1,3x - [1 - x] =10,5
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\(X+30\%X=-1,3\\ X=\dfrac{-1,3}{100\%+30\%}=-1\\ ---\\ X-25\%X=\dfrac{1}{2}\\ X=\dfrac{\dfrac{1}{2}}{100\%-25\%}=\dfrac{2}{3}\)
\(1,\dfrac{3x-1}{4}+\dfrac{6x-2}{8}=\dfrac{1-3x}{6}\\ \Leftrightarrow1,\dfrac{3x-1}{4}+\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow2.\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{3x-1}{4}=\dfrac{1-3x}{12}\\ \Leftrightarrow12\left(3x-1\right)=4\left(1-3x\right)\\ \Leftrightarrow3\left(3x-1\right)=1-3x\\ \Leftrightarrow9x-3-1+3x=0\\ \Leftrightarrow12x-4=0\\ \Leftrightarrow x=\dfrac{1}{3}\)
\(2,\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(C=2\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(C=2\cdot\left(1-\frac{1}{64}\right)\)
\(C=2\cdot\frac{63}{64}\)
\(C=\frac{63}{32}\)
Ta có : \(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+.........+\frac{2}{61.64}\)
\(\Rightarrow C=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{61.64}\right)\)
\(\Rightarrow C=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Rightarrow C=\frac{2}{3}\left(1-\frac{1}{64}\right)\)
\(\Rightarrow C=\frac{2}{3}\frac{63}{64}=\frac{21}{32}\)
|-1,3.x+1/2| = |-5/6+0,2.x|
TH1: -1,3.x+1/2 = -5/6+0,2.x
=> -1,3.x - 0,2.x = -5/6 - 1/2
-1,5.x = -4/3
x = -8/9
TH2: -1,3.x+1/2 = 5/6-0,2.x
=> -1,3.x + 0,2.x = 5/6-1/2
-1,1.x = 1/3
x = -10/33
KL:...
\(3x\left(x-2020\right)-x+2020=0\)
\(3x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\left(3x-1\right)\left(x-2020\right)=0\)
\(\orbr{\begin{cases}x=\frac{1}{3}\left(TM\right)\\x=2020\left(TM\right)\end{cases}}\)
\(b,4-9x^2=0\)
\(2^2-\left(3x\right)^2=0\)
\(\left(2-3x\right)\left(2+3x\right)=0\)
\(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}\orbr{\begin{cases}x=\frac{2}{3}\left(TM\right)\\x=-\frac{2}{3}\left(TM\right)\end{cases}}}\)
\(c,x^2-x+\frac{1}{4}=0\)
\(x^2-x+\left(\frac{1}{2}\right)^2=0\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(d,x\left(x-3\right)+\left(x-3\right)=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)
\(e,9x\left(x-7\right)-x+7=0\)
\(9x\left(x-7\right)-\left(x-7\right)=0\)
\(\left(9x-1\right)\left(x-7\right)=0\)
\(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}\orbr{\begin{cases}x=\frac{1}{9}\left(TM\right)\\x=7\left(TM\right)\end{cases}}}\)
a) 3x(x - 2020) - x + 2020 = 0
<=> 3x(x - 2020) - (x - 2020) = 0
<=> (3x - 1)(x - 2020) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2020\end{cases}}\)
Vậy tập nghiệm phương trình là \(S=\left\{\frac{1}{3};2020\right\}\)
b) \(4-9x^2=0\)
<=> \(\left(2-3x\right)\left(2+3x\right)=0\)
<=> \(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};-\frac{2}{3}\right\}\)là nghiệm phương trình
c) \(x^2-x+\frac{1}{4}=0\)
<=> \(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x-\frac{1}{2}=0\)
<=> \(x=\frac{1}{2}\)
d) x(x - 3) + (x - 3) = 0
<=> (x + 1)(x - 3) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy \(x\in\left\{-1;3\right\}\)là nghiệm phương trình
e) 9x(x - 7) - x + 7 = 0
<=> (9x - 1)(x - 7) = 0
<=> \(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{9};7\right\}\)là nghiệm phương trình
1)\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}=\frac{2x+3y}{8+9}=\frac{34}{17}=2\)
\(\Rightarrow x=4.2=8;y=2.3=6\)
2)\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2+5}=\frac{7}{7}=1\)
\(\Rightarrow x=2;y=-5\)
a) Ta có :
\(\frac{x}{4}=\frac{y}{3}\)
\(=\frac{2x}{8}=\frac{3y}{9}\)
Theo tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{2x}{8}=\frac{3y}{9}=\frac{2x+3y}{8+9}=\frac{34}{17}=2\)
\(\Rightarrow\hept{\begin{cases}\frac{2x}{8}=2\\\frac{3x}{9}=2\end{cases}\Rightarrow}\hept{\begin{cases}x=2.8\div2\\y=2.9\div3\end{cases}\Rightarrow\hept{\begin{cases}x=8\\y=6\end{cases}}}\)
Vậy ....
b) Theo tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-7}{2-\left(-5\right)}=\frac{7}{7}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=1\\\frac{y}{-5}=1\end{cases}\Rightarrow\hept{\begin{cases}x=1.2\\y=1.\left(-5\right)\Rightarrow\end{cases}}\hept{\begin{cases}x=2\\y=-5\end{cases}}}\)
Vậy ...
1,3x - (1-x)=10,5
<=>1,3x-1+x=10,5
<=>2,3x-1=10,5
<=>2,3x=10,5+1=11,5
<=>x=11,5:2,3
=>x=5
1 3x - { 1 -x ) = 10 ,5
=> 13x - 1 +x = 10,5
=> 14x = 11,5
=> x = 11,5/14