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Jungkook Jeon
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Lê Ng Hải Anh
16 tháng 7 2018 lúc 10:11

dài wa,lm xong chắc đến năm sau

1)A=3(x-1)^2-(x+1)^2+2(x-3)(x+3)-(2x+3)^2-(5-20x)

=3(x^2-2x+1)-(x^2+2x+1)+2(x^2-9)-(4x^2+12x+9)-(5-20x)

=3x^2-6x+3-x^2-2x-1+2x^2-18-4x^2-12x-9-5+20x

=-30

2)B=5x(x-7)(x+7)-x(2x-1)^2-(x^3+4x^2-246x)-175

=5x(x^2-49)-x(4x^2-4x+1)-x^3-4x^2+246x-175

=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175

=-175

cn lại lm tg tự nha bn

Bich Hong
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Nguyễn Lê Phước Thịnh
19 tháng 5 2022 lúc 22:08

Bài 1:

a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-5+20x-4x^2-12x-9\)

\(=-30\)

b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

\(=-175\)

d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)

\(=-11x^2-32x+3-48+32x+11x^2-44\)

=-89

Son Nguyen
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Nguyễn Lê Phước Thịnh
22 tháng 5 2022 lúc 0:12

a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

Do đó: A=C+D

\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-4x^2-12x-9-5+20x\)

\(=-30\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

=-175

A=C+D=-30-175=-205

b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

Do đó: B=E+F

\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)

\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)

\(=-18x^3+8x^2+7x-4\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)

\(=-95\)

\(B=-18x^3+8x^2+7x-99\)

trà my Đinh
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Trần Thảo Linh
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Nguyễn Lê Phước Thịnh
4 tháng 8 2021 lúc 11:41

a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x+15-3=2x+2+7\)

\(\Leftrightarrow3x+12=2x+9\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

\(\Leftrightarrow\left(x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

c) Ta có: \(75-\left(2x+1\right)^3=11\)

\(\Leftrightarrow\left(2x+1\right)^3=64\)

\(\Leftrightarrow2x+1=4\)

hay \(x=\dfrac{3}{2}\)

i love rosé
4 tháng 8 2021 lúc 12:31

a) Ta có: 3(x+5)−3=2(x+1)+7

⇔3x+15−3=2x+2+7

⇔3x+12=2x+9

hay x=-3

b) Ta có: 15−(x+2)2=−1

⇔(x+2)2=16

⇔[x+2=4x+2=−4⇔[x=2x=−6

c) Ta có: 

_linh never die_
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Trần Ái Linh
4 tháng 8 2021 lúc 11:56

`3(x+5)-3=2(x+1)+7`

`3x+15-3=2x+2+7`

`x=-3`

.

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

`(x+2)^2=16`

`(x+2)^2=4^2=(-4)^2`

`[(x+2=4),(x+2=-4):}`

`[(x=2),(x=-6):}`

.

`75-(2x+1)^3=11`

`(2x+1)^3=64`

`(2x+1)^2=4^3`

`2x+1=4`

`x=3/2`

i love rosé
4 tháng 8 2021 lúc 12:30

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

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

x=-3

.

15-(x+2)2=-1

(x+2)2=16

(x+2)2=42=(-4)2

[x+2=4x+2=-4

Nguyễn Lê Phước Thịnh
4 tháng 8 2021 lúc 12:39

a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)

\(\Leftrightarrow3x-2x=2+7-15+3\)

hay x=-3

b) Ta có: \(15-\left(x+2\right)^2=-1\)

\(\Leftrightarrow\left(x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vy trần
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Nguyễn Hoàng Minh
8 tháng 9 2021 lúc 14:32

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Nguyễn Lê Phước Thịnh
8 tháng 9 2021 lúc 14:29

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

Tô Mì
8 tháng 9 2021 lúc 14:41

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

Son Nguyen
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Nguyễn Lê Phước Thịnh
22 tháng 5 2022 lúc 1:35

 

a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

Do đó: A=C+D

\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)

\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)

\(=4x^2-8x-16-4x^2-12x-9-5+20x\)

\(=-30\)

\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)

\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)

\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)

=-175

A=C+D=-30-175=-205

b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

Do đó: B=E+F

\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)

\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)

\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)

\(=-18x^3+8x^2+7x-4\)

\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)

\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)

\(=-95\)

\(B=-18x^3+8x^2+7x-99\)

Chanhh
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Nguyễn Lê Phước Thịnh
23 tháng 9 2021 lúc 21:08

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

Nguyễn Hoàng Minh
23 tháng 9 2021 lúc 21:11

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)