=4x^2-4x+1+x^3-27-4(x^2-16)

=4x^2-4x+1+x^3-27-4x^2+64

=x^3-4x+38

Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)

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

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

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

\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)

\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)

\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)

\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)

\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)

\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)

mà \(5x^2+7x+5>0\forall x\)

nên x-1=0

hay x=1

a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)

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

\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)

\(< =>-x^2+5x-6-x^2-x+5=0\)

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

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

đến đây thì pt bậc 2 dể rồi

\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)

\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)

\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)

\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)

\(< =>2+3x-3+x^2-2x+1=0\)

\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)

\(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\left(đkxđ:x\ne0;x\ne2\right)\)

\(< =>\frac{\left(x+2\right)x}{\left(x-2\right)x}-\frac{2}{x\left(x-2\right)}=\frac{x-2}{x\left(x-2\right)}\)

\(< =>\left(x+2\right)x-2=x-2< =>x^2+2x-x-2+2=0\)

\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=-1\left(tm\right)\end{cases}}\)

nhớ kết luận tập nghiệm

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ bên trái khung soạn thảo) để được hỗ trợ tốt hơn.

Bài 4:

\(x^4y-x^4+2x^3-2x^2+2x-y=1\)

\(\Leftrightarrow y(x^4-1)-(x^4-2x^3+2x^2-2x+1)=0\)

\(\Leftrightarrow y(x^2+1)(x^2-1)-[x^2(x^2-2x+1)+(x^2-2x+1)]=0\)

\(\Leftrightarrow y(x^2+1)(x-1)(x+1)-(x-1)^2(x^2+1)=0\)

\(\Leftrightarrow (x^2+1)(x-1)[y(x+1)-(x-1)]=0\)

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

Với $(1)$ ta thu được $x=1$, và mọi $ý$ nguyên.

Với $(2)$

\(y(x+1)=x-1\Rightarrow y=\frac{x-1}{x+1}\in\mathbb{Z}\)

\(\Rightarrow x-1\vdots x+1\)

\(\Rightarrow x+1-2\vdots x+1\Rightarrow 2\vdots x+1\)

\(\Rightarrow x+1\in\left\{\pm 1; \pm 2\right\}\Rightarrow x\in\left\{-2; 0; -3; 1\right\}\)

\(\Rightarrow y\left\{3;-1; 2; 0\right\}\)

Vậy \((x,y)=(-2,3); (0; -1); (-3; 2); (1; t)\) với $t$ nào đó nguyên.

Bài 1:

\(x^2+y^2-8x+3y=-18\)

\(\Leftrightarrow x^2+y^2-8x+3y+18=0\)

\(\Leftrightarrow (x^2-8x+16)+(y^2+3y+\frac{9}{4})=\frac{1}{4}\)

\(\Leftrightarrow (x-4)^2+(y+\frac{3}{2})^2=\frac{1}{4}\)

\(\Rightarrow (x-4)^2=\frac{1}{4}-(y+\frac{3}{2})^2\leq \frac{1}{4}<1\)

\(\Rightarrow -1< x-4< 1\Rightarrow 3< x< 5\)

Vì \(x\in\mathbb{Z}\Rightarrow x=4\)

Thay vào pt ban đầu ta thu được \(y=-1\) or \(y=-2\)

Vậy.......

Bài 2:

Ta có: \(x+y+xy=x^2+y^2\)

\(\Leftrightarrow 2x^2+2y^2=2x+2y+2xy\)

\(\Leftrightarrow 2x^2+2y^2-2x-2y-2xy=0\)

\(\Leftrightarrow (x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)=2\)

\(\Leftrightarrow (x-y)^2+(x-1)^2+(y-1)^2=2(*)\)

\(\Rightarrow (y-1)^2\leq 2<4\Rightarrow -2< y-1< 2\)

\(\Rightarrow -1< y< 3\Rightarrow y\in\left\{0;1;2\right\}\)

Thay $y$ với các giá trị trên vào pt ban đầu ta thu được:

\(y=0\Rightarrow x=0, x=1\)

\(y=1\Rightarrow x=0; x=2\)

\(y=2\Rightarrow x=1;x=2\)

1) tôi giải theo kt lớp 9 nhé nếu theo lp 8 thì phần tích theo đk trong gttđ
   lập bảng xét dấu  
    

với x <1  => x=1   x=-2
với x>1   >x=1      x=-2
vậy  pt có 2 ng phân bịt  x =1 và x=-2
các câu còn lại lm tương tự w nhé 
 chúc bn hc giỏi !!

@trần tuấn phát giải giúp mik kiểu lớp 8 với! Mik k hỉu!

