a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)

\(\left(2x+1\right)^2=\frac{25}{16}\)

\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)

\(x^3-\frac{9}{16}.x=0\)

\(x\left(x^2-\frac{9}{16}\right)=0\)

\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)

\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

   =  \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

    = \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

     = \(\frac{1}{4}+\frac{1}{2}\)

      =  \(\frac{3}{4}\)

b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)

    =\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)

   = \(-\frac{35}{27}+\frac{47}{21}\)

   =        \(\frac{178}{189}\)

c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)

  = \(\frac{117}{13}-\frac{311}{65}\)

 =       \(\frac{274}{65}\)

d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)

= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)

=     \(\frac{1}{3}+\frac{5}{2}\)

=         \(\frac{17}{6}\)

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

\(\frac{5}{7}+\frac{3}{4}=\frac{41}{28}\)\(Đ\)

\(3+\frac{2}{5}=1\)\(S\)

\(\frac{11}{9}-\frac{5}{6}=\frac{7}{18}\)\(Đ\)

\(4-\frac{3}{7}=\frac{25}{7}\)\(Đ\)

\(x+\frac{4}{11}=\frac{3}{5}\).

          \(x=\frac{3}{5}-\frac{4}{11}\)

          \(x=\frac{13}{55}\)

Vậy ta chọn câu C

dung hay sai

5/7 + 3/4 = 41/28 .....Đ...

3 + 2/5 = 1....S....

11/9 - 5/6 = 7/18....Đ...

4 - 3/7 = 25/7.....

tim x

x + 4/11 =3/5

Đáp án : C .13/55

Bài Đúng hay Sai :

5/7 + 3/4 = 41/28 là đúng 

3 + 2/5 = 1 là sai 

11/9 - 5/6 = 7/18 là đúng 

4 - 3/7 = 25/7 là đúng 

Bài tìm x :

x + 4/11 = 3/5 

           x  = 3/5 - 4/11 

           x  = 13/55 

Vậy chọn đáp án C .

~ Chúc em học giỏi ! ~

a) 89-(73-x) = 20

=> 73+x =89-20

=> 73+x=69

=> x=69-73

=> x=-4

b) (x+7)-25=13

=> x+7=13+25

=> x+7=38

=> x=38-7

=> x=31

c) 98-(x+4)=20

=> x+4=98-20

=> x+4=78

=> x=78-4

=> x=73

d) 140:(x-8)=7

=> x-8=140:7

=> x-8=20

=> x=20+8

=> x=28

e) 4(x+41)=400

=> x+41=400:4

=> x+41=100

=> x=100-41

=> x=59

f) x-[42+(-28)]=-8

=> x-14=-8

=> x=-8+14

=> x=6

g) x+5=20-(12-7)

=> x+5=20-5

=> x+5=15

=> x=15-5

=> x=10

h) \(x-11=2.2^3+20:5\)

=> x-11=2.8+4

=> x-11=16+4

=> x-11=20

=> x=20+11

=> x=31

i) \(4\left(x-3\right)=7^2-1^3\)

=> 4(x-3)=49-1

=> 4(x-3)=48

=> x-3=48:4

=> x-3=12

=> x=12+3

=> x=15

mình cần gấp

a, 28 + 2x = 35 - ( -13 )

=> 2x = 35 + 13 - 28

=>2x = 20

=> x = 10. vậy x = 10

a, \(28+2x=35+13\Leftrightarrow28+2x=48\Leftrightarrow2x=20\Leftrightarrow x=10\)

b, \(-25-x-5=415+5x-415\Leftrightarrow-6x=30\Leftrightarrow x=-5\)

\(1)x+\frac{5}{6}\times2\frac{2}{5}-1\frac{1}{4}=35\%\)

 \(x+\frac{5}{6}\times\frac{12}{5}-\frac{5}{4}=\frac{7}{12}\)

\(x+\frac{5}{6}\times\frac{12}{5}=\frac{7}{12}+\frac{5}{4}\)

\(x+\frac{5}{6}.\frac{12}{5}=\frac{8}{5}\)

\(x+\frac{5}{6}=\frac{8}{5}:\frac{12}{5}\)

\(x+\frac{5}{6}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{5}{6}\)

\(x=-\frac{1}{6}\)

HỌC TỐT !

\(2\)) \(\left|x-\frac{1}{2}\right|-\frac{3}{4}=0\)

         \(\left|x-\frac{1}{2}\right|\)       \(=0+\frac{3}{4}\)

         \(\left|x-\frac{1}{2}\right|\)        \(=\frac{3}{4}\)

           \(x-\frac{1}{2}\)          \(=\frac{3}{4}\)hoặc \(-\frac{3}{4}\)

Ta xét 2 trường hợp :

  Trường hợp 1 :         \(x-\frac{1}{2}=\frac{3}{4}\)

                                    \(x\)         \(=\frac{3}{4}+\frac{1}{2}\)

                                    \(x\)           \(=\frac{5}{4}\)

Trường hợp 2 :           \(x-\frac{1}{2}=-\frac{3}{4}\)

                                    \(x\)          \(=-\frac{3}{4}+\frac{1}{2}\)

                                   \(x\)             \(=-\frac{1}{4}\)

Vậy \(x\in\text{{}\frac{5}{4};-\frac{1}{4}\)}

có giải chi tiết ko bạn