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NM
1
17 tháng 6 2020
nhìn cái đề con hơi bị ''sốc'' , thế này ạ ???
Sửa đề \(4+\frac{1}{3}x\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}x\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(4+\frac{1}{3}x\left(-\frac{1}{3}\right)\le x\le\frac{2}{3}x\left(-\frac{11}{12}\right)\)
\(4-\frac{1}{9}x\le x\le-\frac{11}{18}x\)
D
0
H
3
7 tháng 5 2019
\(a,x+\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow x=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}\)
\(b,\frac{2}{3}x-\frac{3}{7}=\frac{1}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{7}+\frac{3}{7}\)
\(\Rightarrow\frac{2}{3}x=\frac{4}{7}\)
\(\Rightarrow x=\frac{4}{7}:\frac{2}{3}=\frac{4}{7}\cdot\frac{3}{2}=\frac{6}{7}\)
\(c,1\frac{2}{3}x-50\%x=-1\frac{1}{6}\)
\(\Rightarrow\frac{5}{3}x-\frac{50}{100}x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{50}{100}\right]x=-\frac{7}{6}\)
\(\Rightarrow\left[\frac{5}{3}-\frac{1}{2}\right]x=-\frac{7}{6}\)
\(\Rightarrow\frac{7}{6}x=-\frac{7}{6}\Leftrightarrow x=-\frac{7}{6}:\frac{7}{6}=-\frac{7}{6}\cdot\frac{6}{7}=-1\)
O
7 tháng 5 2019
a,x+\(\frac{3}{4}\)=\(\frac{1}{4}\)
=>x =\(\frac{1}{4}\)-\(\frac{3}{4}\)
=> x =\(\frac{-1}{2}\)
Vậy x=\(\frac{-1}{2}\)
b,\(\frac{2}{3}\)x-\(\frac{3}{7}\)=\(\frac{1}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{1}{7}\)+\(\frac{3}{7}\)
=>\(\frac{2}{3}\)x =\(\frac{4}{7}\)
=> x =\(\frac{4}{7}\) :\(\frac{2}{3}\)
=> x =\(\frac{4}{7}\).\(\frac{3}{2}\)
=> x =\(\frac{6}{7}\)
Vậy x=\(\frac{6}{7}\)
c,\(\frac{12}{3}\)x-50%x =\(\frac{-11}{6}\)
=>\(\frac{12}{3}\)x-\(\frac{1}{2}\)x=\(\frac{-11}{6}\)
=>x(\(\frac{12}{3}\)-\(\frac{1}{2}\))=\(\frac{-11}{6}\)
=>x(\(\frac{24}{6}\)-\(\frac{3}{6}\))=\(\frac{-11}{6}\)
=>x\(\frac{21}{6}\) =\(\frac{-11}{6}\)
=>x =\(\frac{-11}{6}\):\(\frac{21}{6}\)
=>x =\(\frac{-11}{6}\).\(\frac{6}{21}\)
=>x =\(\frac{-11}{21}\)
Vậy x=\(\frac{-11}{21}\)
LT
2
7 tháng 1 2018
a)
\(|3x+1|=4\)
\(\Rightarrow\orbr{\begin{cases}3x+1=4\\3x+1=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4-1\\3x=-4-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\div3\\x=-5\div3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1,6667\end{cases}}\)
Vậy x = 1
PA
7 tháng 1 2019
1) 14x-8x=10+5
x(14-8)=15
x6=15
x=15/6
2)5x-3x=30-15
2x=15
x=15/2
3)làm tương tự
GT
26 tháng 3 2020
a) \(x-2=-6\)
\(x=-6+2\)
\(x=-4\)
b) \(15-\left(x-7\right)=-21\)
\(x-7=36\)
\(x=43\)
c) \(4.\left(3x-4\right)-2=18\)
\(4\left(3x-4\right)=20\)
\(3x-4=5\)
\(3x=9\)
\(x=3\)
d) \(\left(3x-6\right)+3=32\)
\(3x-6=29\)
\(3x=29+6\)
\(3x=35\)
\(x=\frac{35}{3}\)
e) \(\left(3x-6\right).3=32\)
\(3x-6=\frac{32}{3}\)
\(3x=\frac{32}{3}+6\)
\(3x=\frac{50}{3}\)
\(x=\frac{50}{9}\)
f) \(\left(3x-6\right):3=32\)
\(3x-6=96\)
\(3x=102\)
\(x=34\)
g) \(\left(3x-6\right)-3=32\)
\(3x-6=35\)
\(3x=41\)
\(x=\frac{41}{3}\)
h) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-2^4\right)=2.7=14\)
\(\left(3x-16\right)=14\)
\(3x=14+16=30\)
\(x=10\)
i) \(\left|x\right|=\left|-7\right|\)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
k) \(\left|x+1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
l) \(\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
m) \(x+\left|-2\right|=0\)
\(x+2=0\)
\(x=-2\)
o) \(72-3\left|x+1\right|=9\)
\(3\left|x-1\right|=63\)
\(\left|x-1\right|=21\)
\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)
p) Ta có: \(\left|x-1\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)
mà \(x+1< 0\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-2\)
q) \(\left(x-2\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
hok tốt!!
HM
0
\(\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x^3-6\text{x}^2+12\text{x}-8\right)+\left(9\text{x}^2-1\right)-\left(x^3+3\text{x}^2+3\text{x}+1\right)=0\)
\(\Leftrightarrow x^3-6\text{x}^2+12\text{x}-8+9\text{x}^2-1-x^3-3\text{x}^2-3\text{x}-1=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-6\text{x}^2+9\text{x}^2-3\text{x}^2\right)+\left(12\text{x}-3\text{x}\right)+\left(-8-1-1\right)=0\)
\(\Leftrightarrow9\text{x}-10=0\)
\(\Leftrightarrow9\text{x}=10\Leftrightarrow x=\frac{10}{9}\)
Vậy x = \(\frac{10}{9}\)