a)\(1,2-x+0,8=-1,8-2x\)

\(2-x=-1,8-2x\)

\(2x-x=-1,8-2\)

\(x=-3,8\)

Vậy S={-3,8}

b)\(2,3x-1,4-4x=3,6-1,7x\)

\(2,3x-4x+1,7x=3,6+1,4\)

0=5(vô lí)

Vậy S={\(\varnothing\)}

c)\(6,6-0.9=2,6+0,1x-4\)

\(5,7=0,1x-1,4\)

\(-4,3=0,1x\)

\(x=-43\)

Câu c đáng lẽ là như thế này chứ.

c, 3(2.2-0.3x)=2.6+(0.1x-4)

<=> 6.6-0.9x=2.6+0.1x-4

<=> 6.6-0.9x=0.1x-1.4

<=>-0.9x -0.1x =-8

<=> -x=-8

<=> x=8

Mình trả lời câu d luôn nhé.

d, 3.6-0.5(2x+1)=x-0.25(2-4x)

<=> 3.6-x-0.5=x-0.5+x

<=> 3.1-x=2x-0.5

<=>-x-2x=-3.6

<=> -3x=-3.6

<=> x= 1.2

5.[3/4-2/3x]-13/4=-2{(5/4)²-5/2.1/4+(1/16)²

5.[3/4-2/3x]-13/4=2{25/16-5/8+1/256}

........................=2{400/256-160/250+1/256}

........................=2.241/256

........................=241/128

5.[3/4-2/3x]=241/128+13/4

5.[3/4-2/3x]=241/128+416/128

[3/4-2/3x]=657/128:5

[3/4-2/3x]=657/640

vậy 3/4-2/3x=657/640 và 3/4-2/3x=-657/640

th1:3/4-2/3x va th2:3/4-2/3x=-657/640

còn lại bạn chỉ cần tính kết quả mà dấu[..] là giá tri tuyệt đối đó

=> -3,25;(-1/3x)=28/3

=> -1/3x=28/3*(-3,25)

=> -1/3x=-91/3

=>x=-91/3:(-1/3)

=>  x=91

  vậy x=91

     study well

   k nha 

 mk xin cảm ơn mọi ng nhiều

(-3,25) : (\(-\frac{1}{3}x\)) = (\(-2\frac{1}{3}\)) : (-0,25)

<=> \(-\frac{1}{3}x\)= \(-\frac{39}{112}\)

<=>x = \(\frac{117}{112}\)

\(\left(3,25\right).\left(-3x\right)=\left(-\frac{5}{3}\right).\left(-4\right)\)

\(\left(-3,25\right).\left(-3x\right)=\frac{20}{3}\)

\(\left(-3x\right)=\frac{20}{3}:\left(-3,25\right)\)

\(\left(-3x\right)=-\frac{80}{39}\)

\(x=-\frac{80}{39}:3\)

\(x=-\frac{80}{117}\)

\(\frac{0,25\cdot x}{x}=\frac{5}{6}:0,125\)

\(\Leftrightarrow\frac{\frac{25}{100}\cdot x}{x}=\frac{5}{6}:\frac{125}{1000}\)

\(\Leftrightarrow\frac{\frac{1}{4}\cdot x}{x}=\frac{5}{6}:\frac{1}{8}\)

\(\Leftrightarrow\frac{\frac{1}{4}\cdot x}{x}=\frac{40}{6}\)

\(\Leftrightarrow\frac{1}{4}\cdot x=\frac{40}{6}\cdot x\)

=> x không thỏa mãn

làm 1 bài mẫu nha

a) \(\left(\frac{1}{3}.x\right):\frac{2}{3}=\frac{25}{6}\)

\(\Rightarrow\frac{1}{3}x=\frac{25}{9}\)

\(\Rightarrow x=\frac{25}{9}:\frac{1}{3}\)

\(\Rightarrow x=\frac{25}{3}\)

các bài sau dễ lắm

a) \(\left(\frac{1}{3}x\right):\frac{2}{3}=1\frac{2}{3}:\frac{2}{5}\)

\(\Rightarrow\frac{1}{3}x:\frac{2}{3}=\frac{25}{6}\)

\(\Rightarrow\frac{1}{3}x=\frac{5}{3}\)

\(\Rightarrow x=\frac{5}{3}:\frac{1}{3}\)

\(\Rightarrow x=5\)

a) \(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=1\frac{2}{3}:\frac{2}{5}\)

\(\left(\frac{1}{3}\cdot x\right):\frac{2}{3}=\frac{25}{6}\)

\(\frac{1}{3}x=\frac{25}{6}\cdot\frac{2}{3}\)

\(\frac{1}{3}x=\frac{25}{9}\)

\(x=\frac{25}{9}:\frac{1}{3}\)

\(x=\frac{25}{3}\)

b) \(4,5:0,3=2,25:\left(0,1x\right)\)

\(15=2,25:\left(0,1x\right)\)

\(2,25:15=0,1x\)

\(0,15=0,1x\)

\(0,15:0,1=x\)

\(1,5=x\)

c) \(8:\left(\frac{1}{4}\cdot x\right)=2:0,02\)

\(8:\left(\frac{1}{4}\cdot x\right)=100\)

\(\frac{1}{4}\cdot x=8:100\)

\(\frac{1}{4}\cdot x=0,08\)

\(x=0,08:\frac{1}{4}\)

\(x=0,32\)

d) \(3:2\frac{1}{4}=\frac{3}{4}:\left(6\cdot x\right)\)

\(\frac{4}{3}=\frac{3}{4}:\left(6\cdot x\right)\)

\(\frac{3}{4}:\frac{4}{3}=6\cdot x\)

\(\frac{9}{16}=6\cdot x\)

\(\frac{9}{16}:6=x\)

\(\frac{3}{32}=x\)

e) \(3,8:\left(2\cdot x\right)=\frac{1}{4}:2\frac{2}{3}\)

\(3,8:\left(2\cdot x\right)=\frac{3}{32}\)

\(2\cdot x=3,8:\frac{3}{32}\)

\(2\cdot x=\frac{608}{15}\)

\(x=\frac{608}{15}:2\)

\(x=\frac{304}{15}\)

f) \(\left(0,25\cdot x\right):3=\frac{5}{6}:0,125\)

\(\left(0,25\cdot x\right):3=\frac{20}{3}\)

\(0,25\cdot x=\frac{20}{3}\cdot3\)

\(0,25\cdot x=20\)

\(x=20:0,25\)

\(x=80\)

g) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1\cdot x\right)\)

\(\frac{5}{3}=\frac{2}{3}:\left(0,1\cdot x\right)\)

\(\frac{2}{3}:\frac{5}{3}=0,1\cdot x\)

\(\frac{2}{5}=0,1x\)

\(\frac{2}{5}:0,1=x\)

\(4=x\)

(Nhớ thử lại kết quả nhé)

a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)

b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)

c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)

d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)

Theo đề ta có :

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\left(x+17\right)-\left(x+2\right)=x\)

\(\Rightarrow x=15\)