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a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
a)= \(\frac{\left(2x+3\right)^2}{2x^2+3x-4x-6}\)
=\(\frac{\left(2x+3\right)^2}{x\left(2x+3\right)-2\left(2x+3\right)}\)
= \(\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(2x+3\right)}\)
=\(\frac{2x+3}{x-2}\)
b) = \(\frac{3\left|x-4\right|}{3x^2-3x-1296}\)
= \(\frac{3\left|x-4\right|}{3\left(x^2-x-432\right)}\)
=\(\frac{\left|x-4\right|}{x^2-x-432}\)
a) ĐKXĐ: \(\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\\2-x\ne0\end{cases}}\) => \(\hept{\begin{cases}x\ne-2\\x\ne\pm2\\x\ne2\end{cases}}\) => \(x\ne\pm2\)
Ta có:Q = \(\frac{x-1}{x+2}+\frac{4x+4}{x^2-4}+\frac{3}{2-x}\)
Q = \(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Q = \(\frac{x^2-2x-x+2+4x+4-3x-6}{\left(x+2\right)\left(x-2\right)}\)
Q = \(\frac{x^2-2x}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)
b) ĐKXĐ P: x - 3 \(\ne\)0 => x \(\ne\)3
Ta có: P = 3 => \(\frac{x+2}{x-3}=3\)
=> x + 2 = 3(x - 3)
=> x + 2 = 3x - 9
=> x - 3x = -9 - 2
=> -2x = -11
=> x = 11/2 (tm)
Với x = 11/2 thay vào Q => Q = \(\frac{\frac{11}{2}}{\frac{11}{2}+2}=\frac{11}{15}\)
c) Với x \(\ne\)\(\pm\)2; x \(\ne\)3
Ta có: M = PQ = \(\frac{x+2}{x-3}\cdot\frac{x}{x+2}=\frac{x}{x-3}=\frac{x-3+3}{x-3}=1+\frac{3}{x-3}\)
Để M \(\in\)Z <=> 3 \(⋮\)x - 3
=> x - 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng:
| x - 3 | 1 | -1 | 3 | -3 |
| x | 4 | 2 (ktm) | 6 | 0 |
Vậy ...
\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)
\(C=\frac{2x+3y}{3x+4y}=\frac{2\cdot\frac{3}{4y}+3y}{3\cdot\frac{3y}{4}\cdot4y}\)
\(=\frac{2\cdot\frac{3}{4}+3}{3\cdot\frac{3}{4}+4}=\frac{\frac{9}{2}}{\frac{25}{4}}\)
\(=\frac{9}{2}\cdot\frac{4}{25}=\frac{18}{25}\)
\(\frac{3}{x}=\frac{4}{y}\Rightarrow x=\frac{3y}{4}\)
\(\Rightarrow C=\frac{\frac{3y}{2}+3y}{\frac{9y}{4}+4y}=\frac{\frac{9y}{2}}{\frac{25y}{4}}=\frac{18}{25}\)