Không biết nãy bị lỗi ở đâu, mình gửi lại:<

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

1) Ta có: \(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1-21+x=0\)

\(\Leftrightarrow x^2-3x-20=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)

1)\(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow21-x=x^2-2x+1\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

2)\(\sqrt{8-x}+2=x\)

\(\Leftrightarrow8-x=\left(x-2\right)^2\)

\(\Leftrightarrow8-x=x^2-4x+4\)

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

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

3)\(1+\sqrt{3x+1}=3x\)

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

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

Xét pt \(\left|x-1\right|^{2010}+\left|x-2\right|^{2011}=1\) (1)

Nhận thấy \(x=1\) và \(x=2\) là 2 nghiệm của pt

- Với \(x>2\Rightarrow\left\{{}\begin{matrix}\left|x-2\right|>0\\\left|x-1\right|>1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|^{2010}>1\\\left|x-2\right|^{2011}>0\end{matrix}\right.\)

\(\Rightarrow\left|x-1\right|^{2010}+\left|x-2\right|^{2011}>1\) nên pt vô nghiệm

- Với \(x< 1\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=\left|1-x\right|>0\\\left|x-2\right|=\left|2-x\right|>1\end{matrix}\right.\)

Tương tự như trên ta có \(\left|x-1\right|^{2010}+\left|x-2\right|^{2011}>1\) \(\Rightarrow\) pt vô nghiệm

- Với \(1< x< 2\Rightarrow\left\{{}\begin{matrix}0< \left|x-1\right|< 1\\0< \left|2-x\right|< 1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|^{2010}< \left|x-1\right|\\\left|2-x\right|^{2011}< \left|2-x\right|\end{matrix}\right.\)

\(\Rightarrow\left|x-1\right|^{2010}+\left|x-2\right|^{2011}< \left|x-1\right|+\left|2-x\right|=x-1+2-x=1\)

\(\Rightarrow\) Pt vô nghiệm

Vậy pt có đúng 2 nghiệm \(x=1\); \(x=2\)

Lần lượt thế vào \(x^2+y^2-2x=11\) để tìm y

ĐKXĐ ; \(x\ne\pm1\)

Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)

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

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

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

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

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

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

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

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

=> X = 3

Vậy ..

\(\dfrac{x^4-2x^2+1}{x^3+2x^2+x}=\dfrac{\left(x^2-1\right)^2}{x\left(x^2+2x+1\right)}=\dfrac{\left(x-1\right)^2\left(x+1\right)^2}{x\left(x+1\right)^2}=\dfrac{\left(x-1\right)^2}{x}\)