Mình nhầm đoạn Delta':

\(\Delta'=\left(m-1\right)^2-\left(m-5\right)=m^2-3m+6>0\)

ĐKXĐ: \(x\neq 2;x\neq 4\)

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

\(\Rightarrow-2-\left(x-1\right)\left(x-4\right)=\left(x+3\right)\left(x-2\right)\)

\(\Leftrightarrow-2-\left(x^2-5x+4\right)=x^2+x-6\Leftrightarrow2x^2-4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(l\right)\end{matrix}\right.\)

Vậy x = 0

\(x^2+x+\dfrac{3}{x^2+x+1}=3\)

\(\Leftrightarrow x^2+x+1+\dfrac{3}{x^2+x+1}=4\)

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

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

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

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

Vậy...

ĐKXĐ: \(\left[{}\begin{matrix}x\ne-3\\x\ne1\end{matrix}\right.\).

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

\(\Rightarrow\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-7x+5-2x^2-6x=4\)

\(\Leftrightarrow13x=1\Leftrightarrow x=\dfrac{1}{13}\). (TMĐK)

Vậy..

Câu 5: \(\left\{{}\begin{matrix}4x-5y=-5\\4x-7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-5y=-5\\2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=\dfrac{-15}{4}\end{matrix}\right.\)

Câu 6:

1: Ta có \(\Delta'=\left(m-1\right)^2-4\left(m-5\right)=m^2-6m+21=\left(m-3\right)^2+12>0\) nên phương trình đã cho luôn có 2 nghiệm phân biệt.

2: Theo hệ thức Viète ta có \(\left\{{}\begin{matrix}x_1x_2=m-5\\x_1+x_2=2\left(m-1\right)\end{matrix}\right.\).

Ta có \(10=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=4\left(m-1\right)^2-2\left(m-5\right)\Leftrightarrow4m^2-10m+4=0\Leftrightarrow2\left(m-2\right)\left(2m-1\right)=0\Leftrightarrow\left[{}\begin{matrix}m=2\\m=\dfrac{1}{2}\end{matrix}\right.\).

Vậy...

a) (42 - 98) - (42 - 12) - 12 = 42 - 98 - 42 + 12 - 12 = (42 - 42) - 98 + (12 - 12) = -98

b) (-5) . 4 . (-2) . 3 . (-25) = (-5) . (-2) . 4 . (-25) . 3 = 10 . (-100) . 3 = -3000

\(A=4\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)=4\left(\dfrac{1}{1}-\dfrac{1}{2015}\right)=\dfrac{8056}{2015}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\).

\(PT\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow x\left(x+25\right)-2\left(x+5\right)^2=\left(5-x\right)\left(x-5\right)\)

\(\Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=10x-x^2-25\)

\(\Leftrightarrow-5x=25\Leftrightarrow x=-5\) (loại)

Vậy PT vô nghiệm

Gọi số cần tìm là A.

Số ta được khi viết thêm chữ số 0 bên phải số đó là \(\overline{A0}\).

Theo bài ra ta có \(A+\overline{A0}=297\)

\(A+10\times A=297\)

\(11\times A=297\)

\(A=27\)

Vậy số cần tìm là 27.