a) 3 x 1/5 + 2/5 = 3/5 + 2/5 = 1

b) 3 : 1/5 - 2/5 = 15 - 2/5 = 73/5

bài 2 :
a) X x 1/2 = 1 - 1/5

    X x 1/2 = 4/5 

           X   = 4/5 : 1/2

           X   = 8/5 

vậy x =...

b) x : 1/3 = 1 + 1/5

    x : 1/3 = 6/5

          x   = 6/5 x 1/3

          x   = 2/5

vậy x =...

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

1. Thu gọn biểu thức - Hoc24 làm rồi mà bạn?

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) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

1,a,=x2−6x+8+3x2−15x=4x2−21x+8b,=9x2+12x+4−x2+9=8x2+12x+132,a,⇔x2+8x+16−x2+4=5⇔8x=−15⇔x=−158b,⇔9x2−6x+1−8x2−2x+12x+3−x2=5⇔4x=1⇔x=14

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)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)

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) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)

c)\(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}+\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}+\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{2}{3}.\dfrac{4}{3}=\dfrac{8}{9}\)

d)\(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}-\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{1}{3}.\dfrac{4}{3}=\dfrac{4}{9}\)

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)

\(\Rightarrow\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\)

\(\Rightarrow\dfrac{5}{7}:x=\dfrac{5}{30}-\dfrac{24}{30}\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{19}{30}\)

\(\Rightarrow x=\dfrac{5}{7}:\left(-\dfrac{19}{30}\right)\)

\(\Rightarrow x=\dfrac{5}{7}.\left(-\dfrac{30}{19}\right)\)

\(\Rightarrow x=-\dfrac{150}{133}\)

Bài 3 là hỗn số hả em?

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:\(x=\dfrac{15}{19}\)