\(a,81\cdot2022+25\cdot2022-6\cdot2022=2022\cdot\left(81+25-6\right)=2022\cdot100=202200\)

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

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

\(x-1=\frac{9}{10}\)

\(x=\frac{19}{10}\)

Vậy \(x=\frac{19}{10}\)

( Nếu phần b là hỗn số thì mình làm thế kia , còn nếu là nhân thì bạn tham khảo Câu hỏi của lương bảo ngọc - Toán lớp 5 - Học trực tuyến OLM nhé )

81 x 2022 + 25 x 2022 - 6 x 2022

= ( 81 + 25 - 6 ) x 2022

= 100 x 2022

= 202 200

b) \(\left(\text{x - 1}\right)\frac{\text{2}}{\text{3}}-\frac{\text{1}}{\text{5}}=\frac{\text{2}}{\text{5}}\)

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

=> \(\text{3 x ( x - 1 ) }+\text{2}=\frac{\text{3}}{\text{5}}\text{ x 3 = }\frac{\text{9}}{\text{5}}\)

=> \(\text{3 x ( x - 1 ) }=\frac{\text{9}}{\text{5}}-\text{2}=\frac{\text{-1}}{\text{5}}\)

=> \(\text{ x-1}=\frac{\text{-1}}{\text{5}}:3=\frac{\text{-1}}{\text{15}}\)

=> \(\text{x}=\frac{\text{-1}}{\text{15}}+\text{1 = }\frac{\text{14}}{\text{15}}\)

Bài 2:

a, |x-1| -x +1=0

|x-1| = 0-1+x

|x-1| = -1 + x

 \(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)

 \(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)

 \(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)

x = 2-x

2x = 2

x = 2:2

x=1

b, |2-x| -2 = x

|2-x| = x+2

\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)

2-x = x+2

x+x = 2-2

2x = 0

x = 0

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a, 11 - ( 15 + 11 ) = x - ( 25 - 9 )

11-15-11=x-25+9

-15=x-16

x=1

b, 2 - x = 17 - ( - 5 )

2-x=22

x=-20

c, x - 12 = ( - 9 ) - 15

x-12=-24

x=-12

d, 9 - 25 = ( 7 - x ) - ( 25 + 7 )

(7-x)-32=-16

7-x=16

x=-9

#H

Lời giải:

a. 

$(25-2x)^3:5-3^2=4^2$

$(25-2x)^3:5=4^2+3^2=25$

$(25-2x)^3=25.5=5^3$

$\Rightarrow 25-2x=5$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

b.

$2.3^x=10.3^{12}+8.27^4=10.3^{12}+8.3^{12}=18.3^{12}=2.3^{14}$

$\Rightarrow 3^x=3^{14}$

$\Rightarrow x=14$

a)

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

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{5}{7}\) x \(\dfrac{1}{2}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{5}{14}\)

\(\Rightarrow x=\dfrac{5}{14}+\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{15}{42}+\dfrac{2}{42}\)

\(\Rightarrow x=\dfrac{17}{42}\)

b)

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

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

\(x\) x \(\dfrac{1}{2}=\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:\dfrac{1}{2}\)

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

c)

\(\dfrac{26}{5}-x=\dfrac{9}{15}\) x \(\dfrac{25}{3}\)

\(\dfrac{26}{5}-x=5\)

\(\Rightarrow x=\dfrac{26}{5}-5\)

\(\Rightarrow x=\dfrac{26}{5}-\dfrac{25}{5}\)

\(\Rightarrow x=\dfrac{1}{5}\)

a) x - 1845 = 115 x 2 - 25 x 3

x - 1845 = 230 - 75

x - 1845 = 155

x = 155 + 1845

\(b,230+35x4=x+\frac{700}{5}+30\)

230 + 140 = x + 140 + 30

370 = x + 170

x = 370 - 170

x = 200

a) x - 1845 = 115 x 2 - 25 x 3

x - 1845 = 230 - 75

x - 1845 = 155

x            = 155 + 1845

x            = 2000

          Vậy x = 2000

b) 230 + 35 x 4 = x + 700 / 5+30

    370               = x + 140 + 30

    370               = x + 170

    x                   = 370 - 170

    x                   = 200

             Vậy x = 200

1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)

\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)

2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)

3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)

1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)