1.      \(2x^2-3x-5=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2,5\\x=-1\end{cases}}\)

Vậy tập ngiệm của phương trình là \(S=\left\{2,5;-1\right\}\)

2x²-3x-5=0

2x²+2x-5x-5=0

2x(x+1)+5(x+1)=0

(x+1)(2x+5)=0

TH1 x+1=0 <=>x=-1

TH2 2x+5=0<=>2x=-5<=>x=-5/2

2. ta có:

2(x-2y)-(2x+y)=-1.2-8

2x-4y-2x-y=-2-8

-5y=-10

y=2

thay vào 

x-2y=-1 ( với y=2)

<=> x-2.2=-1

x-4=-1

x=3

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

Bài 2: 

b) Phương trình hoành độ giao điểm của (P) và (d) là:

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

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2-2x+3x-3=0\)

\(\Leftrightarrow2x\left(x-1\right)+3\left(x-1\right)=0\)

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

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

Thay x=1 vào hàm số \(y=2x^2\), ta được:

\(y=2\cdot1^2=2\)

Thay \(x=-\dfrac{3}{2}\) vào hàm số \(y=2x^2\), ta được:

\(y=2\cdot\left(-\dfrac{3}{2}\right)^2=2\cdot\dfrac{9}{4}=\dfrac{9}{2}\)

Vậy: Tọa độ giao điểm của (p) và (D) là (1;2) và \(\left(-\dfrac{3}{2};\dfrac{9}{2}\right)\)

Ta có hệ \(\hept{\begin{cases}\left(4x^2+1\right)x+\left(y-3\right)\sqrt{5-2y}=0\left(1\right)\\4x^2+y^2+2\sqrt{3-4x}=7\left(2\right)\end{cases}}\)

ĐK \(\hept{\begin{cases}y\ge\frac{5}{2}\\x\le\frac{3}{4}\end{cases}}\)

Đặt \(\hept{\begin{cases}2x=a\\\sqrt{5-2y}=b\ge0\end{cases}\Rightarrow\hept{\begin{cases}4x^2=a^2\\5-2y=b^2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}4x^2=a^2\\y-3=\frac{5-b^2}{2}-3=\frac{-1-b^2}{2}\end{cases}}\)

Thế vào (1) ta có \(\left(a^2+1\right)\frac{a}{2}+\frac{-1-b^2}{2}b=0\)

\(\Leftrightarrow\frac{a^3+a}{2}+\frac{-b^3-b}{2}=0\Leftrightarrow a^3-b^3+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)vì \(a^2+ab+b^2+1>0\forall a,b\)

\(\Rightarrow2x=\sqrt{5-2y}\Rightarrow4x^2=5-2y\Rightarrow y=\frac{5-4x^2}{2}\)

Thế y vào (2) ta có \(4x^2+\left(\frac{5-4x^2}{2}\right)^2+2.\sqrt{3-4x}=7\)

\(\Leftrightarrow16x^2+\left(5-4x^2\right)^2+8\sqrt{3-4x}=28\)\(\Leftrightarrow16x^2+25-40x^2+16x^4+8\sqrt{3-4x}-28=0\)

\(\Leftrightarrow16x^4-24x^2+8\sqrt{3-4x}-3=0\)

\(\Leftrightarrow\left(16x^4-1\right)-\left(24x^2-6\right)+\left(8\sqrt{3-4x}-8\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+\left(8\sqrt{3-4x}-8\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(4x^2+1\right)-6\left(4x^2-1\right)+8.\frac{2-4x}{\sqrt{3-4x}+1}=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)\left(4x^2+1\right)-6\left(2x+1\right)\left(2x-1\right)-8.2.\frac{2x-1}{\sqrt{3-4x}+1}=0\)

\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2+1\right)-6\left(2x+1\right)-\frac{16.1}{\sqrt{3-4x}+1}\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left[\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}\right]=0\)

\(\Leftrightarrow2x-1=0\)

Vì với \(y=\frac{5-4x^2}{2}\ge\frac{5}{2}\Rightarrow4x^2-5< 0\Rightarrow\left(2x+1\right)\left(4x^2-5\right)-\frac{16}{\sqrt{3-4x}+1}< 0\)

\(\Leftrightarrow x=\frac{1}{2}\Rightarrow y=\frac{5-4\left(\frac{1}{2}\right)^2}{2}=2\)

Vậy hệ có nghiệm \(\left(x;y\right)=\left(\frac{1}{2};2\right)\)

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:

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

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

\(x-y-5=0\Rightarrow x=y+5\)

Ta có:

\(\left(y+5+y\right)^2+3\left(y+5+y\right)+2=0\)

\(\Leftrightarrow\left(2y+5\right)^2+3\left(2y+5\right)+2=0\)

\(\Leftrightarrow4y^2+20y+25+6y+15+2=0\)

\(\Leftrightarrow4y^2+26y+42=0\)

\(\Leftrightarrow\left(y+3\right)\left(2y+7\right)=0\)

\(\Leftrightarrow y=-3;y=-\frac{7}{2}\)

Thay vào tìm x nốt

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+24=10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)