Có: \(a-b=5\)

=> \(a^2-2ab+b^2=25\)

\(\Rightarrow a^2+b^2=25+2ab=25+2\cdot4=33\)

\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)=5\cdot\left(33+4\right)=5\cdot37=185\)

\(2x\left(x-5\right)-x\left(2x+3\right)=26\)

\(\Leftrightarrow2x^2-10x-2x^2-3x=26\)

\(\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

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

Vì: \(\left(x-1\right)^2\ge0\)

=> \(\left(x-1\right)^2+2\ge2\)

Vậy giá trị của biểu thức trên là 2 khi x=1

a) \(\left(x-3\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\4-5x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{4}{5}\end{array}\right.\)

b) \(\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=5\\x-1=-5\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-4\end{array}\right.\)

a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)

\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)

b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)

\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)

c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)

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

Tìm x:

\(7x\left(x-2\right)=x-2\)

\(\Leftrightarrow7x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{7}\end{array}\right.\)

Phân tích đa thức thành nhân tử:

\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

1) \(M=\frac{x-1}{x-5}=\frac{\left(x-5\right)+4}{x-5}=1+\frac{4}{x-5}\)

Vậy để M nguyên thì \(x-5\inƯ\left(4\right)\)

Mà Ư(4)={1;-1;2;-2;4;-4}

Ta có bảng sau:
 

Vậy x={1;3;4;6;7;9}

2) Để M âm

\(\Leftrightarrow\)\(\frac{x-1}{x-5}< 0\)

\(\Leftrightarrow\begin{cases}x-1>0\\x-5< 0\end{cases}\) hoặc \(\begin{cases}x-1< 0\\x-5>0\end{cases}\)

\(\Leftrightarrow1< x< 5\)

\(\left(x-1\right)x-x\left(x+3\right)=-48\)

\(\Leftrightarrow x^2-x-x^2-3x=-48\)

\(\Leftrightarrow-4x=-48\Leftrightarrow x=12\)

\(A=x^2-x+31=\left(x^2-x+\frac{1}{4}\right)+\frac{123}{4}=\left(x-\frac{1}{2}\right)^2+\frac{123}{4}\)

Vì; \(\left(x-\frac{1}{2}\right)^2\ge0\)

=> \(\left(x-\frac{1}{2}\right)^2+\frac{123}{4}\ge\frac{123}{4}\)

Vậy GTNN của A là \(\frac{123}{4}\) khi \(x=\frac{1}{2}\)