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30 tháng 11 2021

𝑥2+4𝑥+3𝑥2+6𝑥+9

1

Kết hợp các số hạng đồng dạng

𝑥2+4𝑥+3𝑥2+6𝑥+9

4𝑥2+4𝑥+6𝑥+9

2

Kết hợp các số hạng đồng dạng

4𝑥2+4𝑥+6𝑥+9

4𝑥2+10𝑥+9

Đáp án:

4𝑥2+10𝑥+9 .

Đây nha!

`@` `\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`

16 tháng 6 2023

Hello các bạn còn đó ko?

13 tháng 7 2017

a, \(\left(x+3\right)^3-\left(x+2\right)\left(x-2\right)-6x^2-20\)

\(=x^3+9x^2+27x+27-\left(x^2-4\right)-6x^2-20\)

\(=x^3+9x^2+27x+27-x^2+4+6x^2+20\)

\(=x^3+14x^2+27x+51\)

b, \(\left(2x+3\right)\left(4x^2-6x+9\right)-\left(2x-3\right)\left(4x^2+6x+9\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+18-\left(8x^3+12x^2+18x-12x^2-18x-18\right)\)

\(=8x^3+18-8x^3+18=36\)

c, \(\left(2x-1\right)\left(4x^2+2x+1\right)\left(2x+1\right)\left(4x^2-2x+1\right)\)

\(=\left(8x^3+4x^2+2x-4x^2-2x-1\right)\left(8x^3-4x^2+2x+4x^2-2x+1\right)\)

\(=\left(8x^3-1\right)\left(8x^3+1\right)=\left(8x^3\right)^2-1\)

\(=64x^5-1\)

d, \(\left(x+4\right)\left(x^2-4x+16\right)-\left(50+x^2\right)\)

\(=x^3-4x^2+16x+4x^2-16x+64-50-x^2\)

\(=x^3-x^2+14\)

Chúc bạn học tốt!!!

13 tháng 7 2017

Cảm ơn nha !!!

a) \(\frac{4x+3}{6x-4}+\frac{5x-9}{6x-4}\)

\(=\frac{4x+3+5x-9}{2\left(3x-2\right)}=\frac{9x-6}{2\left(3x-2\right)}\)

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

b) \(\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{x^2-1}\)

\(=\frac{2\left(x+1\right)+3\left(x-1\right)-4x+2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x-1}\)

28 tháng 12 2019

a) \(\frac{4x+3}{6x-4}+\frac{5x-9}{6x-4}\)

\(=\frac{4x+3+5x-9}{6x-4}\)

\(=\frac{9x-6}{6x-4}\)

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

\(=\frac{3}{2}.\)

b) \(\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{x^2-1}\)

\(=\frac{2}{x-1}+\frac{3}{x+1}-\frac{4x-2}{\left(x-1\right).\left(x+1\right)}\)

\(=\frac{2.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{3.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{4x-2}{\left(x-1\right).\left(x+1\right)}\)

\(=\frac{2x+2}{\left(x-1\right).\left(x+1\right)}+\frac{3x-3}{\left(x-1\right).\left(x+1\right)}+\frac{-\left(4x-2\right)}{\left(x-1\right).\left(x+1\right)}\)

\(=\frac{2x+2+3x-3-4x+2}{\left(x-1\right).\left(x+1\right)}\)

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

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

Chúc bạn học tốt!

21 tháng 5 2023

a) `sqrt(x^2-6x _9) = 4-x`

`<=> sqrt[(x-3)^2] =4-x`

`<=> |x-3| =4-x ( đk :x<=4)`

`<=> |x-3| = |4-x|`

`<=> [(x-3 =4-x),(x-3 = x-4):}`

`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`

Vậy `S = {7/2}`

b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`

`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `

`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`

`<=> x^2 -9 = x^2 -6x +9`

`<=> 6x = 9+9 =18`

`<=> x=3(t//m)`

Vậy `S={3}`

 

21 tháng 5 2023

c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`

`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`

`<=> |x-1| +|x-2| =3`

xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`

`=> 1-x +2-x =3`

`=> x = 0(t//m)`

xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`

`=> x-1 +2-x =3`

`=>1=3 (vl)`

xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`

`=> x-1+x-2 =3`

`=> x=3(t//m)`

Vậy `S = {0;3}`

31 tháng 7 2023

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

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

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

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

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

a: Ta có: \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)\)

\(=4x^2+12x+9+4x^2-12x+9-8x^2+18\)

\(=36\)

Bài 2: 

a: \(\left(y^2+6x^2\right)\left(y^2-6x^2\right)=y^4-36x^4\)

b: \(\left(4x+5\right)\left(16x^2-20x+25\right)=\left(16x^2-25\right)\left(4x-5\right)\)

\(=64x^3-16x^2-100x+125\)

26 tháng 7 2021

\(A=\left(2x+5\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(8x^3-12x^2+18x+20x^2-30x+45-8x^3+2=8x^2-12x+47\)

Vậy biểu thức phụ thuộc biến x 

\(B=\left(x+3\right)^3-\left(x+9\right)\left(x^2+27\right)\)

\(=x^3+9x^2+27x+27-x^3-27x-9x^2-243=27-243=-216\)

Vậy biểu thức ko phụ thuộc biến x 

AH
Akai Haruma
Giáo viên
26 tháng 7 2021

Lời giải:
$A=(2x+5)(4x^2-6x+9)-2(4x^3-1)$

$=(2x+3)(4x^2-6x+9)+2(4x^2-6x+9)-(8x^3-2)$

$=(2x)^3+3^3+8x^2-12x+18-8x^3+2=48x^2-12x+47$ vẫn phụ thuộc  vào giá trị của biến. Bạn xem lại.

$B=(x+3)^3-(x+9)(x^2+27)$

$=x^3+9x^2+27x+27-(x^3+27x+9x^2+243)$

$=x^3+9x^2+27x+27-x^3-9x^2-27x-243$

$=-216$ không phụ thuộc vào giá trị của biến (đpcm)

`@` `\text {Ans}`

`\downarrow`

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

`= (4x^3 - 4x^2) - (9x - 9)`

`= 4x^2(x - 1) - 9(x - 1)`

`= (4x^2 - 9)(x - 1)`

____

`x^3 + 6x^2 + 11x + 6`

`= x^3 + x^2 + 5x^2 + 5x + 6x + 6`

`= (x^3 + x^2) + (5x^2 + 5x) + (6x + 6)`

`= x^2*(x + 1) + 5x(x + 1) + 6(x + 1)`

`= (x^2 + 5x + 6)(x+1)`

____

`x^2y - x^3 - 9y + 9x`

`= (x^2y - 9y) - (x^3 - 9x)`

`= y(x^2 - 9) - x(x^2 - 9)`

`= (y - x)(x^2 - 9)`

b: =x^3+x^2+5x^2+5x+6x+6

=(x+1)(x^2+5x+6)

=(x+1)(x+2)(x+3)

c: =x^2(y-x)-9(y-x)

=(y-x)(x^2-9)

=(y-x)(x-3)(x+3)

a: =(4x^3-4x^2)-(9x-9)

=4x^2(x-1)-9(x-1)

=(x-1)(4x^2-9)

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