14/3-2x.9/4=3/2
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5 tháng 7 2023
4:
=>(2x+3,5)=7/12*3/14=21/168=1/8
=>2x=1/8-7/2=1/8-28/8=-27/8
=>x=-27/16
5: =>1/3:3x=-21/4
=>3x=-1/3:21/4=-1/3*4/21=-4/63
=>x=-4/189
6: =>2+7/9-3/4(x+1)=7/9
=>2-3/4(x+1)=0
=>3/4(x+1)=2
=>x+1=2:3/4=2*4/3=8/3
=>x=5/3
18 tháng 1 2024
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: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)
=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)
=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)
=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)
11 tháng 7 2017
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
11 tháng 7 2017
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
4 tháng 10 2021
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
18 tháng 5 2020
a) \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2+5=\frac{61}{9}\)
=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{61}{9}-5\)
=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}\)
=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}:4\)
=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9\cdot4}=\frac{16}{36}=\frac{4}{9}\)
=> \(\frac{1}{2}x-\frac{1}{3}=\pm\frac{2}{3}\)
Trường hợp 1 : \(\frac{1}{2}x-\frac{1}{3}=\frac{2}{3}\)
=> \(\frac{1}{2}x=1\)
=> \(x=1:\frac{1}{2}=2\)
Trường hợp 2 : \(\frac{1}{2}x-\frac{1}{3}=-\frac{2}{3}\)
=> \(\frac{1}{2}x=-\frac{2}{3}+\frac{1}{3}=-\frac{1}{3}\)
=> \(x=\left(-\frac{1}{3}\right):\frac{1}{2}=\left(-\frac{1}{3}\right)\cdot2=-\frac{2}{3}\)
b) \(9\left(2x-\frac{1}{3}\right)^3-1=-\frac{2}{3}\)
=> \(9\left(2x-\frac{1}{3}\right)^3=-\frac{2}{3}+1=\frac{1}{3}\)
=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3}:9\)
=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3\cdot9}=\frac{1}{27}\)
=> \(2x-\frac{1}{3}=\frac{1}{3}\)
=> \(2x=\frac{2}{3}\)
=> \(x=\frac{2}{3}:2=\frac{1}{3}\)
Bài cuối tương tự
25 tháng 2 2022
1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)
\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)
2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)
\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)
=>-8x+8=0
hay x=1(nhận)
c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)
\(\Leftrightarrow x^2+x-6=0\)
=>(x+3)(x-2)=0
=>x=-3(nhận) hoặc x=2(nhận)