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Sang Nguyễn Xuân
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『Kuroba ム Tsuki Ryoo...
16 tháng 6 2023 lúc 9:02

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hà Quỳnh Chi
16 tháng 6 2023 lúc 10:04

Hello các bạn còn đó ko?

Nguyễn Anh Thư
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Ngô Hải Đăng
1 tháng 9 2020 lúc 13:54

\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)

\(\Rightarrow-x=5\)

\(\Rightarrow x=-5\)

     \(\text{Vậy x=-5}\)

\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)

\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)

\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)

\(\Rightarrow-16x-8=7\)

\(\Rightarrow-16x=15\)

\(\Rightarrow x=\frac{-15}{16}\)

      \(\text{Vậy }x=\frac{-15}{16}\)

\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)

\(\Rightarrow-9+8x-1=8\)

\(\Rightarrow8x=18\)

\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)

      \(\text{Vậy }x=\frac{9}{4}\)

\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)

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Hoàng Huy
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👁💧👄💧👁
29 tháng 7 2021 lúc 8:50

\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)

\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)

\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)

hải trung
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Akai Haruma
12 tháng 3 2022 lúc 17:46

Lời giải:
a.

$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$

$=5x-(-4x+3)=5x+4x-3=9x-3$

b.

$(-3x^2+2x-1)+(4x^2-2x+3)$

$=-3x^2+2x-1+4x^2-2x+3=x^2+2$

Thùy Linh
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Akai Haruma
28 tháng 10 2023 lúc 17:50

Lời giải:
PT 

\(\Leftrightarrow 4x(x^2-1)-4x^3-(4x^2+4x-3)=2\\ \Leftrightarrow 4x^3-4x-4x^3-4x^2-4x+3=2\\ \Leftrightarrow -4x^2-8x+1=0\\ \Leftrightarrow 4x^2+8x-1=0\\ \Leftrightarrow (2x+2)^2-5=0\\ \Leftrightarrow 2x+2=\sqrt{5} \text{ hoặc } 2x+2=-\sqrt{5}\)

$\Leftrightarrow x=\frac{\sqrt{5}-2}{2}$ hoặc $x=\frac{-\sqrt{5}-2}{2}$

 

Sỹ Tiền
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HT.Phong (9A5)
31 tháng 7 2023 lúc 6:15

p) \(\left(9-x\right)\left(x^2+2x-3\right)\)

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

\(=9x^2+18x-27-x^3-2x^2+3x\)

\(=-x^3+7x^2+21x-27\)

n) \(\left(-x+3\right)\left(x^2+x+1\right)\)

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

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

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

o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)

\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)

\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)

\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)

q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)

\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)

\(=6x^3-12x^2-18x+x^2-2x-3\)

\(=6x^3-11x^2-20x-3\)

r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)

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

\(=-2x^3-6x^2+2x-x^2-3x+1\)

\(=-2x^3-7x^2-x+1\)

u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)

\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)

\(=-2x^3+2x^2+12x+3x^2-3x-18\)

\(=-2x^3+5x^2+9x-18\)

s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)

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

\(=-4x^3-12x^2+8x+5x^2+15x-10\)

\(=-4x^3-7x^2+23x-10\)

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)

\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)

\(=-x^2-3+2x^4+6x+18-12x^3\)

\(=2x^4-12x^3-x^2+6x+15\)

Nguyễn Lê Phước Thịnh
30 tháng 7 2023 lúc 22:55

p: (-x+9)(x^2+2x-3)

=-x^3-2x^2+3x+9x^2+18x-27

=-x^3+7x^2+21x-27

n: (-x+3)(x^2+x+1)

=-x^3-x^2-x+3x^2+3x+3

=-x^3+2x^2+2x+3

o: (-6x+1/2)(x^2-4x+2)

=-6x^3+24x^2-12x+1/2x^2-2x+1

=-64x^3+49/2x^2-14x+1

q: (6x+1)(x^2-2x-3)

=6x^3-12x^2-18x+x^2-2x-3

=6x^3-11x^2-20x-3

r: (2x+1)(-x^2-3x+1)

=-2x^3-6x^2+2x-x^2-3x+1

=-2x^3-7x^2-x+1

u: =-2x^3+2x^2+12x+3x^2-3x-18

=-2x^3+5x^2+9x-18

s: =-4x^3-12x^2+8x+5x^2+15x-10

=-4x^3-7x^2+23x-10

Sỹ Tiền
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HT.Phong (9A5)
31 tháng 7 2023 lúc 5:39

v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)

\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)

\(=-x^2-3x+2c^3x+6x+18-12c^3\)

\(=-x^2+3x+2c^3x+18-12c^3\)

f) \(\left(2x-5\right)\left(x^2-x+3\right)\)

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

\(=2x^3-2x^2+6x-5x^2+5x-15\)

\(=2x^3-7x^2+11x-15\)

w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)

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

\(=3x^3-6x^2-15x+x^2-2x-5\)

\(=3x^3-5x^2-17x-5\)

x) \(\left(6x-3\right)\left(x^2+x-1\right)\)

\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)

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

\(=6x^3+3x^2-9x+3\)

y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)

\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)

\(=15x^2+5x-5x^3-6x-2+2x^2\)

\(=-5x^3+17x^2-x-2\)

z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)

\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)

\(=3x^3+3x^2+3x+4x^2+4x+4\)

\(=3x^3+7x^2+7x+4\)

Nguyễn Lê Phước Thịnh
31 tháng 7 2023 lúc 0:29

f: =2x^3-2x^2+6x-5x^2+5x-15

=2x^3-7x^2+11x-15

w: =3x^3-6x^2-15x+x^2-2x-5

=3x^3-5x^2-17x-5

x: =6x^3+6x^2-6x-3x^2-3x+3

=6x^3+3x^2-9x+3

y: =(5x-2)(-x^2+3x+1)

=-5x^3+15x^2+5x+2x^2-6x-2

=-5x^3+17x^2-x-2

z: =3x^3+3x^2+3x+4x^2+4x+4

=3x^3+7x^2+7x+4

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