a: ĐKXĐ: x>=0

M nguyên khi \(\sqrt{x}+1\) ⋮2

=>\(\sqrt{x}\) là số lẻ

=>x là số chính phương lẻ

=>\(x=\left(2k+1\right)^2\left(k\in Z\right)\)

b: \(\sqrt{x}+1\ge1\forall x\) thỏa mãn ĐKXĐ

=>\(M=\frac{\sqrt{x}+1}{2}\ge\frac12\forall x\) thỏa mãn ĐKXĐ

=>M không có giá trị lớn nhất

a: Thay x=-4 vào B, ta được:

\(B=\dfrac{-4+3}{-4}=\dfrac{-1}{-4}=\dfrac{1}{4}\)

b: \(P=A\cdot B=\dfrac{x^2-3x+2x-9+3x+9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}\)

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

c: Để P nguyên thì \(x-3\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{4;2;8;-2\right\}\)

cảm on tiên sinh

a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

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

a: A=−4x/(x+4)(x−1)

a: \(A=\left(\frac{x}{x+2}+\frac{2}{x-2}+\frac{4x}{4-x^2}\right):\frac{2x+1}{8x+16}\)

\(=\frac{x\left(x-2\right)+2\left(x+2\right)-4x}{\left(x-2\right)\left(x+2\right)}\cdot\frac{8\left(x+2\right)}{2x+1}\)

\(=\frac{x^2-2x+2x+4-4x}{x-2}\cdot\frac{8}{2x+1}=\frac{\left(x-2\right)^2}{\left(x-2\right)}\cdot\frac{8}{2x+1}\)

\(=\frac{8\left(x-2\right)}{2x+1}=\frac{8x-16}{2x+1}\)

b: Thay \(x=-2\frac12=-2,5\) vào A, ta được:

\(A=\frac{8\cdot\left(-2,5\right)-16}{2\cdot\left(-2,5\right)+1}=\frac{-20-16}{-5+1}=\frac{-36}{-4}=9\)

c: Để A nguyên thì 8x-16⋮2x+1

=>8x+4-20⋮2x+1

=>-20⋮2x+1

=>2x+1∈{1;-1;5;-5}

=>2x∈{0;-2;4;-6}

=>x∈{0;-1;2;-3}

Kết hợp ĐKXĐ, ta được: x∈{0;-1;-3}

mk cần gấp lắm các bạn ạk

BÀI 1:

a)  \(ĐKXĐ:\)          \(x-3\)\(\ne\)\(0\)

                          \(\Leftrightarrow\)\(x\)\(\ne\)\(3\)

b)   \(A=\frac{x^3-3x^2+4x-1}{x-3}\)

\(=\frac{\left(x^3-3x^2\right)+\left(4x-12\right)+11}{x-3}\)

\(=\frac{x^2\left(x-3\right)+4\left(x-3\right)+11}{x-3}\)

\(=x^2+4+\frac{11}{x-3}\)

Để  \(A\)có giá trị nguyên thì  \(\frac{11}{x-3}\)có giá trị nguyên

hay  \(x-3\)\(\notinƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta lập bảng sau

\(x-3\)    \(-11\)         \(-1\)             \(1\)           \(11\)

\(x\)             \(-8\)               \(2\)              \(4\)           \(14\)

Vậy....

tach 14-x = 10-4-x roi sau do chac ban cung phai tu biet lam

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

=\(\frac{-3+x}{\left(x-2\right)\left(x-3\right)}.\frac{x+1}{1-x}\)

=\(\frac{1}{\left(x-2\right)}.\frac{x+1}{1-x}\)

=\(\frac{x+1}{\left(x-2\right)\left(1-x\right)}\)

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...