1:

a: \(35\cdot16+35\cdot28-44\cdot15\)

\(=35\left(16+28\right)-44\cdot15\)

\(=44\left(35-15\right)\)

\(=44\cdot20=880\)

b: \(240-2\left(3\cdot5^2-20:2^2\right)\)

\(=240-2\left(3\cdot25-20:4\right)\)

\(=240-150+10=10+90=100\)

2:

b: \(\left(8-3x\right)^4-1=15\)

=>\(\left(3x-8\right)^4=16\)

=>\(\left[{}\begin{matrix}3x-8=2\\3x-8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=6\end{matrix}\right.\)

=>x=10/3 hoặc x=2

c: \(218-5\left(x-8\right)=2^5:2^2\)

=>\(218-5\left(x-8\right)=2^3=8\)

=>5(x-8)=210

=>x-8=42

=>x=50

d: \(\left(5-3x\right)^4-1=15\)

=>\(\left(3x-5\right)^4-1=15\)

=>\(\left(3x-5\right)^4=16\)

=>\(\left[{}\begin{matrix}3x-5=-4\\3x-5=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)

a. 5x + 3(x² - x - 1)

= 5x + 3x² - 3x - 3

= 3x² + 5x - 3x - 3

= 3x² + 2x - 3

b. (5 - x)(5 + x) - (2x - 1)²

25 - x² - (4x² - 4x + 1)

= 25 - x² - 4x² + 4x - 1

= 25 - 1 - x² - 4x² + 4x 

= 24 - 5x² + 4x

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

b) \(\left(5-x\right)\left(5+x\right)-\left(2x-1\right)^2=25-x^2-4x^2+4x-1=-5x^2+4x+24\)

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

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

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

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

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

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

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

a: \(=15x^5-25x^4+15x^3\)

b: \(=2x^3+10x^2-8x-x^2-5x+4\)

\(=2x^3+9x^2-13x+4\)

Bài này đề bài phải là khai triển biểu thức, chứ không phải là tính em nhé.

Lời giải:

Ta áp dụng hằng đẳng thức đáng nhớ thôi.

a. $(3+2x)^3=3^3+3.3^2.2x+3.3.(2x)^2+(2x)^3$

$=8x^3+36x^2+54x+27$

b.

$(\frac{1}{2}-y)^3=(\frac{1}{2})^3-3.(\frac{1}{2})^2.y+3.\frac{1}{2}y^2-y^3$

$=-y^3+\frac{3}{2}y^2-\frac{3}{4}y+\frac{1}{8}$

c.

$(x-5)(x^2+5x+25)=(x-5)^2(x^2+5x+5^2)$

$=x^3-5^3=x^3-125$

d.

$(3x+\frac{1}{2})(9x^2-\frac{3}{2}x+\frac{1}{4})$

$=(3x+\frac{1}{2})[(3x)^2-3x.\frac{1}{2}+(\frac{1}{2})^2]$

$=(3x)^3+(\frac{1}{2})^3=27x^3+\frac{1}{8}$

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

Bài 2: 

a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

a: \(2x^3-50x=0\)

=>\(2x\left(x^2-25\right)=0\)

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

=>x∈{0;5;-5}

b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

=>2x(3x-5)+(3x-5)=0

=>(3x-5)(2x+1)=0

=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)

c: \(9\left(3x-2\right)=x\left(2-3x\right)\)

=>9(3x-2)-x(2-3x)=0

=>9(3x-2)+x(3x-2)=0

=>(3x-2)(x+9)=0

=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)

d: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

=>(x-3)(x+2)=0

=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

e: \(25x^2-2=0\)

=>\(25x^2=2\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

f: \(x^2-25=6x-9\)

=>\(x^2-6x-16=0\)

=>(x-8)(x+2)=0

=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)

g: 5x(x-3)-2x+6=0

=>5x(x-3)-2(x-3)=0

=>(x-3)(5x-2)=0

=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)

h: 3x(x-7)-2(x-7)=0

=>(x-7)(3x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)

i: \(7x^2-28=0\)

=>\(7x^2=28\)

=>\(x^2=4\)

=>x=2 hoặc x=-2

j: 2x+1+x(2x+1)=0

=>(2x+1)(x+1)=0

=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)

k: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

=>4(x+2)=0

=>x+2=0

=>x=-2

l: \(x^3+5x^2-4x-20=0\)

=>\(x^2\left(x+5\right)-4\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(x^2-4\right)=0\)

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

=>x∈{-5;2;-2}

m: \(x^2-25+2\left(x+5\right)=0\)

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

=>(x+5)(x-3)=0

=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)

n: \(x^2-3x+2=0\)

=>\(x^2-x-2x+2=0\)

=>x(x-1)-2(x-1)=0

=>(x-1)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

o: \(x^2-6x+8=0\)

=>\(\left(x-2\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

p: \(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

=>(x-7)(x+2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)

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

=>\(x^2-4x+4-x^2+9=6\)

=>-4x+13=6

=>-4x=6-13=-7

=>x=7/4

r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)

=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)

=>-4x+26=18