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a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
| x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 1 | 1 | -1 | 3 | -3 |
| y + 1 | 3 | -3 | 1 | -1 |
| x | 2 | 0 | 4 | -2 |
| y | 2 | -4 | 0 | -2 |
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
Bài này dễ nên nãy không có định làm, nhưng mà nghĩ lại thôi làm vậy:Đ
a/ \(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{2}{4}\)
Ta có: \(\dfrac{-4}{y}=\dfrac{2}{4}\Rightarrow y=\dfrac{-4.4}{2}=-8\)
\(\dfrac{x}{-2}=\dfrac{-4}{y}=\dfrac{-4}{-8}=\dfrac{1}{2}\Rightarrow x=\dfrac{-2.1}{2}=-1\)
b/\(\dfrac{2}{x}=\dfrac{y}{-3}\Rightarrow xy=-3.2=-6\)
\(\Rightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Ta có bảng giá trị của x, y như sau:
| x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 1 |
c/ \(\dfrac{x+1}{2}=\dfrac{8}{x+1}\Rightarrow\left(x+1\right)^2=2.8=16\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=4^2\\\left(x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
\(a.\)
\(\dfrac{x}{2}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)
\(b.\)
\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)
\(c.\)
\(\dfrac{x}{4}=\dfrac{y}{5}\)
\(\Leftrightarrow\dfrac{2x}{8}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)
mà x+y=35
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(10;25)
b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)
nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)
hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)
mà y-3x=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
Vậy: (x,y)=(1;5)
c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)
nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)
mà 2x-y=15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)
Vậy: (x,y)=(20;25)
a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)
Áp dụng bổ đề trên ta có:
\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
a) \(\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{x+y}{2+3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{2+3}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{5}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{3\left(x+y\right)}{15}-\dfrac{5y}{15}\)
\(\dfrac{x}{2}=\dfrac{3x-2y}{15}\)
\(\Rightarrow15x=2\left(3x-2y\right)\)
\(15x=6x-4y\)
\(15x-6x=4y\)
\(9x=4y\)
(CÒN LẠI MÌNH KHÔNG BIẾT LÀM)
b) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{4}{y}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{20}{5y}\)
\(\dfrac{x}{3}=\dfrac{1+4}{y+1}\)
\(\Rightarrow x\left(y+1\right)=15\)
(CÒN NHIÊU TỰ LÀM NHÉ)
Giải:
a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)
\(\Rightarrow x=\dfrac{12.-4}{16}=-3\)
\(\Rightarrow y=\dfrac{16.21}{12}=28\)
\(\Rightarrow z=\dfrac{12.80}{16}=60\)
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\) =0
\(\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\) \(=0+\dfrac{2}{5}\)
\(x.\dfrac{11}{15}\) \(=\dfrac{2}{5}\)
x \(=\dfrac{2}{5}:\dfrac{11}{15}\)
x \(=\dfrac{6}{11}\)
c) (2x-3)(6-2x)=0
⇒2x-3=0 hoặc 6-2x=0
x=3/2 hoặc x=3
d) \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\)
\(2x-5=\dfrac{-13}{2}\)
\(2x=\dfrac{-13}{2}+5\)
\(2x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:2\)
\(x=\dfrac{-3}{4}\)
e) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}:2\)
\(\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\) hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\)
\(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)
x=-4
y=-1
Mình k biết cách trình bày nên b thôg cảm ạ:( Có thể tick cho mình đko:)?
Tặng coin mình đi!!!
x=-4
y=-1
Chúc bạn học tốt, 10 điểm nha