$\begin{cases}5x-y=5\\y-4x=1\end{cases}$

`<=>` $\begin{cases}\5x-y+y-4x=6\\y-4x=1end{cases}$

`<=>` $\begin{cases}x=6\\y=1+4x=25\end{cases}$

Vậy HPT có nghiệm `(x,y)=(6,25)`

$\begin{cases}5x-y=5\\y-4x=1\end{cases}$

`<=>` $\begin{cases}\5x-y+y-4x=6\\y-4x=1\end{cases}$

`<=>` $\begin{cases}x=6\\y=1+4x=25\end{cases}$

Vậy HPT có nghiệm `(x,y)=(6,25)`

\(\left\{{}\begin{matrix}5x-y=5\\y-4x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-y=5\\-4x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\-4x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\-4.6+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\-24+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=25\end{matrix}\right.\)

Vậy...

mình vừa lên lớp 9 , chưa học phương trình bậc 2 

a)2x³ + 7x² - x - 12 =0

=>2x³+x²-4x+6x²+3x-12=0

=>x(2x²+x-4)+3(2x²+x-4)=0

=>(x+3)(2x²+x-4)=0

=>x+3=0 hoặc 2x²+x-4=0

Xét x+3=0 <=>x=-3

Xét 2x²+x-4=0 ta dùng delta

\(\Delta=1^2-\left(-4\left(2.4\right)\right)=33>0\)

=>pt có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{33}}{4}\)

b)- x^3 + x^2 + 7x + 2 =0

=>-x³+3x²+x-2x²+6x+2=0

=>-x(x²-3x-1)+(-2)(x²-3x-1)=0

=>-(x+2)(x²-3x-1)=0

=>-(x+2)=0 hoặc x²-3x-1=0

Xét -(x+2)=0 <=>x=-2

Xét x²-3x-1=0 theo delta ta có:

\(\Delta=\left(-3\right)^2-\left(-4\left(1.1\right)\right)=13>0\)

=>pt cũng có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\)

xài hóc ne đi

\(2x^3+7x^2+7x+2=0\\ 2\left(x^3+1\right)+7x\left(x+1\right)=0\\ 2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\\ \left(x+1\right)\left(2x^2-2x+2+7x\right)=0\\ \left(x+1\right)\left(2x^2+5x+2\right)=0\\ \left(x+1\right)\left(2x+1\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;\frac{-1}{2};-2\right\}\)

vì x=0 không là nghiệm của pt => chia cả 2 vế cho x²≠0

2x²-7x+9-\(\dfrac{7}{x}\)+\(\dfrac{2}{x^2}\)=0

<=>\(\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+9=0\)

<=>\(2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+9=0\)

đặt \(x+\dfrac{1}{x}\)=y =>\(x^2+\dfrac{1}{x^2}=y^2-2\) ta đc

2(y²-2)-7y+9=0

<=> 2y²-4-7y+9=0

<=>2y²-7y+5=0

<=> 2y²-2y-5y+5=0

<=> (2y²-2y)-(5y-5)=0

<=> 2y(y-1)-5(y-1)=0

<=>(y-1)(2y-5)=0

<=>\(\left\{{}\begin{matrix}y-1=0\\2y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\y=\dfrac{5}{2}\end{matrix}\right.\)

Với y=1 ta có

\(x+\dfrac{1}{x}=1\) =>x²-x+1=0 (vô nghiệm)

Với y=5/2

\(x+\dfrac{1}{x}=\dfrac{5}{2}\) => x=2 và x=\(\dfrac{1}{2}\)

vậy pt có S=\(\left\{2;\dfrac{1}{2}\right\}\)

\(2x^4-7x^3+9x^2-7x+2=0\)

\(\Leftrightarrow2x^4-2x^3-x^3-4x^3+2x^2+x^2+4x^2+2x^2-x-4x-2x+2=0\)

\(\Leftrightarrow\left(2x^4-2x^3+2x^2\right)-\left(x^3-x^2+x\right)-\left(4x^3-4x^2+4x\right)+\left(2x^2-2x+2\right)=0\)

\(\Leftrightarrow2x^2\left(2x^2-2x+2\right)-\dfrac{1}{2}x\left(2x^2-2x+2\right)-2x\left(2x^2-2x+2\right)+\left(2x^2-2x+2\right)=0\)

\(\Leftrightarrow\left(2x^2-2x+2\right)\left(x^2-\dfrac{1}{2}x-2x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-2x+2\right)\left[x\left(x-\dfrac{1}{2}\right)-2\left(x-\dfrac{1}{2}\right)\right]=0\)

\(\Leftrightarrow\left(2x^2-2x+2\right)\left(x-\dfrac{1}{2}\right)\left(x-2\right)=0\)

Vì: \(2x^2-2x+2=\left(\sqrt{2}x-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}>0\forall x\)

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

Vậy..................

p/s: 1 cách khác :))

$2x^4-7x^3+10x^2-7x+2=0$

\(\Leftrightarrow x^2\left(2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}\right)=0\)

\(\Leftrightarrow2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}=0\)

\(\Leftrightarrow\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+10=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+10=0\)

Đặt \(x+\dfrac{1}{x}\) là a ,thì \(x^2+\dfrac{1}{x^2}\) là a²-2 ta được

2(a²-2)-7a+10=0

⇔2a²-4-7a+10=0

⇔2a²-7a+6=0

⇔2a²-4a-3a+6=0

⇔(2a²-4a)-(3a-6)=0

⇔2a(a-2)-3(a-2)=0

⇔(a-2)(2a-3)=0

\(\Rightarrow\left[{}\begin{matrix}a-2=0\\2a-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\a=\dfrac{3}{2}\end{matrix}\right.\)

ta được a=2 => \(x+\dfrac{1}{x}=2\) => x=1

a=\(x+\dfrac{1}{x}=\dfrac{3}{2}\)(vô nghiệm)

vậy S={1}

ĐKXĐ: \(3\le x\le5\)

\(2x^2-7x-2-\sqrt{x-3}-\sqrt{5-x}=0\)

\(\Leftrightarrow2x^2-7x-4+1-\sqrt{x-3}+1-\sqrt{5-x}=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+1\right)-\dfrac{x-4}{1+\sqrt{x-3}}+\dfrac{x-4}{1+\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+1-\dfrac{1}{1+\sqrt{x-3}}+\dfrac{1}{1+\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+\dfrac{\sqrt{x-3}}{1+\sqrt{x-3}}+\dfrac{1}{1+\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow x-4=0\) (ngoặc to luôn dương)

\(\Leftrightarrow x=4\)