Bài 1: Phương trình bậc nhất hai ẩn

\(D=\dfrac{\sqrt{2}+\sqrt{3}+2-\sqrt{3}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1-\sqrt{3}\)

`2sqrt(x-1) + 3sqrt(x-1) = 5`

`-> sqrt(x-1) = 1`

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

`->` \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

`sqrt(4x-4)+sqrt(9x-9)=5`. `ĐKXĐ: x-1 >=0 <=> x>=1`

`=> 2sqrt(x-1)+3sqrt(x-1)=5`

`=>sqrt(x-1)=1`

`=>`\(\left[\begin{array}{} x-1=1\\ x-1=-1 \end{array} \right.\)

`=>`\(\left[\begin{array}{} x=2\\ x=0 \end{array} \right.\)

Mà `x>=1 => x=2`

\(3x+1=7x-11\)

\(3x=7x-11-1\)

\(3x=7x-12\)

\(\left(3x-7x\right)=-12\)

\(-4x=-12\)

\(x=\dfrac{-12}{-4}\)

`=>x=3`

Vậy ptr có nghiệm là: \(\left\{3\right\}\)

=>-4x=-12

hay x=3

bbn có thể tham khảo:

https://hoc247.net/hoi-dap/toan-9/giai-phuong-trinh-x-3-x-2-x-1-3-faq101745.html

refer

`=> 4x^3 - x^3 - 3x^2 - 3x = 1`

`=> 4x^3 = (x+1)^3`

`=> x - x/(3^sqrt 4) = 1/(3^sqrt 4)`

`=> x = 1/(3^sqrt4 - 1)`

lỗi r bn

lỗi

;lx

lx

lỗi r bn 

a: x^2-7x+13=0

Δ=(-7)^2-4*1*13=49-52=-3<0

=>PTVN

b: -5x^2+5x-1.25=0

=>4x^2-4x+1=0

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

=>2x-1=0

=>x=1/2

d: 2x^2+3x+1=0

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

=>x=-1 hoặc x=-1/2