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\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
1.
\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)
2. Xem lại đề nha!
4.
\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)
5.
\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)
6.
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
=> 2(2x+1) = 6.7
4x+2=42
4x=40
x=10
Vậy x=10
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)
b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)
d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)
f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)
a. \(\dfrac{6}{2x+1}=\dfrac{2}{7}\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{21}\Rightarrow2x+1=21\)
\(\Rightarrow2x=21-1=20\Rightarrow x=\dfrac{20}{2}=10\)
Vậy x = 10
b. \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\Rightarrow\dfrac{24}{7x-3}=\dfrac{24}{150}\Rightarrow7x-3=150\)
\(\Rightarrow7x=150+3=153\Rightarrow x=\dfrac{153}{7}\)
Vậy \(x=\dfrac{153}{7}\)
c. \(\dfrac{4}{x-6}=\dfrac{-12}{18}\Rightarrow-12\cdot\left(x-6\right)=4\cdot18=72\)
\(\Rightarrow x-6=\dfrac{72}{-12}=-6\Rightarrow x=-6+6=0\)
\(\dfrac{y}{24}=\dfrac{-12}{18}\Rightarrow y=\dfrac{-12\cdot24}{18}=-16\)
Vậy x = 0 ; y = -16
Bài 1:
c)\(\left(x-1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left(x-1\right)^2-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1+3x-2\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\)
\(\Leftrightarrow\left(1-2x\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=0\\4x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
\(a.\dfrac{1}{x}-\dfrac{y}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{x}=\dfrac{1}{3}+\dfrac{y}{6}\)
\(\Rightarrow\dfrac{1}{x}=\dfrac{2+y}{6}\)
\(\Rightarrow x\left(2+y\right)=6\)
\(\Rightarrow x\inƯ\left(6\right);2+y\inƯ\left(6\right)\)
...
b) \(b.\dfrac{x}{2}+\dfrac{3}{y}=\dfrac{5}{4}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{5}{4}-\dfrac{3}{y}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{5y-12}{4y}\)
\(\Rightarrow4xy=10y-24\)
\(\Rightarrow2y\left(2x-5\right)=24\)
mà 2x - 5 lẻ nên 2x - 5 thuộc tập hợp ước của 24 bao gồm n~ số lẻ
\(\Rightarrow2x\inƯ\left(24\right);2x-5\in\left\{...\right\}\)
...
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
