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bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)
b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))
\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)
\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)
d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)
\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)
\(\Leftrightarrow2x^2+2x=2x^2+1\)
\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
1) Ta có\(\frac{x+2}{5}=\frac{1}{x-2}\)
=> (x + 2)(x - 2) = 5
=> x2 + 2x - 2x - 4 = 5
=> x2 - 4 = 5
=> x2 = 9
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
2) \(\frac{3}{x-4}=\frac{x+4}{3}\)
=> (x - 4)(x + 4) = 9
=> x2 + 4x - 4x - 16 = 9
=> x2 - 16 = 9
=> x2 = 25
=> \(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
a, \(\frac{x+2}{5}=\frac{1}{x-2}ĐK:x\ne2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{5\left(x-2\right)}=\frac{5}{5\left(x-2\right)}\Leftrightarrow\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^2-2x+2x-4=5\Leftrightarrow x^2=9\Leftrightarrow x\pm3\)
b, \(\frac{3}{x-4}=\frac{x+4}{3}ĐK:x\ne4\)
\(\Leftrightarrow\frac{9}{\left(x-4\right)3}=\frac{\left(x+4\right)\left(x-4\right)}{3\left(x-4\right)}\Leftrightarrow9=x^2-4x+4x-16\)
\(\Leftrightarrow x^2-16=9\Leftrightarrow x^2=25\Leftrightarrow x=\pm5\)
c, \(\frac{x+2}{x+6}=\frac{3}{x}=1ĐK:x\ne0;-6\)
Xét : \(\frac{x+2}{x+6}=1\Leftrightarrow x+2=x+6\Leftrightarrow-4\ne0\)
Xét : \(\frac{3}{x}=1\Leftrightarrow3=x\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
bạn có thể viết rõ đề ra được không?