a: 2x-3/2+3/4=-4

=>2x-3/4=-4

=>2x=-13/4

hay x=-13/8

b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)

=>2/3x+3/5=12/145

=>2/3x=-15/29

hay x=-45/58

c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)

=>1/2x-3/5x+13/5=-7/10x-7/5

=>-1/10x+7/10x=-7/5-13/5

=>3/5x=-2

hay x=-2:3/5=-10/3

Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)

=> \(2x-2=-\frac{1}{2}\)

=> \(2x=\frac{3}{2}\)

=> \(x=\frac{3}{4}\)

\(\Leftrightarrow\dfrac{233}{286}\left(2x-x^2\right)=\dfrac{-233}{572}\\ \Leftrightarrow x\left(2-x\right)=\dfrac{-1}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2-x=\dfrac{-1}{2}\Leftrightarrow x=\dfrac{5}{2}\end{matrix}\right.\)

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!!

|x| - 7 = 11

<=> |x| = 18

<=> x = 18

hoặc x = -18

Vậy...

trả lời 

(3/2-5/11-3/13)+(3/-4+5/22+3/26)=0,407...

chúc bn

 hc tốt

trả lời 

(3/2-5/11-3/13)+(3/-4+5/22+3/26)=0,40734265734

chúc bn 

hc tốt

(3/2-5/11-3/13) +(3/4+/5/22+3/26)

=0,4734265734

học tốt

lớp 1 mà cậu

4.24.5²-(3³.18+3³.12)

=4.24.25-[27.(18+12)]

=(4.25).24-[27.30]

=100.24-810

=2400-810

=1590

4.24.5^2-(3^3.18+3^3.12)

31.15.7^2.4-31.49.40

1+2+3+4+5+6+7+8+9+10

1+3+5+7+9+11+13+15+19

2+6+10+14+22+23+26+34

5+8+11+14+17+20+23+26+29

1+6+11+16+21+26+31+36+41+47+51

10+13+16+19+22+25+28+31+34+37+40

5+7+9+11+13+15+17+3+8+13+18+23+28

4+7+10+13+16+19+5+9+13+17+21+25 = 1590

#HT#

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

kb lqmb vs mk ko mk là P.A.D8a1

Làm hết thì đã lên " THÁNH "