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\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{2x}{2.3}=\frac{5y}{5.2}=\frac{2x}{6}=\frac{5y}{10}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{2x}{6}=\frac{5y}{10}=\frac{2x+5y}{6+10}\)\(=\frac{32}{16}=2\)
\(\frac{2x}{6}=2\Rightarrow2x=12\Rightarrow x=6\)
\(\frac{5y}{10}=2\Rightarrow5y=20\Rightarrow y=4\)
Vậy ..
ta có: x/3 =y/2 => 2x/6 = 5y/10
áp dụng tính chất dãy tỉ số bằng nhau ta có:
2x/6 = 5y/10 = 2x + 5y/ 6 + 10 = 32/16 = 2
=> x = 3 . 2 = 6 ; y = 2 . 2 = 4
vậy ( x , y ) = ( 6 ; 4 )
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
Phá ngoặc ra là ok rồi mà:
\(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Rightarrow4x-x-\dfrac{1}{2}=2x-\dfrac{1}{2}+5\)
\(\Rightarrow3x-\dfrac{1}{2}=2x+\dfrac{9}{2}\)
\(\Rightarrow3x-\dfrac{1}{2}-\dfrac{9}{2}=2x\)
\(\Rightarrow3x-5=2x\)
\(\Rightarrow x=5\)
Ta :có
\(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow4x-2x=\left(\dfrac{1}{2}-5\right)+\left(x+\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x=-4+x\)
\(\Leftrightarrow2x-x=-4\)
\(\Rightarrow x=-4\)