Tìm x biết:

b/\(\left(2x+3\right)^2-\left(5x-4\right)\left(5x+4\right)=\left(x+5\right)^2-\left(3x-1\right)\left(7x+2\right)-\left(x^2-x+1\right)\)

<=> \(4x^2 +12x+9-25x^2+16-x^2-10x-25+21x^2+6x-7x-2+x^2-x+1=0\)

<=>0x-1=0

<=>0x=1 (vô lí) (dòng này không cần ghi thêm cũng được)

=> Không có giá trị x nào thỏa mãn

c/ \((1-3x)^2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)^2\)

<=>\(1-6x+9x^2-9x^2-x+18x+2-9x^2+16+9x^2+54x+81=0\)

<=> 65x+100=0

<=> x=\(\dfrac{-20}{13}\)

d/\((3x+4)(3x-4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2\)

<=> \(9x^2-16-4x^2-20x-25-x^2+10x-25-4x^2-4x-1+x^2+2x-x^2+2x-1=0\)

<=> -10x-68=0

<=> x=\(\dfrac{-34}{5}\)

a) ta có : \(x^4+3x^3-2x^2+3x+1=0\)

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

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

\(\Leftrightarrow\left(x^2+4x+1\right)\left(x^2-x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+1=0\\x^2-x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}-2+\sqrt{3}\\-2-\sqrt{3}\end{matrix}\right.\\x\in\varnothing\end{matrix}\right.\) vậy \(x=-2+\sqrt{3};x=-2-\sqrt{3}\)

b) ta có : \(x^4-2x^3-5x^2+2x+1=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{3+\sqrt{13}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

vậy \(x=\dfrac{3+\sqrt{13}}{2};x=\dfrac{3-\sqrt{13}}{2};x=\dfrac{-1+\sqrt{5}}{2};x=\dfrac{-1-\sqrt{5}}{2}\)

\(\begin{cases}x^5-3x^4+2x^2-2x+2\ge0\\x^4-2x^3-x+2=0\\x^2-3x+2=0\\\left(x^2-1\right)\left(x-2\right)=0\end{cases}\)  (*)

\(x^5-3x^4+2x^2-2x+2\ge0\) (1)

\(x^4-2x^3-x+2=0\) (2)

\(x^2-3x+2=0\)  (3)

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

Từ 

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

x=1 thỏa mãn tất cả các phương trình, bất phương trình còn lại nên là nghiệm của hệ

x=2 không thỏa mãn (1) nên x=2 không là nghiệm của hệ

Vậy hệ phương trình (*) có nghiệm duy nhất là x=1

\(1,\left|2x-3\right|=x-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\\left[{}\begin{matrix}2x-3=x-5\\2x-3=-x+5\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}5\\\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)

=> pt vô nghiệm

\(2,\left|3x+2\right|=x+1\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\)

\(3,\left|2x+1\right|=7-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}7-x\text{≥}0\\\left[{}\begin{matrix}2x+1=7-x\\2x+1=x-7\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}7\\\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\end{matrix}\right.\) (loại)

=> pt vô nghiệm

\(4,\left|2x-5\right|=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\)

\(5,\left|6x-2\right|=3x-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-4\text{≥}0\\\left[{}\begin{matrix}6x-2=3x-4\\6x-2=-3x+4\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}\frac{4}{3}\\\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(6,\left|3x-2\right|=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\text{≥}0\\\left[{}\begin{matrix}3x-2=x-2\\3x-2=-x+2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}2\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(7,\left|2x+3\right|=1\)

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

\(8,\left|2-x\right|=2x-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\left[{}\begin{matrix}2-x=2x-1\\2-x=-2x+1\end{matrix}\right.\end{matrix}\right.\)

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

\(9,\left|2x-1\right|=x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\\left[{}\begin{matrix}2x-1=x-3\\2x-1=-x+3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm

\(10,2\left|x-1\right|=x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left[{}\begin{matrix}2x-2=x+2\\2x-2=-x-2\end{matrix}\right.\end{matrix}\right.\)

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

a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\x^2+6x+9=21-x^2-4x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\2x^2+10x-12=0\end{matrix}\right.\Leftrightarrow x=1\)

b: \(\left|x^2+5x+4\right|-4=x\)

=>|x^2+5x+4|=x+4

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+5x+4-x-4\right)\left(x^2+5x+4+x+4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-4\\\left(x^2+4x\right)\left(x^2+6x+8\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-2;-4\right\}\)

c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-5x+4-2x+1\right)\left(2x^2-5x+4+2x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x^2-7x+5\right)\left(2x^2-3x+3\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{2};1\right\}\)