Question

[Diệu's Khói's]

1. Cho P = \(\frac{x}{x+2}\)+ \(\frac{x+3}{x-2}\)+\(\frac{6-9x}{4-x^2}\)

a) Rút gọn P

b) Tìm x để P=3

2. Cho B= \(\frac{2a^2}{a^2-1}\)+\(\frac{a}{a+1}\)-\(\frac{a}{a-1}\)

a) Rút gọn B

b) Tìm a nguyên để B nguyên

3. Rút gọn Q= \(\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

4. Cho P=\(\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)

a) Rút gọn P

b) Tìm x để P=-4

\(\Leftrightarrow\)Giúp với mình cần gấp\(\Leftrightarrow\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a) Ta có: \(P=\frac{x}{x+2}+\frac{x+3}{x-2}+\frac{6-9x}{4-x^2}\)

\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6-9x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2-2x+x^2+5x+6-6+9x}{\left(x-2\right)\left(x+2\right)}\)

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

b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Để P=3 thì \(\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}=3\)

\(\Leftrightarrow2x^2+12x=3\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow2x^2+12x=3\left(x^2-4\right)\)

\(\Leftrightarrow2x^2+12x=3x^2-12\)

\(\Leftrightarrow2x^2+12x-3x^2+12=0\)

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

\(\Leftrightarrow x^2-12x-12=0\)

\(\Leftrightarrow x^2-12x+36-24=0\)

\(\Leftrightarrow\left(x-6\right)^2=24\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=\sqrt{24}\\x-6=-\sqrt{24}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6+2\sqrt{6}\left(nhận\right)\\x=6-2\sqrt{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: khi P=3 thì \(x\in\left\{6+2\sqrt{6};6-2\sqrt{6}\right\}\)

Bài 2:

a) Ta có: \(B=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)

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

\(=\frac{2a^2+a^2-a-a^2-a}{\left(a+1\right)\cdot\left(a-1\right)}=\frac{2a^2-2a}{\left(a+1\right)\left(a-1\right)}\)

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

b) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)

Để B là số nguyên thì \(2a⋮a+1\)

\(\Leftrightarrow2a+2-2⋮a+1\)

\(\Leftrightarrow-2⋮a+1\)

\(\Leftrightarrow a+1\inƯ\left(-2\right)\)

\(\Leftrightarrow a+1\in\left\{1;-1;2;-2\right\}\)

hay \(a\in\left\{0;-2;1;-3\right\}\)

mà \(a\notin\left\{1;-1\right\}\)

nên \(a\in\left\{0;-2;-3\right\}\)

Vậy: khi B có giá trị nguyên thì \(a\in\left\{0;-2;-3\right\}\)

Bài 3:

Ta có: \(Q=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x+2\right)\left(x-2\right)}\)

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

Bài 4:

a) Ta có: \(P=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)

\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right)\)

\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}+2+3\sqrt{x}-6}{\sqrt{x}-2}\)

\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\cdot4\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Để P=-4 thì \(\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}=-4\)

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

\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(x-4\sqrt{x}+4\right)\)

\(\Leftrightarrow-16x+16\sqrt{x}=-4x+16\sqrt{x}-16\)

\(\Leftrightarrow-16x+16\sqrt{x}+4x-16\sqrt{x}+16=0\)

\(\Leftrightarrow-12x+16=0\)

\(\Leftrightarrow-12x=-16\)

hay \(x=\frac{4}{3}\)(nhận)

Vậy: Khi P=-4 thì \(x=\frac{4}{3}\)

Answer 2

[Diệu's Khói's]

Bạn giải thích kĩ hơn về phần bài 2 câu b được ko ạk