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c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)
=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)
=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12
=>-16x-1=-15x+12
=>-x=13
=>x=-13
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow x+1-\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
=>-x-4+4x+4=2x+6+3x+6
=>3x=5x+12
=>-2x=12
hay x=-6(nhận)
\(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)\(\Leftrightarrow\)\(\dfrac{x^2+2x+1+1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow\) \(x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = x + 2 + x + 3 - x - 1 - x - 4
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = 0
\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) = \(\dfrac{2}{x+2}\) + \(\dfrac{3}{x+3}\)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}\) + \(\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}\) = \(\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}\) + \(\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\Leftrightarrow\) \(\dfrac{x+4+4x+4}{x^2+5x+4}\) = \(\dfrac{2x+6+3x+6}{x^2+5x+6}\)
\(\Leftrightarrow\) \(\dfrac{5x+8}{x^2+5x+4}\) = \(\dfrac{5x+12}{x^2+5x+6}\)
Đặt 5x + 8 = y; x2 + 5x + 4 = t, ta có:
\(\dfrac{y}{t}\) = \(\dfrac{y+4}{t+2}\)
\(\Leftrightarrow\) \(\dfrac{y\left(t+2\right)}{t\left(t+2\right)}\) = \(\dfrac{t\left(y+4\right)}{t\left(t+2\right)}\)
\(\Leftrightarrow\) yt + 2y = yt + 4t
\(\Leftrightarrow\) 2y = 4t
\(\Leftrightarrow\) 2(5x + 8) = 4(x2 + 5x + 4)
\(\Leftrightarrow\) 10x + 16 = 4x2 + 20x + 16
\(\Leftrightarrow\) 16 - 16 = 4x2 + 20x - 10x
\(\Leftrightarrow\) 0 = 4x2 + 10x
\(\Leftrightarrow\) 2x(2x + 5) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
CHÚC BN HOK TỐT...
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
Hai câu là hoàn toàn giống nhau, mình làm câu a, câu b bạn tự làm tương tự:
ĐKXĐ: ...
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1\)
Đặt \(4x+\frac{7}{x}-10=t\)
\(\Leftrightarrow\frac{4}{t+2}+\frac{3}{t}=1\Leftrightarrow4t+3\left(t+2\right)=t\left(t+2\right)\)
\(\Leftrightarrow t^2-5t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x+\frac{7}{x}-10=-1\\4x+\frac{7}{x}-10=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-9x+7=0\\4x^2-16x+7=0\end{matrix}\right.\) (bấm casio)
điều kiện xác định \(x\ne0\)
ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
\(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-\left(x^3+2x^2+4x-2x^2-4x-8\right)}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{x^3-2x^2+4x+x^2-2x+4-x^3-2x^2-4x+2x^2+4x+8}{x^4-2x^3+4x^2+2x^3-4x^2+8x+4x^2-8x+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\) \(\Leftrightarrow\dfrac{-x^2+2x+12}{x^4+4x^2+16}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)
\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)
tới đây bn bấm máy tính nha
Lời giải:
a)
\(\frac{x-2}{6x^2-6x}-\frac{1}{4x^2-4}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x^2-1)}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x-1)(x+1)}\)
\(=\frac{2(x+1)(x-2)}{12x(x-1)(x+1)}-\frac{3x}{12x(x-1)(x+1)}=\frac{2(x+1)(x-2)-3x}{12x(x-1)(x+1)}\)
\(=\frac{2x^2-5x-4}{12x(x-1)(x+1)}=\frac{2x^2-5x-4}{12x^3-12x}\)
b) ĐK: \(x\neq \pm 1\)
\(\frac{(x+1)(x^2-2x+1)}{6x^3+6}:\frac{x^2-1}{4x^2-4x+4}\)
\(=\frac{(x+1)(x-1)^2}{6(x^3+1)}.\frac{4x^2-4x+4}{x^2-1}\)
\(=\frac{4(x+1)(x-1)^2(x^2-x+1)}{6(x+1)(x^2-x+1)(x^2-1)}\)
\(=\frac{2(x-1)}{3(x+1)}\)
\(\Leftrightarrow\dfrac{3x-1}{\left(6x-7\right)\left(3x+4\right)}-\dfrac{4x}{\left(8x-3\right)\left(3x+4\right)}=\dfrac{3}{\left(8x-3\right)\left(6x-7\right)}\)
=>(3x-1)(8x-3)-4x(6x-7)=3(3x+4)
=>24x^2-9x-8x+3-24x^2+28x=9x+12
=>11x+3=9x+12
=>2x=9
=>x=9/2
bạn thử phẩn tích mẫu thành nhân tử xem băng phuowg pháp tách hoạch nhẩm nghiệm cx đc