a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

=>-4x=5

hay x=-5/4

b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

=>42x=41

hay x=41/42

`a)(10x+9)x-(5x-1)(2x+3)=8`

`<=>10x^2+9x-10x^2-15x+2x+3=8`

`<=>-4x=5`

`<=>x=-5/4`     Vậy `S={-5/4}`

`b)(3x-5)(7-5x)+(5x+2)(3x-2)-2=0`

`<=>21x-15x^2-35+25x+15x^2-10x+6x-4-2=0`

`<=>42x=41`

`<=>x=41/42`       Vậy `S={41/42}`

a: 

=>-4x=5

hay x=-5/4

b: 

=>42x=41

hay x=41/42

a) (3x - 1)(2x + 7) - (x + 1)(6x - 5) = 16

6x² + 21x - 2x - 7 - 6x² + 5x - 6x + 5 = 16

(6x² - 6x²) + (21x - 2x + 5x - 6x) + (-7 + 5) = 16

18x - 2 = 16

18x = 18

x = 1

Vậy x = 1

b) (10x + 9)x - (5x - 1)(2x + 3) = 8

10x² + 9x - 10x² - 15x + 2x + 3 = 8

(10x² - 10x²) + (9x - 15x + 2x) + 3 = 8

-4x + 3 = 8

-4x = 5

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

Vậy x = \(\frac{-5}{4}\)

c) x(x + 1)(x + 6) - x³ = 5x

(x² + x)(x + 6) - x³ = 5x

x³ + 7x² + 6x - x³ = 5x

7x² + 6x = 5x

x(7x + 6) = 5x

=> 7x + 6 = 5

7x = -1

x = \(\frac{-1}{7}\)

Vậy x = \(\frac{-1}{7}\)

d) (3x - 5)(7 - 5x) + (5x + 2)(3x - 2) - 2 = 0

21x - 15x² - 35 + 25x + 15x² - 10x + 6x - 4 - 2 = 0

(-15x² + 15x²) + (21x + 25x - 10x + 6x) + (-35 - 4 - 2) = 0

42x - 41 = 0

42x = 41

x = \(\frac{41}{42}\)

Vậy x = \(\frac{41}{42}\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a/ pt đãcho tương đương với

6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16

<=>18x=18

=> x=1

b/ pt đã cho tương đương với

10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8

<=> -4x=5

<=.> x=-\(\frac{5}{4}\)

c/ pt đã cho tương đương với

21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0

<=>42x=41

<=> x= \(\frac{41}{42}\)

d/ pt đã cho tương đương với

( x\(^2\)+x )(x+6)-x\(^3\)=5x

<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x

<=> 8x\(^2\)+6x-5x=0

<=>8x\(^2\)+16x-10x-5x=0

<=> (x+2)2x-5(x+2)=0

<=> (x+2)(2x-5)=0

<=>x+2=0 hoặc 2x+5=0

=> x=-2 hoặc x= -\(\frac{5}{2}\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

a.

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)

\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)

\(18x-2=16\)

\(18x=16+2\)

\(18x=18\)

\(x=\frac{18}{18}\)

\(x=1\)

b.

\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3=8\)

\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)

\(-4x=8-3\)

\(-4x=5\)

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

c.

\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)

\(42x=41\)

\(x=\frac{41}{42}\)

- mọi người ơi giúp mình với ạ. ai mình cx cho đúng

A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)

B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

Câu C) bạn xem lại đề nha mik tính ko đc

D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?