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a) 4x + 5(x - 3) = 3
<=> 4x + 5x - 15 = 3
<=> 9x = 3 + 15
<=> 9x = 18
<=> x = 2
b) -3(x - 5) + 6(x + 2) = 9
<=> -3x + 15 + 6x + 12 = 9
<=> 3x + 27 = 9
<=> 3x = -18
<=> x = -6
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
a) 3x2-7x=0
<=> x(3x-7)=0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{3}\end{cases}}}\)
b) làm tương tự
c) \(\left(x^2-1\right)^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=3\\x^2-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
\(6-2\left|1+3x\right|\le6\)'
Max \(A=6\Leftrightarrow1+3x=0\)
\(\Rightarrow3x=-1\)
\(\Rightarrow x=\frac{-1}{3}\)
\(\left|x-2\right|+\left|x-5\right|\ge0\)
Max \(B=0\Leftrightarrow\hept{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)
a) \(\frac{x}{4}=\frac{9}{12}\)
\(\Rightarrow x=\frac{9}{12}\cdot4\)
\(\Rightarrow x=3\)
b) \(-\frac{8}{3}:x=\frac{5}{6}\)
\(\Rightarrow x=-\frac{8}{3}:\frac{5}{6}\)
\(\Rightarrow x=-\frac{16}{5}=-3,2\)
c) \(\left(\frac{3}{4}-x\right)\cdot\frac{10}{3}=\frac{2}{5}\)
\(\Rightarrow\frac{3}{4}-x=\frac{2}{5}:\frac{10}{3}\)
\(\Rightarrow x=\frac{3}{4}-\frac{3}{25}=\frac{63}{100}=0,63\)
Nhớ k mk đó nkoa!@^_^@! Love you
a) ĐKXĐ: \(x\ne\text{5}\)
\(\frac{x-3}{x-5}=\frac{5}{9}\)
\(\Rightarrow9\left(x-3\right)=5\left(x-5\right)\)
\(\Leftrightarrow9x-27=5x-25\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)
Vậy \(x=\frac{1}{2}\)
b) ĐKXĐ: \(x\ne5;x\ne9\)
\(\frac{x-3}{x-5}=\frac{x-8}{x-9}\)
\(\Rightarrow\left(x-3\right)\left(x-9\right)=\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow x^2-9x-3x+27=x^2-8x-5x+40\)
\(\Leftrightarrow x^2-12x+27=x^2-13x+40\)
\(\Leftrightarrow x=13\left(TM\right)\)
Vậy \(x=13\)