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a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
a) x = 6 ; y = 15.
x = -6 ; y = -15.
b) x = 2 ; y = 2.
x = -2 ; y = -2.
\(\frac{3}{2}X\)\(-\)\(\frac{1}{3}\)= \(\frac{1}{6}\)
\(\frac{3}{2}X\)= \(\frac{1}{6}\)+ \(\frac{1}{3}\)
\(\frac{3}{2}X\)= \(\frac{1}{2}\)
\(X\)= \(\frac{3}{2}\): \(\frac{1}{2}\)
\(X\)= \(\frac{3}{2}\)x \(\frac{2}{1}\)
\(X\)= 3
k mình nha
Chúc bạn học giỏi
Mình cảm ơn bạn nhiều
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
=> (x-1)(x+3) = (x+2)(x-2) ( tích chéo bạn nhé :)))))
=> x^2 - x + 3x - 3 = x^2 + 2x - 2x - 4 ( bước này là mình nhân tung ra bạn nhé ^.^)
=> x^2 + 2x - 3 = x^2 - 4
=> x^2 +2x = x^2 - 4 +3
=> x^2 + 2x = x^2 - 1
=> 2x = -1 ( bớt cả 2 vế đi x^2)
=> x = \(\frac{-1}{2}\)
đúng 100% nha
ủng hộ mk đi bạn
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(x=0-2004\)
\(x=-2004\)