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\(a)\dfrac{1}{3}x+\dfrac{2}{5}\left(x+1\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\left(\dfrac{5}{15}+\dfrac{6}{15}\right)=\dfrac{-2}{5}\)
\(\Leftrightarrow x.\dfrac{11}{15}=\dfrac{-2}{5}\)
\(\Leftrightarrow x=\dfrac{-2}{5}.\dfrac{15}{11}\)
\(\Leftrightarrow x=\dfrac{-6}{11}\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
\(\Leftrightarrow x\left(-\dfrac{7}{2}-\dfrac{5}{4}\right)=\dfrac{1}{4}\cdot\dfrac{-4}{3}+\dfrac{1}{6}+\dfrac{3}{4}\)
\(\Leftrightarrow x\cdot\dfrac{-19}{4}=\dfrac{7}{12}\)
hay x=-7/57
a . x-1/9 = 6/3
=> x-1/9 = 2
=> x - 1 = 2 . 9
=> x - 1 = 18
=> x = 18 + 1
=> x = 19
b . -x /2 = -31/x
=> -x . x = -31 . 2
=> -( x^2 ) = -62
=> x^2 = 62
=> x = 7,874...
c x/4 = 18/( x + 1 )
=> x . ( x + 1 ) = 18 . 4
=> x. ( x + 1 ) = 72
=> x = 8 ( vì 8 . ( 8 + 1 ) = 72 )
chúc học giỏi !!!
a) x= 7 phần 9 :7 phần 5
x= 7 phần 9 nhân 5 phần 7
x= 5 phần 9
b) x nhân 4= 5 nhân (-3)
x nhân 4= -15
x= -15 : 4
x= -15 phần 4
\(-|\frac{1}{2}x+1|-15=15\frac{1}{4}\)
\(\Rightarrow-|\frac{1}{2}x+1|=15+\frac{1}{4}+15\)
\(\Rightarrow-|\frac{1}{2}x+1|=\frac{121}{4}\)
\(\Rightarrow|\frac{1}{2}x+1|=\frac{-121}{4}\)
NX: vì giá trị tuyệt đối của mọi số đều ko âm\(\Rightarrow∄x\)
-|1/2 . x + 1| - 15 = 61
-|1/2 . x + 1| = 61 + 15
-|1/2 . x + 1| = 76
-|1/2 . x| = 76 - 1
-|1/2 . x| = 75
-|x|= 75 : 1/2
-|x|= 150
-x = -150
x = 150
Ta có: \(\frac{a-x}{b-y}=\frac{a}{b}\Rightarrow\left(a-x\right)b=\left(b-y\right)a\)
\(\Rightarrow ab-bx=ab-ay\Rightarrow bx=ay\)
\(\Rightarrow\frac{x}{y}=\frac{a}{b}\left(ĐPCM\right)\)
a) \(\dfrac{x}{2}+\dfrac{y}{3}=\dfrac{x+y}{2+3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{2+3}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{x+y}{5}-\dfrac{y}{3}\)
\(\dfrac{x}{2}=\dfrac{3\left(x+y\right)}{15}-\dfrac{5y}{15}\)
\(\dfrac{x}{2}=\dfrac{3x-2y}{15}\)
\(\Rightarrow15x=2\left(3x-2y\right)\)
\(15x=6x-4y\)
\(15x-6x=4y\)
\(9x=4y\)
(CÒN LẠI MÌNH KHÔNG BIẾT LÀM)
b) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{4}{y}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{20}{5y}\)
\(\dfrac{x}{3}=\dfrac{1+4}{y+1}\)
\(\Rightarrow x\left(y+1\right)=15\)
(CÒN NHIÊU TỰ LÀM NHÉ)
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
cảm ơn nhiều nha