a: \(B=\frac{1}{x+4}-\frac{3x}{4-x}-\frac{25x-4}{x^2-16}\)

\(=\frac{x-4+3x\left(x+4\right)-25x+4}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3x^2+12x-24x}{\left(x-4\right)\left(x+4\right)}=\frac{3x^2-12x}{\left(x-4\right)\left(x+4\right)}=\frac{3x\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\frac{3x}{x+4}\)

b: |3-2x|=5

=>|2x-3|=5

=>\(\left[\begin{array}{l}2x-3=5\\ 2x-3=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=8\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)

Thay x=-1 vào B, ta được:

\(B=\frac{3\cdot\left(-1\right)}{-1+4}=\frac{-3}{3}=-1\)

c: A<=2/3B

=>\(\frac{2x-5}{x+4}\le\frac23\cdot\frac{3x}{x+4}=\frac{2x}{x+4}\)

=>\(-\frac{5}{x+4}\le0\)

=>x+4>0

=>x>-4

Kết hợp ĐKXĐ, ta được: x>-4 và x<>4

x=24 nên x+1=25

C=x^4-x^3(x+1)+x^2(x+1)-x(x+1)+30

=x^4-x^4-x^3+x^3+x^2-x^2-x+30

=-x+30

=30-24

=6

Chọn B.

Ta có: 

1) \(4-25x^2=0\)

\(\Rightarrow\left(2-5x\right)\left(2+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

2) Tính thì phải cho giá trị của x.

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

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

\(=\left[{}\begin{matrix}\left(\dfrac{5}{2}-1\right)^3=\dfrac{27}{8}\\\left(-\dfrac{5}{2}-1\right)^3=-\dfrac{343}{8}\end{matrix}\right.\)

Câu 2:

a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)

b: Sửa đề: \(A=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\left(\dfrac{2}{x^2}-\dfrac{x-1}{x}\right)\)

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

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

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

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

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

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

c: Khi x=2024 thì \(A=\dfrac{2024+1}{2\cdot2024}=\dfrac{2025}{4048}\)

Câu 1:

a: \(25x^2\left(x-3y\right)-15\left(3y-x\right)\)

\(=25x^2\left(x-3y\right)+15\left(x-3y\right)\)

\(=\left(x-3y\right)\left(25x^2+15\right)\)

\(=\left(x-3y\right)\cdot5\cdot\left(5x^2+3\right)\)

b: \(x^4-5x^2+4\)

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

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

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

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

dap an: 2000

TL
 

3y²+3y+25x²-10x+4

HT

TL:

3y² + 3y + 25x² - 10x + 4

~HT~

\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)

\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)

\(C=\left|5x-2\right|+\left|5x\right|=\left|2-5x\right|+\left|5x\right|\)

\(C\ge\left|2-5x+5x\right|=2\)

Dấu " = " xảy ra \(\Leftrightarrow\)( 2 - 5x ) . 5x \(\ge\)0

\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge0\\2-5x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x\le0\\2-5x\le0\end{cases}}\)

\(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)

Vậy GTNN của C là 2 \(\Leftrightarrow\)\(0\le x\le\frac{2}{5}\)

\(C=\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)

\(C=\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x\right)^2}\)

\(C=\left|5x-2\right|+\left|5x\right|\)

\(C=\left|2-5x\right|+\left|5x\right|\ge\left|2-5x+5x\right|=2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2-5x\ge0\\5x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le\frac{2}{5}\\x\ge0\end{cases}\Leftrightarrow0\le}x\le\frac{2}{5}}\)