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Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a) \(\left(x-2\right)\left(y+1\right)=14\)
Do \(x,y\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)
\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)
Vô lí => không có x,y thỏa mãn
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)
nên \(\dfrac{x}{y}=\dfrac{3}{7}\)
b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)
hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)
2:
a: =>2(x+1)=26
=>x+1=13
=>x=12
b: =>(6x)^3=125
=>6x=5
=>x=5/6(loại)
c: =>\(7\cdot3^x\cdot\dfrac{1}{3}+11\cdot3^x\cdot3=318\)
=>3^x=9
=>x=2
d: -2x+13 chia hết cho x+1
=>-2x-2+15 chia hết cho x+1
=>15 chia hết cho x+1
=>x+1 thuộc {1;3;5;15}
=>x thuộc {0;2;4;14}
e: 4x+11 chia hết cho 3x+2
=>12x+33 chia hết cho 3x+2
=>12x+8+25 chia hết cho 3x+2
=>25 chia hết cho 3x+2
=>3x+2 thuộc {1;-1;5;-5;25;-25}
mà x là số tự nhiên
nên x=1
1:
a: Đặt A=2^2024-2^2023-...-2^2-2-1
Đặt B=2^2023+2^2022+...+2^2+2+1
=>2B=2^2024+2^2023+...+2^3+2^2+2
=>B=2^2024-1
=>A=2^2024-2^2024+1=1
c: \(=\dfrac{3^{12}\cdot2^{11}+2^{10}\cdot3^{12}\cdot5}{2^2\cdot3\cdot3^{11}\cdot2^{11}}=\dfrac{2^{10}\cdot3^{12}\left(2+5\right)}{2^{13}\cdot3^{12}}\)
\(=\dfrac{7}{2^3}=\dfrac{7}{8}\)
a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
\(a,\text{Vì }x,y\in N\Leftrightarrow x+2\ge2;y+3\ge3\\ \Leftrightarrow\left(x+2\right)\left(y+3\right)=6=2\cdot3=3\cdot2\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;0\right)\)
\(b,\Leftrightarrow\left(x-3\right)\left(y+1\right)=7\cdot1=1\cdot7\\ \left\{{}\begin{matrix}x-3=7\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x-3=1\\y+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;0\right);\left(4;6\right)\right\}\)