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a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)
\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)
\(\Leftrightarrow\left\{\left(x^2-x+1\right)^2+2x^2\right\}=\left(3x^2\right)^2\)
\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)
\(\Leftrightarrow\left(x^2-x+1\right)=x^2\)
\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)
Vậy...
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 \(x=\frac{4}{3}\)
