a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-1\\2x-3y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}6x+4y=-2\\6x-9y=12\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}13y=-14\\2x-3y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=-\frac{14}{13}\\2x-3.\left(-\frac{14}{13}\right)=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=-\frac{14}{13}\\x=\frac{5}{13}\end{matrix}\right.\)

Vậy phương trình trên có nghiệm ( x;y ) = ( \(\frac{5}{13};-\frac{14}{13}\) )

b, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)

- Ta có : \(\frac{5}{x-2}-\frac{4}{x-1}=3\)

=> \(\frac{5\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{4\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=3\)

=> \(5\left(x-1\right)-4\left(x-2\right)=3\left(x-2\right)\left(x-1\right)\)

=> \(5x-5-4x+8-3x^2+6x+3x-6=0\)

=> \(10x-3x^2-3=0\)

=> \(\left(3x-1\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right.\) ( TM )

Vậy phương trình trên có tập nghiệm là \(S=\left\{3;\frac{1}{3}\right\}\)

a)

∆'=3+9=12

x1=(√3-2√3)/3=-√3/3

x2=(√3+2√3)/3=√3

b.

<=>

x+y=-3(1)

2x-3y=-1(2)

(1).2-(2)<=>5y=-5;y=-1

=>(x,y)=(-2;-1)

nếu là lớp 8 thì rất hoan nghênh

a) \(\Delta'=\left(\sqrt{3}\right)^2-3\cdot\left(-3\right)=12>0\)

phương trình có 2 nghiệm phân biệt:

\(\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+\sqrt{12}}{3}=\sqrt{3}\\x=\dfrac{\sqrt{3}-\sqrt{12}}{3}=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

kết luận: \(x=\sqrt{3}\), \(x=-\dfrac{\sqrt{3}}{3}\)

b) \(\left\{{}\begin{matrix}x\left(x-1\right)+y=\left(x+1\right)\left(x-3\right)\\2x-3y=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\5x+9=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

kết luận: \(\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

a) lập delta giải bình thường

b) rút y từ 1 trong 2 pt thế vào pt còn lại

@Leo :) giỏi wa ha . ghen tị ghê

Bài 2: 

a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)

\(=5m^2-2m+9>0\forall m\)

Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m

Bài 1:

ĐKXĐ \(2x\ne y\)

Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)

HPT trở thành

\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)

1) \(-2x^2+x+1-2\sqrt[]{x^2+x+1}=0\)

\(\Leftrightarrow2\sqrt[]{x^2+x+1}=-2x^2+x+1\left(1\right)\)

Ta có :

\(2\sqrt[]{x^2+x+1}=2\sqrt[]{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge\sqrt[]{3}\)

Dấu "=" xảy ra khi và chỉ khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow-2x^2+x+1=\sqrt[]{3}\)

\(\Leftrightarrow2x^2-x+\sqrt[]{3}-1=0\)

\(\Delta=1-8\left(\sqrt[]{3}-1\right)=9-8\sqrt[]{3}\)

\(pt\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\\x=\dfrac{1-\sqrt[]{9-8\sqrt[]{3}}}{4}\left(loại\right)\end{matrix}\right.\) \(\left(vì.x=-\dfrac{1}{2}\right)\)

Vậy phương trình cho vô nghiệm

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

a) Đặt t = \(x^2\)

ta có p/t : \(3x^2-4x+1=0\)

Bn lập Δ rồi giải

b)⇔ \(\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\)

tự giải