a ) Ta có : 4(x - 5) - 3(x + 7) = -19

<=> 4x - 20 - 3x - 21 = -19

=> x - 41 = -19

=> x = -19 + 41

=> x = 22

b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5

<=> 7x - 21 - 15 + 5x = 11x - 5

<=> 12x - 36 = 11x - 5

=> 12x - 11x = -5 + 36

=> x = 31 

típ đi bạn

a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)

\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)

b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)

\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)

c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)

\(=\dfrac{-56}{95}+\dfrac{7}{15}\)

\(=\dfrac{-7}{57}\)

c: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)

=>7/5x+2/3=4/3x-1/4 hoặc 7/5x+2/3=1/4-4/3x

=>1/15x=-11/12 hoặc 41/15x=-5/12

=>x=-55/4 hoặc x=-25/164

d: |7/8x+5/6|=|1/2x+5|

=>|42x+40|=|24x+240|

=>42x+40=24x+240 hoặc 42x+40=-24x-240

=>18x=200 hoặc 66x=-280

=>x=100/9 hoặc x=-140/33

cái câu a) kiểu j đấy

Bài 1

a, 2³ + ( x - 3² ) = 1

x - 3² = 1 - 2³ = -7

x = -7 + 3²

x = 2

b, 5 . (x+7) -10 = 40

5 . (x+7) = 50

x+7 = 50 :5 =10

x = 10 - 7

x = 3

Bài 6:

a: \(x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14-3}{21}=\dfrac{-17}{21}\)

d: \(x=\dfrac{9}{10}\cdot\dfrac{-5}{9}=\dfrac{-1}{2}\)

e: \(\Leftrightarrow x\cdot\dfrac{1}{3}=\dfrac{14}{21}-\dfrac{3}{21}=\dfrac{11}{21}\)

=>x=11/7

a. 

| x | + \(\left|-\frac{2}{5}\right|=-\frac{5}{3}\)

| x | + \(\frac{2}{5}=-\frac{5}{3}\)

| x | = \(-\frac{5}{3}-\frac{2}{5}\)

| x | = \(-\frac{31}{15}\)

\(\Rightarrow x\in\varnothing\)vì trị đối \(\ge\)0

Vậy x \(\in\varnothing\)

b.

| x - 3 | = \(\frac{4}{5}\)

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

\(\Rightarrow\)x = \(\frac{4}{5}+3\)hoặc \(-\frac{4}{5}+3\)

\(\Rightarrow\)x = \(\frac{19}{5}\)hoặc \(\frac{11}{5}\)

Vậy x \(\in\){ \(\frac{19}{5}\); \(\frac{11}{5}\)} 

c.

| x - 7 | = \(\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(\frac{5}{3}\)hoặc \(-\frac{5}{3}\)

\(\Rightarrow\)x = \(\frac{5}{3}+7\)hoặc \(-\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{26}{3}\)hoặc \(\frac{16}{3}\)

Vậy x \(\in\){ \(\frac{26}{3}\); \(\frac{16}{3}\)}

d.

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

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

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

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

Vậy x \(\in\){ \(\frac{3}{4}\); \(\frac{1}{4}\)}

e.

| x - 7 | = \(-\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(-\frac{5}{3}\)hoặc \(\frac{5}{3}\)

\(\Rightarrow\)x = \(-\frac{5}{3}+7\)hoặc \(\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{16}{3}\)hoặc \(\frac{26}{3}\)

Vậy x \(\in\){ \(\frac{16}{3}\); \(\frac{26}{3}\)}

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

a)

\(\dfrac{7}{5}+\dfrac{5}{6}:5-\dfrac{3}{8}\cdot\left(-3\right)\\ =\dfrac{7}{5}+\dfrac{1}{6}+\dfrac{9}{8}\\ =\dfrac{168+20+135}{120}\\ =\dfrac{323}{120}\)

1, Tính tổng:

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)

2, Tìm x:

\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)

- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(=\frac{5}{7}\cdot-\frac{7}{11}\)

\(=-\frac{5}{11}\)