Số?
a) 2 x ? = 2
? x 1 = 5
b) 5 : ? = 1
? : 2 = 1
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1
22 tháng 1
a: |3x-1|>5
=>\(\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x>6\\3x< -4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
b: \(\left|x^3+1\right|>=x+1\)
=>\(\left|x^3+1\right|-x-1>=0\)(1)
TH1: x>=-1
BPT (1) sẽ tương đương với: \(x^3+1-x-1>=0\)
=>\(x^3-x>=0\)
=>\(x\left(x^2-1\right)>=0\)
=>\(x\left(x-1\right)\left(x+1\right)>=0\)
=>x(x-1)>=0
=>\(\left[{}\begin{matrix}x>=1\\x< =0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x>=1\\-1< =x< =0\end{matrix}\right.\)
TH2: x<-1
BPT (1) sẽ tương đương với:
\(-x^3-1-x-1>=0\)
=>\(x^3+x+2< =0\)
=>\(x^3+x^2-x^2-x+2x+2< =0\)
=>\(\left(x+1\right)\left(x^2-x+2\right)< =0\)
=>x+1<=0
=>x<=-1
c: \(\left|x+1\right|>\left|x-2\right|\)
=>|x+1|-|x-2|>0(2)
TH1: x<-1
BPT (2) sẽ tương đương:
-x-1-(2-x)>0
=>-x-1-2+x>0
=>-3>0(vô lý)
=>\(x\in\varnothing\)
TH2: -1<=x<2
BPT (2) sẽ tương đương:
\(x+1-\left(2-x\right)>0\)
=>x+1-2+x>0
=>2x-1>0
=>2x>1
=>\(x>\dfrac{1}{2}\)
=>\(\dfrac{1}{2}< x< 2\)
TH3: x>=2
BPT (2) sẽ tương đương:
\(x+1-\left(x-2\right)>0\)
=>x+1-x+2>0
=>3>0(luôn đúng)
=>x>=2
d: |x-1|>|x+2|-3
=>|x-1|-|x+2|+3>0(3)
TH1: x<-2
BPT (3) sẽ tương đương:
1-x-(-x-2)+3>0
=>4-x+x+2>0
=>6>0(luôn đúng)
=>x<-2
TH2: -2<=x<1
BPT (3) sẽ tương đương:
\(1-x-\left(x+2\right)+3>0\)
=>\(4-x-x-2>0\)
=>-2x+2>0
=>-2x>-2
=>x<1
=>-2<=x<1
TH3: x>=1
BPT (3) sẽ tương đương:
\(x-1-x-2+3>0\)
=>0>0(sai)
=>\(x\in\varnothing\)
e: |x-1|+|x-5|>8(4)
TH1: x<1
BPT (4) sẽ tương đương:
1-x+5-x>8
=>6-2x>8
=>-2x>-2
=>x<1
TH2: 1<=x<5
BPT (4) sẽ tương đương:
x-1+5-x>8
=>4>8(vô lý)
=>\(x\in\varnothing\)
TH3: x>=5
BPT (4) sẽ tương đương:
\(x-1+x-5>8\)
=>2x>8+6=14
=>x>7
f: |x-3|+|x+1|<8(5)
TH1: x<-1
BPT (5) sẽ tương đương:
3-x-x-1<8
=>-2x<6
=>x>-3
=>-3<x<-1
TH2: -1<=x<3
BPT (5) sẽ tương đương:
x+1+3-x<8
=>4<8(luôn đúng)
=>-1<=x<3
TH3: x>=3
BPT (5) sẽ tương đương:
x-3+x+1<8
=>2x-2<8
=>2x<10
=>x<5
=>3<=x<5
29 tháng 10 2021
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
MH
14 tháng 9 2021
a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)
b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)
14 tháng 9 2021
a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)
b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)
c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)
d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
NT
1
N
28 tháng 2 2022
a)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\Rightarrow\dfrac{47}{10}.x=\dfrac{13}{15}\)
\(\Rightarrow x=\dfrac{13}{15}:\dfrac{47}{10}\)
\(\Rightarrow x=\dfrac{13}{10}.\dfrac{10}{47}\)
\(\Rightarrow x=\dfrac{13}{47}\)
b)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)
\(\Rightarrow\dfrac{5}{7}:x=\dfrac{5}{30}-\dfrac{24}{30}\)
\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{19}{30}\)
\(\Rightarrow x=\dfrac{5}{7}:\left(-\dfrac{19}{30}\right)\)
\(\Rightarrow x=\dfrac{5}{7}.\left(-\dfrac{30}{19}\right)\)
\(\Rightarrow x=-\dfrac{150}{133}\)
31 tháng 3 2021
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
a, 2 x 1 = 2
5 x 1 = 5
b, 5 : 5 = 1
2 : 2 = 1
a, 2 x 1 = 2 , 5 x 1 =5
b, 5 : 5 =1 , 2 : 2 =1