2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

câu bc hình bị lỗi

1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)

\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)

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

b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)

\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)

mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)

Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)

\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)

\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)

Bài 1Bài 2

2.1

Bài 4Bạn tham khảo nha. Chúc bạn học tốt

a: Khi m=4 thì (1) sẽ là:

x^2-6x-7=0

=>x=7 hoặc x=-1

b: Sửa đề: 2x1+3x2=-11

x1+x2=2m-2

=>2x1+3x2=-11 và 2x1+2x2=4m-4

=>x2=-11-4m+4=-4m-7 và x1=2m-2+4m+7=6m+5

x1*x2=-2m+1

=>-24m^2-20m-42m-35+2m-1=0

=>-24m^2-60m-34=0

=>\(m=\dfrac{-15\pm\sqrt{21}}{12}\)

a) PT bậc nhất một ẩn là: x-2=0; 4-0,2x=0
b) Giải:
x-2=0     (*)
⟺ x=-2
Vậy tập nghiệm của pt (*) là S={-2}
 4-0,2x=0    (**)
⟺-0,2x=-4
⟺x=-4/-0,2=20
Vậy tập nghiệm của pt (**) là S={20}

\(a,\left\{{}\begin{matrix}x+2y=5\\3x+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\3\left(5-2y\right)+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\15-6y+4y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2y\\15-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=5-2.5\\y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=5\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3-3x\\2y-\left(3-3x\right)=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3-3x\\2y-3+3x=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3-3x\\5x-3=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=3-3.2\\x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+2y=5\\3x+4y=5\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x+6y=15\\3x+4y=5\end{matrix}\right.\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=10\\x+2y=5\end{matrix}\right.\Leftrightarrow\)\(\left\{{}\begin{matrix}y=5\\x+2\times5=5\end{matrix}\right.\Leftrightarrow\)\(\left\{{}\begin{matrix}y=5\\x+10=5\end{matrix}\right.\Leftrightarrow\)\(\left\{{}\begin{matrix}y=5\\x=-5\end{matrix}\right.\)vậy.....

c: ĐKXĐ: x<>8

\(\dfrac{3}{2x-16}+\dfrac{3x-20}{x-8}+\dfrac{1}{8}=\dfrac{13x-102}{3x-24}\)

=>\(\dfrac{9}{6\left(x-8\right)}+\dfrac{18x-120}{6\left(x-8\right)}-\dfrac{26x-204}{6\left(x-8\right)}=\dfrac{-1}{8}\)

=>\(\dfrac{18x-111-26x+204}{6\left(x-8\right)}=\dfrac{-1}{8}\)

=>\(\dfrac{-8x+93}{6x-48}=\dfrac{-1}{8}\)

=>\(\dfrac{8x-93}{6x-48}=\dfrac{1}{8}\)

=>8(8x-93)=6x-48

=>64x-744-6x+48=0

=>58x=696

=>x=12

d: ĐKXĐ: x<>1; x<>-1

\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x+4}+\dfrac{12x-1}{4x-4}\)

=>\(\dfrac{24}{4\left(x-1\right)\left(x+1\right)}+\dfrac{20\left(x^2-1\right)}{4\left(x-1\right)\left(x+1\right)}=\dfrac{\left(8x-1\right)\left(x-1\right)+\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}\)

=>8x^2-9x+1+12x^2+12x-x-1=24+20x^2-20

=>20x^2+2x=20x^2+4

=>2x=4

=>x=2(loại)

Vậy hệ phương trình có nghiệm duy nhất là (x; y) = (3; -1)