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\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)
\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)
\(\Rightarrow x=165;y=20;z=25\)
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 \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
\(\frac{x}{-7}=\frac{5}{-35}\)
\(\frac{x.5}{-35}=\frac{5}{-35}\)
=> x . 5 = 5
x = 5 : 5
x = 1
Bài 1:
\(S=4\left(\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+...+\dfrac{1}{43\cdot49}\right)\)
\(=\dfrac{4}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{48}{49}=\dfrac{96}{147}=\dfrac{32}{49}\)
Bài 3:
Theo đề, ta có:
\(\dfrac{a}{b}=\dfrac{a+10}{b+10}\)
=>ab+10a=ab+10b
=>10a=10b
=>a/b=1
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 \(x^2=-9\)
\(TH1:\) \(x^2=9\)
\(\Rightarrow\)\(x=3\)
\(TH2:\)\(x^2=-9\)
\(\Rightarrow\)\(x=-3\)
Vì -24:-6=4
mà -24:-6=x:3=4:y^2=z^3:-2
Suy ra x=4x3=12
y^2=4:4=1; y=1
z^3=4x-2=-8;z=-2
\(b,\frac{z}{7}=-\frac{11}{-28}\)
\(\Leftrightarrow z.\left(-28\right)=-11.7\)
\(\Leftrightarrow z.\left(-28\right)=-77\)
\(\Leftrightarrow z=\frac{11}{4}\)
\(a,-\frac{2}{3}=\frac{x-3}{-6}=\frac{10}{5-y}=\frac{4-2z}{9}\)
Xét :
\(-\frac{2}{3}=\frac{x-3}{-6}\)
\(\Leftrightarrow-2.\left(-6\right)=\left(x-3\right).3\)
\(\Leftrightarrow12=\left(x-3\right).3\)
\(\Leftrightarrow4=x-3\Leftrightarrow x=7\)
Xét
\(-\frac{2}{3}=\frac{10}{5-y}\)
\(\Leftrightarrow-2.\left(5-y\right)=10.3\)
\(\Leftrightarrow-10+2y=30\)
\(\Leftrightarrow2y=40\Leftrightarrow y=20\)
Xét :
\(-\frac{2}{3}=\frac{4-2z}{9}\)
\(\Leftrightarrow-2.9=\left(4-2z\right).3\)
\(\Leftrightarrow-18=\left(4-2z\right).3\)
\(\Leftrightarrow-6=4-2z\)
\(\Leftrightarrow10=2z\Leftrightarrow z=5\)
Vậy \(\left(x;y;z\right)=\left(7;20;5\right)\)
a)\(\frac{-2}{3}=\frac{x-3}{6}=\frac{10}{5-y}=\frac{4-2z}{9}\)
Ta có :
Vậy \(\left(x,y,z\right)\in\left(7,-20,5\right)\)
b) \(\frac{z}{7}=\frac{-11}{-28}\)
Ta có : \(\frac{z}{7}=\frac{11}{28}\)
=> \(28z=77\)
=> \(z=\frac{77}{28}=\frac{11}{4}\)
\(-\frac{2}{3}=\frac{x-3}{-6}=\frac{10}{5-y}=\frac{4-2z}{9}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{x-3}{-6}\Leftrightarrow\left(x-3\right)3=-2.\left(-6\right)\)
\(\Leftrightarrow\left(x-3\right)3=12\)
\(\Leftrightarrow x-3=4\)
\(\Leftrightarrow x=7\)
\(\Leftrightarrow-\frac{2}{3}=\frac{10}{5-y}\Leftrightarrow-2.\left(5-y\right)=3.10\)
\(\Leftrightarrow-2.\left(5-y\right)=30\)
\(\Leftrightarrow5-y=30:\left(-2\right)\)
\(\Leftrightarrow5-y=-15\)
\(\Leftrightarrow y=5-\left(-15\right)\)
\(\Leftrightarrow y=10\)
\(\Leftrightarrow-\frac{2}{3}=\frac{4-2z}{9}\Leftrightarrow3.\left(4-2z\right)=-2.9\)
\(\Leftrightarrow3.\left(4-2z\right)=-18\)
\(\Leftrightarrow4-2z=-6\)
\(\Leftrightarrow2z=4-\left(-6\right)\)
\(\Leftrightarrow2z=10\)
\(\Leftrightarrow z=5\)
vậy ...
b)\(\frac{z}{7}=\frac{-11}{-28}\)
\(\frac{z}{7}=\frac{11}{28}\)
\(\Leftrightarrow z=7.11:28=\frac{11}{4}\)
Mun Già
bạn ơi mk ko ỉu chổ bạn làm \(\frac{10}{5-y}\) cho lắm sao lạ =-20
thay vào thử đi sai rồi
Ây da,\(\frac{-2}{3}=\frac{10}{5-y}\)=> \(-2\left(5-y\right)=30\)=> \(5-y=-15\)=> y = \(5-\left(-15\right)=20\)
ღ Không TênッNhạt ᴾ漫·舞тoáɴ๖ۣۜнọc 5 - (-15) = 5 + 15 = 20 nhé chứ không phải bằng 10,như thế nó ra kết quả bằng -2 mà \(-2\ne-\frac{2}{3}\)nhé.
ây da tui cx sai
\(y=5-\left(-15\right)\)
\(y=20\)
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