đề ko có vấn đề nhỉ?

1.

Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)

Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)

\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)

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

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

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

Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):

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

\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)

Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\): 

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)

Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)

\(\Leftrightarrow4x-2+2x=5x-20\)

\(\Leftrightarrow x=-18\)

b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-3x=-1\)

hay \(x=\dfrac{1}{3}\)

c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)

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

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

d: Ta có: \(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)

\(\Leftrightarrow x^2-25+x^2-3x=x-25\)

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

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ne\pm y\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\2u+3v=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u+3v=5\\\end{matrix}\right.\)

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

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

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

b.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+7=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-5x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)

\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)

\(\Leftrightarrow x+2010=0\) vì \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)

=> x=-2010

vậy.....

1)

\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)

\(\Leftrightarrow x=105\)

b)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow50-x=0\)

\(\Leftrightarrow x=50\)

2)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)

\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)

\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)

\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

c. \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

Đặt: \(y=x+4\), ta có:

\(\left(y-1\right)^4+\left(y+1\right)^4=2\)

\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)

\(\Leftrightarrow2y^4+12y^2=0\)

\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)

\(\Leftrightarrow y=0\)

\(\Leftrightarrow x=-4\)

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

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

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

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

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

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

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

\(\Leftrightarrow x=1\)