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\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)
\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)
\(\frac{2}{3}-x=-\frac{7}{6}\)
\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)
\(x=\frac{2}{3}+\frac{7}{6}\)
\(x=\frac{11}{6}\)
a)\(\left|x-5\right|-x=3\)
\(TH1:x-5-x=3\)
\(-5=3\)(ko xảy ra)
\(xkoTM\)
\(TH2:-\left(x-5\right)-x=3\)
\(5-x-x=3\)
\(5-2x=3\)
\(2x=2\)
x=1
Vậy x=1
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\frac{x-2}{5}+\frac{x-3}{4}=\frac{x-4}{3}+\frac{x-5}{2}\)
=> \(\frac{x-2}{5}-1+\frac{x-3}{4}-1=\frac{x-4}{3}-1+\frac{x-5}{2}-1\)
=> \(\frac{x-7}{5}+\frac{x-7}{4}=\frac{x-7}{3}+\frac{x-7}{2}\)
=> \(\frac{x-7}{5}+\frac{x-7}{4}-\frac{x-7}{3}-\frac{x-7}{2}=0\)
=> \(\left(x-7\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
=> x-7 = 0
=> x= 7
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 )
\(\left|x\right|+3=5\)
\(\Rightarrow\left|x\right|=5-3\)
\(\Rightarrow\left|x\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(\left|x\right|+3=5\)
Vì \(5>3\) nên \(\left|x\right|>0\)
\(x+3=5\)
\(x=5-3\)
\(x=2\)