Bài 4:

a: \(216x^3+27y^3=27\left(8x^3+y^3\right)\)

\(=27\left[\left(2x\right)^3+y^3\right]\)

\(=27\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

b: \(64a^3-8=8\left(8a^3-1\right)\)

\(=8\left[\left(2a\right)^3-1^3\right]\)

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

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

d: \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)

\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x\cdot2y+\left(2y\right)^2\right]\)

\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

Bài 5:

a: \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)

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

\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)

\(=2y^2-10xy\)

b: \(\left(x-y\right)^3-3\left(x-y\right)^2\cdot x+3\left(x-y\right)\cdot x^2-x^3\)

\(=\left(x-y-x\right)^3\)

\(=\left(-y\right)^3=-y^3\)

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

\(=27\left(x+1\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)

\(=25\left(x+1\right)^3-25x^2+10x-1\)

\(=25x^3+75x^2+75x+25-25x^2+10x-1\)

\(=25x^3+50x^2+85x+24\)

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

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

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

\(=\left(-3x+2\right)\left(3x^2-9x+13\right)+\left(3x-1\right)^2\)

\(=-9x^3+27x^2-39x+6x^2-18x+26+9x^2-6x+1\)

\(=-9x^3+42x^2-63x+27\)

\(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=4+5+\dfrac{3}{8}+\dfrac{2}{3}\)

\(=9+\dfrac{9}{24}+\dfrac{16}{24}\)

\(=9+\dfrac{25}{24}\)

\(=10\dfrac{1}{24}\)

\(4\dfrac{3}{8}+5\dfrac{2}{3}\)

\(=\dfrac{35}{8}+\dfrac{17}{3}\)

\(=\dfrac{105}{24}+\dfrac{136}{24}\)

\(=\dfrac{241}{24}\)

\(\dfrac{1}{5}.\left(x+2\right)^2+\dfrac{1}{3}.\left(2x-2\right)^3=\dfrac{1}{5}.\left(x+2\right)^2+\dfrac{1}{3}.2^3\)

\(\Rightarrow\left(2x-2\right)^3=2^3\)

\(\Rightarrow2x-2=2\)

\(\Rightarrow2x=2+2\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4\div2\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

`1/5 . (x+2)^2 + 1/3 . (2x - 2)^3 = 1/5 . (x+2)^2 + 1/3 . 2^3`

`<=> 1/5 . (x+2)^2 -  1/5 . (x+2)^2+ 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 0 + 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 1/3 . (2x - 2)^3 = 1/3 . 2^3`

`<=> 1/3 : 1/3 . (2x - 2)^3 = 2^3`

`<=> 1 . (2x - 2)^3 = 2^3`

`<=> (2x - 2)^3 = 2^3`

`<=> 2x - 2 = 2`

`<=> 2x = 2+2 `

`<=> 2x = 4`

`<=> x = 4 : 2`

`<=> x = 2`

Vậy `x = 2`

`A= (x+5)/(x+3 )` 

Điều kiện: `x ≠ -3`

Do `x ∈ Z => x + 5` và `x + 3∈ Z`

Để `A ∈ Z <=> x + 5 ⋮x + 3`

`<=> x + 3 + 2 ⋮ x + 3`

Do `x + 3 ⋮ x + 3`

Nên `2 ⋮ x + 3`

`=> x + 3 ∈ Ư(2) =` {`-2;-1;1;2`}

`=> x ∈` {`-5;-4;-2;-1`} (Thỏa mãn)

Vậy ...

------------------------------

`B =(x-2)/(x+1)`

Điều kiện: `x ≠ -1`

Do `x ∈ Z => x -2` và `x + 1 ∈ Z`

Để `B ∈ Z <=> x -2 ⋮x + 1`

`<=> x + 1 - 3 ⋮x + 1`

Do `x + 1 ⋮x + 1`

Nên `3⋮x + 1`

`=> x + 1 ∈ Ư(3) =` {`-3;-1;1;3`}

`=> x ∈` {`-4;-2;0;2`} (Thỏa mãn)

Vậy ...

\(A=\dfrac{x+5}{x+3}\in Z\)

\(\Rightarrow\left(x+5\right)⋮\left(x+3\right)\)

Mà \(\left(x+3\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x+5\right)-\left(x+3\right)⋮\left(x+3\right)\)

\(\Rightarrow2⋮\left(x+3\right)\)

\(\Rightarrow\left(x+3\right)\inƯ\left(2\right)\)

\(\Rightarrow\left(x+3\right)\in\left\{1;-1;-2;2\right\}\)

Ta có bảng giá trị:

Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)

Những câu còn lại, cách làm tương tự, nếu như còn thắc mắc thì bạn tag mình nhé.

a: AF//BE

AF\(\perp\)AC

Do đó: BE\(\perp\)AC

b: Vì \(\widehat{F}=\widehat{EDC}\left(=75^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên AF//CD

mà AF\(\perp\)AB

nên CD\(\perp\)AB

=>\(\widehat{C_1}=90^0\)

Ta có: BE//AF

=>\(\widehat{E_2}=\widehat{F}=75^0\)

Ta có: \(\widehat{E_1}+\widehat{E_2}=180^0\)(hai góc kề bù)

=>\(\widehat{E_1}=180^0-75^0=105^0\)

Vì BE\(\perp\)AC

nên \(\widehat{B_1}=90^0\)

(\(x+1\))(y + 1) = 6

Lập bảng ta có:

Theo bảng trên ta có 

(\(x;y\)) = (0; 5); (1; 2); (2; 1); (5; 0)

\(0< 25^0< 90^0\Rightarrow cos25^0>0\)

\(\Rightarrow cos25^0=\sqrt{1-sin^225^0}=\sqrt{1-a^2}\)

\(tan25^0=\dfrac{sin25^0}{cos25^0}=\dfrac{a}{\sqrt{1-a^2}}\)

\(cot25^0=\dfrac{1}{tan25^0}=\dfrac{\sqrt{1-a^2}}{a}\)

a: AG\(\perp\)AB

BD\(\perp\)AB

Do đó: AG//BD

b: Ta có: \(\widehat{FEB}=\widehat{FAC}\left(=45^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên ED//AC

c: Vì \(\widehat{CHD}=\widehat{HDG}\left(=65^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên CF//DG

d: Ta có: \(\widehat{EAC}+\widehat{BAC}=\widehat{EAB}\)

=>\(\widehat{A_2}=90^0-45^0=45^0\)

Ta có: \(\widehat{EAC}=\widehat{BAC}\left(=45^0\right)\)

mà tia AC nằm giữa hai tia AB,AE

nên AC là phân giác của góc BAE

e: Xét ΔABC vuông tại B có \(\widehat{BAC}=45^0\)

nên ΔBAC vuông cân tại B

=>\(\widehat{C_1}=\widehat{A_2}=45^0\)

f: AC//ED

=>\(\widehat{C_2}=\widehat{CHD}\)(hai góc so le trong)

=>\(\widehat{C_2}=65^0\)

Ta có: \(\widehat{C_1}+\widehat{C_2}+\widehat{C_3}=180^0\)

=>\(\widehat{C_3}=180^0-65^0-45^0=70^0\)

FE//CD

=>\(\widehat{F_1}=\widehat{C_3}\)(hai góc so le trong)

=>\(\widehat{F_1}=70^0\)

CF//GD

=>\(\widehat{G_1}=\widehat{F_1}\)

=>\(\widehat{G_1}=70^0\)