Lời giải:
ĐKXĐ: $x\in\mathbb{R}$

Đặt $\sqrt{x^2+x+1}=a; \sqrt{x^2-x+1}=b(a,b\geq 0)$. PT trở thành:
$a=a^2-b^2+b$

$\Leftrightarrow (a-b)(a+b)-(a-b)=0$

$\Leftrightarrow (a-b)(a+b-1)=0$

$\Rightarrow a=b$ hoặc $a+b=1$

Nếu $a=b\Leftrightarrow a^2=b^2\Leftrightarrow x^2+x+1=x^2-x+1$

$\Leftrightarrow x=0$

Nếu $a+b=1$

$\Leftrightarrow \sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1$

$\Leftrightarrow \sqrt{x^2+x+1}=1-\sqrt{x^2-x+1}$

$\Rightarrow x^2+x+1=x^2-x+2-2\sqrt{x^2-x+1}$

$\Leftrightarrow 1-2x=2\sqrt{x^2-x+1}$

$\Rightarrow (1-2x)^2=4(x^2-x+1)$

$\Leftrightarrow -3=0$ (vô lý)

Vậy pt có nghiệm $x=0$

Bài 1:

a, Thay m=-1 vào (1) ta có:
\(x^2-2\left(-1+1\right)x+\left(-1\right)^2+7=0\\ \Leftrightarrow x^2+1+7=0\\ \Leftrightarrow x^2+8=0\left(vô.lí\right)\)

Thay m=3 vào (1) ta có:

\(x^2-2\left(3+1\right)x+3^2+7=0\\ \Leftrightarrow x^2-2.4x+9+7=0\\ \Leftrightarrow x^2-8x+16=0\\ \Leftrightarrow\left(x-4\right)^2=0\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\)

b, Thay x=4 vào (1) ta có:

\(4^2-2\left(m+1\right).4+m^2+7=0\\ \Leftrightarrow16-8\left(m+1\right)+m^2+7=0\\ \Leftrightarrow m^2+23-8m-8=0\\ \Leftrightarrow m^2-8m+15=0\\ \Leftrightarrow\left(m^2-3m\right)-\left(5m-15\right)=0\\ \Leftrightarrow m\left(m-3\right)-5\left(m-3\right)=0\\ \Leftrightarrow\left(m-3\right)\left(m-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=3\\m=5\end{matrix}\right.\)

c, \(\Delta'=\left[-\left(m+1\right)\right]^2-\left(m^2+7\right)=m^2+2m+1-m^2-7=2m-6\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2m-6\ge0\Leftrightarrow m\ge3\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2+7\end{matrix}\right.\)

\(x_1^2+x_2^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-2\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-2m^2-14=0\\ \Leftrightarrow2m^2+8m-10=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-5\left(ktm\right)\end{matrix}\right.\)

\(x_1-x_2=0\\ \Leftrightarrow\left(x_1-x_2\right)^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-4\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-4m^2-28=0\\ \Leftrightarrow8m=28=0\\ \Leftrightarrow m=\dfrac{7}{2}\left(tm\right)\)

Bài 2:

a,Thay m=-2 vào (1) ta có:

\(x^2-2x-\left(-2\right)^2-4=0\\ \Leftrightarrow x^2-2x-4-4=0\\ \Leftrightarrow x^2-2x-8=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

b, \(\Delta'=\left(-m\right)^2-\left(-m^2-4\right)\ge0=m^2+m^2+4=2m^2+4>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-m^2-4\end{matrix}\right.\)

\(x_1^2+x_2^2=20\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=20\\ \Leftrightarrow2^2-2\left(-m^2-4\right)=20\\ \Leftrightarrow4+2m^2+8-20=0\\ \Leftrightarrow2m^2-8=0\\ \Leftrightarrow m=\pm2\)

\(x_1^3+x_2^3=56\\ \Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=56\\ \Leftrightarrow2^3-3\left(-m^2-4\right).2=56\\ \Leftrightarrow8-6\left(-m^2-4\right)-56\\ =0\\ \Leftrightarrow8+6m^2+24-56=0\\ \Leftrightarrow6m^2-24=0\\ \Leftrightarrow m=\pm2\)

\(x_1-x_2=10\\ \Leftrightarrow\left(x_1-x_2\right)^2=100\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-100=0\\ \Leftrightarrow2^2-4\left(-m^2-4\right)-100=0\\ \Leftrightarrow4+4m^2+16-100=0\\ \Leftrightarrow4m^2-80=0\\ \Leftrightarrow m=\pm2\sqrt{5}\)