}\Rightarrow-8.z=3.160\Leftrightarrow-8.z=480\Leftrightarrow z=-60\)
Vậy: \(a=-88\) ; \(y=3\) ; \(z=-60\)
b, Ta có: \(\frac{x+1}{5}=\frac{y}{20}=\frac{6}{10}=\frac{3}{5}\)
Do: \(\frac{x+1}{5}=\frac{3}{5}\Rightarrow\left(x+1\right)5=3.5\Leftrightarrow x+1=3\Leftrightarrow x=2\)
\(\frac{y}{20}=\frac{3}{5}\Rightarrow y.5=3.20\Leftrightarrow y.5=60\Leftrightarrow y=12\)
Vậy: \(x=2\) ; \(y=12\)
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a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)
\(\Rightarrow x=-25;y=-35;z=-20\)
b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)
\(\Rightarrow x=-25;y=20;z=35\)
a) Ta có \(\frac{x}{x+5}=\frac{x}{x+y}\)
\(\Rightarrow x.\left(x+y\right)=x.\left(x+5\right)\Rightarrow x^2+xy=x^2+5y\Rightarrow xy=5x\)
\(\Rightarrow xy-5x=0\Rightarrow x.\left(y-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\y=5\end{cases}}\)
Vậy x = 0 hoặc y = 5
b) \(\frac{x-7}{y-8}=\frac{7}{8}\)
\(\Rightarrow\left(x-7\right).8=7.\left(y-8\right)\Rightarrow8x-56=7y-56\Rightarrow8x=7y\)(1)
Từ x - y = 4 nên x = y + 4 . Thay x = y + 4 vào (1) ta có :
\(8.\left(y+4\right)=7y\Rightarrow8y+32=7y\Rightarrow y=-32\)
Do đó x = - 28
Vậy x = -28 ; y = -32
a) Ta có: \(\frac{x}{x+5}=\frac{x}{x+y}\)
\(\Leftrightarrow\frac{x}{x+5}=\frac{x}{x+5}\Rightarrow y=5\)
\(\Leftrightarrow\frac{x}{x+5}=\frac{x}{x}.\frac{x}{5}\)
Đặt mẫu số chung là 5. Ta quy đồng phân số:\(\frac{x}{x}=\frac{x.5}{x.5}=\frac{5}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\y=5\end{cases}}\)
b) Ta có: \(\frac{x-7}{y-8}=\frac{7}{8}\Rightarrow\frac{x}{y}=\frac{7}{8}+\frac{7}{8}=\frac{14}{8}\)
Mà \(\frac{x}{y}=\frac{14}{8}\Rightarrow\orbr{\begin{cases}x=14\\y=8\end{cases}}\)
a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...
a, \(\frac{17}{y}=\frac{-7}{11}\)
\(\Rightarrow17\cdot11=-7\cdot y\)
\(\Rightarrow187=-7\cdot y\)
\(\Rightarrow\frac{187}{-7}=y\)
b, \(\frac{-8}{3x-1}=\frac{4}{7}\)
\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)
\(\Rightarrow3x-1=-14\)
\(\Rightarrow3x=-14+1\)
\(\Rightarrow3x=-13\)
\(\Rightarrow x=\frac{-13}{3}\)
c, \(\frac{x}{-3}=\frac{-3}{x}\)
\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x^2=\left(\pm3\right)^2\)
\(\Rightarrow x=\pm3\)
d, \(\frac{-4}{y}=\frac{x}{2}\)
\(\Rightarrow-4\cdot2=x\cdot y\)
\(\Rightarrow-8=x\cdot y\)
\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
ta có bảng :
| x | -1 | -8 | -2 | -4 |
| y | 8 | 1 | 4 | 2 |
a)\(\frac{14}{y}\)\(=\) \(\frac{-7}{11}\)
\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)
\(y=\)\(\frac{14\cdot11}{-7}\)
\(y=22\)
c) \(\frac{x}{-3}\) = \(\frac{-3}{x}\)
\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)
\(\Rightarrow\)\(x^2=9\)
\(\Rightarrow\)\(x^2=9\)hoặc \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)
\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)
=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)
Bài 1:
a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)
\(xy\) = 12
12 = 22.3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}
Lập bảng ta có:
| \(x\) | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| y | -1 | -2 | -3 | -4 | -6 | -12 | 12 | 6 | 4 | 3 | 2 | 1 |
Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)
b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)
\(x\) = \(\dfrac{2}{7}\).y
\(x\) \(\in\)z ⇔ y ⋮ 7
y = 7k;
\(x\) = 2k
Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
a)45/120=3/8=y/8->y=3
3/8=33/88->x=88