dễ zậy 

thật của lớp 7 ko

len google di ban

mk chua hoc bai nay

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

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

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

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

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

a. (x - 2²) - 1 = 0

<=> x - 4 - 1 = 0

<=> x = 5

b. 4 - (x - 2)² = 0

<=> 2² - (x - 2)² = 0

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

<=> x(4 - x) = 0

<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

d. (3x - 2)² - (2x + 3)² = 5(x + 4)(x - 4)

<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x² - 16)

<=> (x - 5)(5x + 1) = 5x² - 80

<=> 5x² + x - 25x - 5 = 5x² - 80

<=> 5x² - 5x² + x - 25x = -80 + 5

<=> -24x = -75

<=> x = \(\dfrac{25}{8}\)

a)\(\left(x-2^2\right)-1=0\Rightarrow x-4-1=0\Rightarrow x=5\)

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

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

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

Bài 4 xem lại đề nhé bác

a: =>|5x+4|=19

=>5x+4=19 hoặc 5x+4=-19

=>5x=15 hoặc 5x=-23

=>x=3 hoặc x=-23/5

b: =>3|2x-9|=33

=>|2x-9|=11

=>2x-9=11 hoặc 2x-9=-11

=>2x=20 hoặc 2x=-2

=>x=10 hoặc x=-1

d: =>|17x-5|=|17x+5|

=>17x-5=17x+5 hoặc 17x-5=-17x-5

=>34x=0

hay x=0

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

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

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha