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d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)
hay A=x-2
a)\(\frac{x^2+5x+4}{x^2-1}=\frac{A}{x^2-2x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=\frac{A}{\left(x-1\right)^2}\)
\(\Leftrightarrow\frac{x+4}{x-1}=\frac{A}{\left(x-1\right)^2}\). Nhân 2 vế ở tử với x-1 ta có:
\(x+4=\frac{A}{x-1}\Leftrightarrow A=\left(x-1\right)\left(x+4\right)=x^2+3x-4\)
b)\(\frac{x^2-3x}{2x^2-7x+3}=\frac{x^2+4x}{A}\)
\(\Leftrightarrow\frac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}=\frac{x\left(x+4\right)}{A}\)
\(\Leftrightarrow\frac{x}{2x-1}=\frac{x\left(x+4\right)}{A}\).Nhân 2 vế ở mẫu với x ta có:
\(2x-1=\frac{x+4}{A}\)\(\Leftrightarrow\left(2x-1\right)\left(x+4\right)=A\Leftrightarrow A=2x^2+7x-4\)
ap dung cong thuc: a/b = c/d <=> ad= bc <=> c = ad/b
A = (4x2-7x+3)(x2+2x+1)/(x2-1)
\(\frac{5x^2-13x+6}{A}=\frac{5x-3}{2x+5}\)
\(\Rightarrow\left(5x^2-13x+6\right)\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow\left[5x\left(x-2\right)-3\left(x-2\right)\right]\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow\left(x-2\right)\left(5x-3\right)\left(2x+5\right)=A.\left(5x-3\right)\)
\(\Rightarrow A=\left(x-2\right)\left(2x+5\right)=2x^2+x-10\)
\(\frac{A}{3x-1}=\frac{12x^2+4x}{9x^2-1}\)
\(\Rightarrow A.\left(9x^2-1\right)=\left(3x-1\right)\left(12x^2+4x\right)\)
\(\Rightarrow A.\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)4x\left(3x+1\right)\)
\(\Rightarrow A=4x\)