bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

\(=> 2x+3=7\)

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)

Bài 2 :

Câu a : \(y\left(y^3+y^2-y-2\right)-\left(y^2-2\right)\left(y^2+y+1\right)\)

\(=y^4+y^3-y^2-2y-y^4-y^3-y^2+2y^2+2y+2\)

\(=2\) \(\Rightarrow\) ko phụ thuộc vào biến .

Câu b : \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)

\(=29\Rightarrow\) ko thuộc vào biến

Câu c : \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)

\(=3x^2+15x-3x^2+3x-18x+18\)

\(=18\) \(\Rightarrow\) ko thuộc vào biến

Câu d : \(\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+5\)

\(=8x^3-24x^2+72x+24x^2-72x+216-8x^3+5\)

\(=221\) \(\Rightarrow\) không thuộc vào biến

câu 1) a) \(\left(x^2+2xy+y^2\right)\left(x+y\right)=\left(x+y\right)^2\left(x+y\right)=\left(x+y\right)^3\)

b) \(y\left(y^3+y^2-3y-2\right)+\left(y^2-2\right)\left(y^2+y-1\right)\)

\(=y^4+y^3-3y^2-2y+y^4+y^3-y^2-2y^2-2y+2\)

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

\(=2y\left(y+2\right)\left(y^2-y-1\right)+2=2\left(y^2+2y\right)\left(y^2-y-1\right)+2\)

\(=2\left(y^2+2y\right)\left(y^2-y-1+1\right)=2\left(y^2+2y\right)\left(y^2-y\right)\)

c) \(6x^2-\left(2x+5\right)\left(3x-2\right)=6x^2-\left(6x^2-4x+15x-10\right)\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-11x+10\)

d) \(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3-2x\right)\)

\(\)\(=6x^2+2x-3x-1+9x-6x^2+12-8x=11\)

e) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)\)

\(=21x-15x^2-35+25x-\left(10x-15x^2+4-6x\right)\)

\(21x-15x^2-35+25x-10x+15x^2-4+6x=42x-39\)

a)(x2 – 2xy + y2)(x – y)

   = (x2 – 2xy + y2).x + (x2 – 2xy + y2).(–y)

   = x2.x + (–2xy).x + y2.x + x2.(–y) + (–2xy).(–y) + y2.(–y)

  = x3 – 2x2y + xy2 – x2y + 2xy2 – y3

   = x3 – (2x2y + x2y) + (xy2 + 2xy2) – y3

   = x3 – 3x2y + 3xy2 – y3.

c)6x^2-(2x+5) (3x-2)

6x^2-(6X²-4x+15x-10)

6x²-6x²+4x-15x+10

-11x+10

d)(2x-1)(3x+1)+(3x+4)(3-2x)

(=)6x2-3x+2x-1+6x-6x2+12-8x

  (=)-4x+11

Bài làm:

a) | 5x - 4 | = | x + 2 |

=> 5x - 4 = x + 2

=> 5x - x = 2 + 4

=> x . (5 - 1) = 6

=> x . 4 = 6

=> x = 6 : 4 = 1,5

b) | x + 2/5 | = 2x

=> x + 2/5 = 2x hoặc x + 2/5 = -2x

* x + 2/5 = 2x

=> x - 2x = -2/5

=> x . (1 - 2) = -2/5

=> x .(-1) = -2/5

=> x = -2/5 : (-1)

=> x = 2/5

* x + 2/5 = -2x

=> x + 2x = 2/5

=> x . (1 + 2) = 2/5

=> x . 3 = 2/5

=> x = 2/5 : 3

=> x = 2/15

mk chỉ làm 2 bài này thôi, còn 2 bài kia mk ko có pít làm. Sorry!

b) Theo bài ra , ta có : 

(2x - 5) - (3x - 7) = x + 3 

(=) 2x - 5 - 3x + 7 = x + 3 

(=) -2x = 1 

(=) x = -1/2 

Vậy x = -1/2 

Chúc bạn học tốt =))

I I  là dấu giá trị tuyệt đối nhé

|7 + 5x| = 1 - 4x

=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)

=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)

|4x² - 2x| + 1 = 2x

=> |4x² - 2x| = 2x - 1

=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)

Vậy ...

Bài 7 :

\(\frac{1}{4}-\left(2x-1\right)^2=0\)

\(\left(2x-1\right)^2=\frac{1}{4}-0\)

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

\(\left(2x-1\right)^2=\left(\frac{1}{2}\right)^2\)

TH1:\(\Rightarrow2x-1=\frac{1}{2}\)

\(2x=\frac{1}{2}+1\)

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

\(x=\frac{3}{4}\)

TH2:\(\Rightarrow2x-1=-\frac{1}{2}\)

\(2x=-\frac{1}{2}+1\)

\(2x=\frac{1}{2}\)

\(x=\frac{1}{4}\)

Vậy x \(\in\left\{\frac{1}{4};\frac{3}{4}\right\}\)

Bài 6 :

\(3^{x+1}=81\)

\(3^{x+1}=3^4\)

\(x+1=4\)

\(\Rightarrow x=3\)

Vậy x = 3

a) Ta có: (4x-2)(x+5)=0

⇔2(2x-1)(x+5)=0

mà 2≠0

nên \(\left[{}\begin{matrix}2x-1=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

Vậy: x=-5

b) Ta có: 2x-9=-8-9

⇔2x-9=-17

hay 2x=-8

⇔x=-4(tm)

Vậy: x=-4

c) Ta có: 3|x-1|-27=0

⇔3|x-1|=27

⇔|x-1|=9

⇔\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-8\left(tm\right)\end{matrix}\right.\)

Vậy: x∈{-8;10}

d) Ta có: 5(3x+8)-7(2x+3)=16

⇔15x+40-14x-21-16=0

⇔x+3=0

hay x=-3(tm)

Vậy: x=-3

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

\(\Rightarrow\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)

Vậy...

b)\(2x-9=-8-9\\ \Leftrightarrow2x-9=-17\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)

Vậy...

c)\(2\left|x-1\right|-27=0\\ \Leftrightarrow\left|x-1\right|=\frac{27}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{27}{2}\\x-1=-\frac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{2}\\x=-\frac{25}{2}\end{matrix}\right.\)

Vậy...

a) (4x – 2)(x + 5) = 0

⇔ \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}4x=2\\x=-5\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)

Vậy ...

b) 2x – 9 = -8 – 9

⇔ 2x = - 8

⇔ x = - 4

Vậy x = - 4

c) 3.| x -1 | - 27 = 0

⇔ 3 . | x - 1| = 27

⇔ | x - 1| = 9

⇔ \(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

Vậy x ∈ { 10 ; - 8}

d) 5.(3x + 8) –7.(2x + 3) = 16

⇔ 15x + 40 - 14x - 21 = 16

⇔ x + 19 = 16

⇔ x = - 3

Vậy ..