1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

ĐK:x\(\ge\)0

Đặt t=x²+3x(t\(\ge\) 0)ta được:

\(\sqrt{t+12}=t\Leftrightarrow t^2=t+12\)

<=>t²-t-12=0

\(\Delta=49\Rightarrow\sqrt{\Delta}=7\)

\(\Delta>0,\text{phương trình có 2 nghiệm phân biệt}\)

\(t_1=4\left(thỏa\right);t_2=-3\left(loại\right)\)

t=4=>x²+3x=4

<=>x²+3x-4=0

\(\Delta=25\Rightarrow\sqrt{\Delta}=5;\Delta>0,pt\text{ có 2 nghiệm phân biệt:}\)

\(x_1=1\left(thỏa\right);x_2=-4\left(loại\right)\)

Vậy S={1}

\(ĐK:x^2-3x+5\ge0\)

Đặt \(\sqrt{x^2-3x+5}=a\ge0\)

\(PT\Leftrightarrow a+a^2-5=7\\ \Leftrightarrow a^2+a-12=0\\ \Leftrightarrow\left(a-3\right)\left(a+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=3\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2-3x+5}=3\\ \Leftrightarrow x^2-3x+5=9\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

đặt \(x^2-3x=y\)

\(pt\Leftrightarrow\sqrt{y+5}+y=7\\ \Leftrightarrow\sqrt{y+5}=7-y\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=\left(7-y\right)^2\\7-y\ge0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y+5=49-14y+y^2\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-15y+44=0\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y^2-11y\right)-\left(4y-44\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-11\right)\left(y-4\right)=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=4\\y=11\end{matrix}\right.\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x=4\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-3x-4=0\\y\le7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)\left(x+1\right)\\y\le7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\\y\le7\end{matrix}\right.\)

Vậy \(x\in\left\{4;-1\right\}\)

\(\left(x^2-2\right)-\left(x-2\right)\left(3x-7\right)=0\\ \Leftrightarrow x^2-2-\left(3x^2-6x-7x+14\right)=0\\ \Leftrightarrow x^2-2-3x^2+13x-14=0\\ \Leftrightarrow-2x^2+13x-16=0\\\Leftrightarrow \left[{}\begin{matrix}x=\dfrac{13+\sqrt{41}}{4}\\x=\dfrac{13-\sqrt{41}}{4}\end{matrix}\right.\)

xem lại đề ạ

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

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

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1