1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2

2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2

7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7

3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0

8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9

4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2

9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9

5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8

10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4

1) 2(x + 5) + 3(x + 7) = 41

2x + 10 + 3x + 21 = 41

5x + 31 = 41

5x = 41 - 31

5x = 10

x = 10 : 5

x = 2

2) 5(x + 6) + 2(x - 3) = 38

5x + 30 + 2x - 6 = 38

7x + 24 = 38

7x = 38 - 24

7x = 14

x = 14 : 7

x = 2

3) 7(5 + x) + 2(x - 10) = 15

35 + 7x + 2x - 20 = 15

9x + 15 = 15

9x = 15 - 15

9x = 0

x = 0

4) 3(x + 4) + (8 - 2x) = 22

3x + 12 + 8 - 2x = 22

x + 20 = 22

x = 22 - 20

x = 2

5) 4(x + 5) + 3(7 - x) = 49

4x + 20 + 21 - 3x = 49

x + 41 = 49

x = 49 - 41

x = 8

6) 7(x - 1) + 5(3 - x) = 11x - 10

7x - 7 + 15 - 5x = 11x - 10

2x - 11x + 8 = -10

-9x = -10 - 8

-9x = -18

x = -18 : (-9)

x = 2

7) 4(2 + x) + 3(x - 2) = 12

8 + 4x + 3x - 6 = 12

7x + 2 = 12

7x = 12 - 2

7x = 10

x = 10/7

8) 5(2 + x) + 4(3 - x) = 10x - 15

10 + 5x + 12 - 4x = 10x - 15

10x - 15 = x + 22

10x - x = 22 + 15

9x = 37

x = 37/9

9) 7(x - 2) + 5(3 - x) = 11x - 6

7x - 14 + 15 - 5x = 11x - 6

11x - 6 = 2x + 1

11x - 2x = 1 + 6

9x = 7

x = 7/9

10) 5(3 - x) + 5(x + 4) = 6 + 4x

15 - 5x + 5x + 20 = 6 + 4x

6 + 4x = 35

4x = 35 - 6

4x = 29

x = 29/4

a: \(\dfrac{x-3}{5-x}=\dfrac{5}{7}\left(x\ne5\right)\)

=>7(x-3)=5(5-x)

=>7x-21=25-5x

=>12x=46

=>x=23/6

b: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)(ĐKXĐ: \(x\notin\left\{1;-7\right\}\))

=>(x-2)(x+7)=(x+4)(x-1)

=>\(x^2+5x-14=x^2+3x-4\)

=>5x-14=3x-4

=>2x=10

=>x=5(nhận)

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

a) 3/8 . x = 9/8 - 1

3/8 . x = 1/8

x = 1/8 : 3/8

x = 1/3

b) 4/5 . x = 7/5 - 1/5

4/5 . x = 6/5

x = 6/5 : 4/5

x = 3/2

c) 12/7 : x + 2/3 = 7/5

12/7 : x = 7/5 - 2/3

12/7 : x = 11/15

x = 12/7 : 11/15

x = 180/77

d) 3.(x + 7) - 15 = 27

3.(x + 7) = 27 + 15

3.(x + 7) = 42

x + 7 = 42 : 3

x + 7 = 14

x = 14 - 7

x = 7

a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)

b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)

\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)

c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)

d) \(3\left(x+7\right)-15=27\)

\(\Leftrightarrow3\left(x+7\right)=42\)

\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

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

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

<=> \(\left(x+\frac{1}{3}\right).\left(-7\right)=\frac{1}{2}.x\)

<=> \(-7x-\frac{7}{3}=\frac{1}{2}.x\)

<=> \(-7x-\frac{1}{2}.x=\frac{7}{3}\)

<=> \(\left(-7-\frac{1}{2}\right).x=\frac{7}{3}\)

<=> \(\frac{-15}{2}.x=\frac{7}{3}\)

<=> \(x=\frac{7}{3}:\frac{-15}{2}=\frac{-14}{45}\)

\(-7x-\frac{7}{3}=\frac{1}{2}.x\)

<=> \(-7x-\frac{1}{2}x=\frac{7}{3}\)

<=> \(\left(-7-\frac{1}{2}\right).x=\frac{7}{3}\)

<=> \(\frac{-15}{2}.x=\frac{7}{3}\)

<=> \(x=\frac{7}{3}:\frac{-15}{2}=\frac{-14}{45}\)

cảm ơn bạn nha!

a,     \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\)    = \(\dfrac{10}{21}\)

    (\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\)  =  \(\dfrac{10}{21}\)

     - \(\dfrac{5}{21}\) \(\times\) \(x\)      = \(\dfrac{10}{21}\)

                 \(x\)      = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))

                 \(x\)      = -2 

b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)

            \(x\) - \(\dfrac{1}{3}\)    =  \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))

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

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

              \(x\)      = - \(\dfrac{13}{6}\)

c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)

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

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

              \(x\)           = - \(\dfrac{47}{10}\)

\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)

\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)

a,73​x−32​x=2110​⇒x(73​−32​)=2110​⇒x.−215​=2110​⇒x=−2b,357​:(x−31​)=−252​⇒51​:(x−31​)=−252​⇒x−31​=−25​⇒x=−613​c,3.(x−21​)−5.(x+53​)=−x+51​⇒3x−23​−5x+5=−x+51​

⇒�(3−5)−32+5=−�+15⇒−2�−132=−�+15⇒−�−135=15⇒�=15−135⇒�=−125.⇒x(3−5)−23​+5=−x+51​⇒−2x−213​=−x+51​⇒−x−513​=51​⇒x=51​−513​⇒x=−512​.

a) `4x(x-5)-(x-1)(x-3)=23`

`<=> 4x^2-20x-x^2+4x-3=23`

`<=>3x^2-16x-26=0`

`<=> x=(8+-\sqrt142)/3

*Nếu đề là: `4x(x-5)-(x-1)(4x-3)=23`

`<=> 4x^2-20x-4x^2+7x-3=23`

`<=>-13x=26`

`<=>x=-2`

b) `(x-5)(x-4)-(x+1)(x-2)=7`

`<=>x^2-9x+20-x^2+x+2=7`

`<=>-8x=-15`

`<=>x=15/8`

\(\text{a) 4x(x-5)-(x-1)(1x-3)=23}\)

\(\Leftrightarrow4x^2-20x-\left(x^2-3x-x+3\right)=23\)

\(\Leftrightarrow4x^2-20x-x^2+3x+x-3=23\)

\(\Leftrightarrow3x^2-16x-26=0\)

Giải phương trình bậc 2 một ẩn ta được :

\(x_1=\dfrac{8-\sqrt{142}}{3};x_2=\dfrac{8+\sqrt{142}}{3}\)