Tìm x :
a) | x - 5/4 | < 1/3
b) 2/5 < | x - 7/5 | < 3/5
c) 1/3 - x < | x + 2/3 | <3
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a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
a) Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\)
\(\Leftrightarrow\dfrac{a}{8}=\dfrac{b}{12}\)(1)
Ta có: \(\dfrac{b}{4}=\dfrac{c}{5}\)
nên \(\dfrac{b}{12}=\dfrac{c}{15}\)(2)
Từ (1) và (2) suy ra \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\)
mà a+b+c=2
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{2}{35}\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{a}{8}=\dfrac{2}{35}\\\dfrac{b}{12}=\dfrac{2}{35}\\\dfrac{c}{15}=\dfrac{2}{35}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{16}{35}\\b=\dfrac{24}{35}\\c=\dfrac{30}{35}=\dfrac{6}{7}\end{matrix}\right.\)
Vậy: \(a=\dfrac{16}{35}\); \(b=\dfrac{24}{35}\); \(c=\dfrac{6}{7}\)
b) Ta có: 2a=3b=5c
nên \(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}\)
mà a+b-c=3
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{\dfrac{1}{2}}=\dfrac{b}{\dfrac{1}{3}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b-c}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{19}{30}}=\dfrac{90}{19}\)
Do đó:
\(\left\{{}\begin{matrix}2a=\dfrac{90}{19}\\3b=\dfrac{90}{19}\\5c=\dfrac{90}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{45}{19}\\b=\dfrac{30}{19}\\c=\dfrac{18}{19}\end{matrix}\right.\)
Vậy: \(a=\dfrac{45}{19}\); \(b=\dfrac{30}{19}\); \(c=\dfrac{18}{19}\)
a: =>x+7/4=6:2/3=9
=>x=29/4
b: =>x:5/3=7/5
=>x=7/5*5/3=7/3
c:=>x+1/6=5/3
=>x=10/6-1/6=3/2
d: =>x+4/5=4/5+3/7+3/5
=>x=3/7+3/5=36/35
e: =>x/35=4/5-5/7=3/35
=>x=3
f: =>13/28+x=1/2
=>x=1/28
g: =>1/3-x=1/9
=>x=2/9
\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)
b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)
=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)
=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)
=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)
=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)
d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
a)\(=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
b)\(=\dfrac{5}{9}\times\dfrac{3}{2}=\dfrac{15}{18}=\dfrac{5}{6}\)
d)\(=\left(\dfrac{12}{8}-\dfrac{3}{8}\right)\times2=\dfrac{9}{8}\times2=\dfrac{18}{8}=\dfrac{9}{4}\)
c)\(=\dfrac{4}{3}-\dfrac{5}{6}=\dfrac{8}{6}-\dfrac{5}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
a) 1 + 4/3 = 7/3
b) 5/9 : 2/3 = 5/6
c ) 4/3 -1/3 x 5/2
= 1 x 5/2
= 5/2
d) ( 3/2 - 3/8) : 1/2
= 9/8 : 1/2
= 9/4
e) 15/16 : 3/8 x 3/4
= 5/2 x 3/4
= 15/8
f) 7/19 x 1/3 x 7/19 x 2/3
= 7/19 x (1/3 x 2/3)
= 7/19 x 2/9
= 14/171
g) 3/5 x 8/27 x 25/3
= 3/5 x 25/3 x 8/27
= 5 x 8/27
= 40/27
h) 1/5 + 4/11 + 4/5 + 7/11
= (1/5 + 4/5) + (4/11 + 7/11)
= 1 + 1
= 2
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)