1 B

2 A

3 B

4 A

5 D

6 A

7 A

8 C

9 B

10 D

11 B

\(\Leftrightarrow n^5+n^2-n^2+1⋮n^3+1\)

\(\Leftrightarrow-n^3+n⋮n^3+1\)

\(\Leftrightarrow n=1\)

Câu 11:

Ta có: \(\left(2x-3\right)\left(3x+2\right)-\left(2x-3\right)^2=-18\)

\(\Leftrightarrow6x^2+4x-9x-6-4x^2+12x-9=-18\)

\(\Leftrightarrow2x^2+7x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Em ơi ảnh đề rất mờ em ạ!

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2

1.2.3+....+88.89.90

=1.2.3.(4-0):4+...+88.89.90.(91-87):4

=(1.2.3.4-0.1.2.3):4+...+(88.89.90.91-87.88.89.90):4 (Ta gạch các số hạng giống nhau)

=(0.1.2.3+88.89.90.91):4

=88.89.90.91:4

=16036020 

a) 

<=> \(3x-12x^2+12x^2-6x=9\)

<=> \(-3x=9\)

<=> \(x=-3\)

b)

<=> \(6x-24x^2-12x+24x^2=6\)

<=> \(-6x=6\)

<=> \(x=-1\)

c) 

<=> \(6x-4-3x+6=1\)

<=> \(3x+2=1\)

<=> \(x=-\frac{1}{3}\)

d) 

<=> \(9-6x^2+6x^2-3x=9\)

<=> \(-3x=0\)

<=> \(x=0\)

e) KO HIỂU ĐỀ

f) 

<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)

<=> \(-17x+1=8\)

<=> \(x=-\frac{7}{17}\)

g) 

<=> \(-6x^2+x+1+6x^2-3x=9\)

<=> \(-2x=8\)

<=> \(x=-4\)

h)

<=> \(x^2-x+2x^2+5x-3=4\)

<=> \(3x^2+4x=7\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)

a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)

\(\Rightarrow3x-12x^2+12x^2-6x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)

\(\Rightarrow6x-24x^2-12x+24x^2=6\)

\(\Rightarrow-6x=6\)

\(\Rightarrow x=-1\)

c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)

\(\Rightarrow6x-4-3x+6=1\)

\(\Rightarrow3x+2=1\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

Để mình làm nốt câu n,m,p,q

n, (x² - 2x + 4)(x + 2) - x(x - 1)(x + 1) + 3 = 0

=> x²(x + 2) - 2x(x + 2) + 4(x + 2) - x(x² - 1) + 3 = 0

=> x³ + 2x² - 2x² - 4x + 4x + 8 - x³ + x + 3 = 0

=> (x³ - x³) + (2x² - 2x²) + (-4x + 4x + x) + (8 + 3) = 0

=> x + 11 = 0

=> x = -11

Vậy x = -11

m) (2x - 1)(x + 3) - (x - 4)(2x - 5) = 4x + 1

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

=> 2x² + 6x - x - 3 - 2x² + 5x + 8x - 20 = 4x +1 

=> (2x² - 2x²) + (6x - x + 5x + 8x) + (-3 - 20) = 4x + 1

=> 18x - 23 = 4x + 1

=> 18x - 23 - 4x - 1 = 0

=> 14x + (-23 - 1) = 0

=> 14x - 24 = 0

=> 14x = 24

=> x = 12/7

Vậy x = 12/7

p) (2x - 1)(2x - 3) - (4x + 3)(x - 2) = 8 - x

=> 2x(2x - 3) - 1(2x - 3) - 4x(x - 2) - 3(x - 2) = 8 - x

=> 4x² - 6x - 2x + 3 - 4x² + 8x - 3x + 6 = 8 - x

=> (4x² - 4x²) + (-6x - 2x + 8x - 3x) + (3 + 6) = 8 - x

=> -3x + 9 = 8 - x

=> -3x + 9 - 8 + x = 0

=> (-3x + x) + 1 = 0

=> -2x + 1 = 0

=> -2x = -1

=> x = 1/2

q, 6x² - 2x(3x + 3/2) = 9

=> 6x² - 6x² - 3x = 9

=> -3x = 9

=> x = -3

\(\Leftrightarrow\left\{{}\begin{matrix}5x=10\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2x-3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2x-3\\3x+2x-3=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-3\\5x=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2.2-3\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)