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a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)
a) A = \(\dfrac{6n+7}{2n+3}\) = \(\dfrac{6n+9}{2n+3}\) − \(\dfrac{2}{2n+3}\) nguyên
⇔ 2n + 3 ∈ Ư(2) = {-2; -1; 1; 2}
⇔ 2n ∈ {-5; -4; -2; -1}
Vì n nguyên nên n ∈ {-2; -1}
a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5
Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2
=-3x^4-x^3+3x^2-10x+2
b: P(x)+Q(x)
=x^4+5-3x^4-x^3+3x^2-10x+2
=-2x^4-x^3+3x^2-10x+7
Q(x)-P(x)
=-3x^4-x^3+3x^2-10x+2-x^4-5
=-4x^4-x^3+3x^2-10x-3
P(x)-Q(x)=-(Q(x)-P(x))
=4x^4+x^3-3x^2+10x+3
với |2x+10|+|3x-1|+|1-x|=3 ta có 2 trường hợp:
trường hợp 1:|2x+10|+|3x-1|+|1-x|=2x+10+3x-1+1-x=3
4x+10=3
4x=-7
x=-7/4
trường hợp 2:|2x+10|+|3x-1|+|1-x|=-(2x+10)+[-(3x-1)]+[-(1-x)]=3
-2x-10-3x+1-1+x=3
-4x-10=3
-4x=13
x=-13/4
/ là dấu phần nhé!
a, \(3\left(2x-1\right)-3x\left(-x+2\right)=5x-\left(1-3x\right)\cdot x\\ 6x-3+3x^2-6x=5x-x+3x^2\\ 3x^2-3=4x+3x^2\\ 3x^2-3x^2=4x+3\\ 4x+3=0\\ 4x=-3\\ x=\frac{-3}{4}\)
Vậy \(x=\frac{-3}{4}\)
b, \(x-\frac{x-3}{4}=3-\frac{x-3}{12}\\ \frac{4x-x-3}{4}=\frac{36-x-3}{12}\\ \frac{3x-3}{4}=\frac{33-x}{12}\\ \Rightarrow12\left(3x-3\right)=4\left(33-x\right)\\ 36x-36=132-4x\\ 36x+4x=132+36\\ 40x=168\\ x=\frac{168}{40}=\frac{21}{5}\)
Vậy \(x=\frac{21}{5}\)
a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
\(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-1=0\\\frac{2x-1}{3}=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x=1\\2x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\2x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{1}{2}\end{array}\right.\)
(5x - 1)(2x - 1/3) = 0
TH1:
5x - 1 = 0
5x = 1
x = 1/5
TH2:
2x - 1/3 = 0
2x = 1/3
x = 1/3 : 2
x = 1/3 . 1/2
x = 1/6
Vậy x = 1/5 hoặc x = 1/6