Vậy hệ phương trình có nghiệm duy nhất (x; y) = (6; 1).

tham khảo 

https://hoidapvietjack.com/q/57243/giai-cac-phuong-trinh-sau-a-2x12-2x-12-b-x2-3x-2-5x2-3x60

b) (2x+1)²-2x-1=2

\(< =>4x^2+4x+1-2x-1=2\)

\(< =>4x^2+2x-2=0\)

\(< =>4x^2+4x-2x-2=0\)

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

\(< =>4x\left(x+1\right)-2\left(x+1\right)=0\)

\(< =>\left(x+1\right)\left(4x-2\right)=0\)

\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\4x-2=0=>x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy....

b) (2x+1)²-2x-1=2

Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

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

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

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

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

`<=>(2x-1-2)(2x-1+2)=0`

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

\(\frac{{x + 5}}{3} = 1 - \frac{{x - 2}}{4}\)

\(\frac{{\left( {x + 5} \right).4}}{{3.4}} = \frac{{12}}{{12}} - \frac{{\left( {x - 2} \right).3}}{{4.3}}\)

\(\frac{{4x + 20}}{{12}} = \frac{{12}}{{12}} - \frac{{3x - 6}}{{12}}\)

\(4x + 20 = 12 - \left( {3x - 6} \right)\)

\(4x + 20 = 12 - 3x + 6\)

\(4x + 3x = 12 + 6 - 20\)

\(7x =  - 2\)

\(x = \left( { - 2} \right):7\)

\(x = \frac{{ - 2}}{7}\)

Vậy phương trình có nghiệm là \(x = \frac{{ - 2}}{7}\).

ĐK: `x>=0 ; x \ne 25/49`

`(3\sqrtx+1)/(7\sqrtx-5)=8/15`

`<=>15(3\sqrtx+1)=8(7\sqrtx-5)`

`<=>45\sqrtx+15=56\sqrtx-40`

`<=>11\sqrtx=55`

`<=>\sqrtx=5`

`<=>x=25`

Vậy `S={25}`.

Ta có: \(\dfrac{3\sqrt{x}+1}{7\sqrt{x}-5}=\dfrac{8}{15}\)

\(\Leftrightarrow56\sqrt{x}-40-45\sqrt{x}-15=0\)

\(\Leftrightarrow11\sqrt{x}=55\)

hay x=25

b) (0,5 - x)² - 3 = 0 ⇔ 0,5² + x² - 2.0,5x - 3 = 0

                               ⇔ x² + 0,25 - x - 3 = 0

                               ⇔ x² - x - 2,75 = 0

                               ⇔ ????

c: \(\Leftrightarrow\left(2x-3\sqrt{2}\right)\left(2x+\sqrt{2}\right)=0\)

hay \(x\in\left\{\dfrac{3\sqrt{2}}{2};-\dfrac{\sqrt{2}}{2}\right\}\)

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

\(\Leftrightarrow3y=3\)

\(\Rightarrow y=1\left(1\right)\)

Thay (1) vào ptr đầu: \(2x+1=5\)

\(\Rightarrow x=2\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

\(PTHH:2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ \text{Số phân tử }KClO_3:\text{Số phân tử }O_2=2:3\)