1)(4x-10)(20+5x)=0
2) $\frac{x+1}{x-1}$ -$\frac{x-1}{x+1}$ =$\frac{4}{x^2-1}$
3)$\frac{2x-8}{6}$ -$\frac{3x+1}{4}$ =$\frac{9x-2}{8}$ +$\frac{3x-1}{12}$
5)(x+2)(3-4x)=x^2 +4x+4
6)(2x+1)(x+1)^2(2x+3)=0
1)(4x-10)(20+5x)=0
2) $\frac{x+1}{x-1}$ -$\frac{x-1}{x+1}$ =$\frac{4}{x^2-1}$
3)$\frac{2x-8}{6}$ -$\frac{3x+1}{4}$ =$\frac{9x-2}{8}$ +$\frac{3x-1}{12}$
5)(x+2)(3-4x)=x^2 +4x+4
6)(2x+1)(x+1)^2(2x+3)=0
Đáp án:
↓↓↓
Giải thích các bước giải:
`a,(4x-10)(20-5x)=0`
⇒ \(\left[ \begin{array}{l}4x-10=0\\20-5x=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=5/2\\x=4\end{array} \right.\)
`b,(x+1)/(x-1)-(x-1)/(x+1)=4/(x^2-1)`
⇒`x`$\neq$ `±1`
⇒`(x+1)/(x-1)-(x-1)/(x+1)-4/(x^2-1)=0`
⇒`((x+1)^2-(x-1)^2-4)/((x-1)(x+1))=0`
⇒`(4(x-1))/((x-1)(x+1))=0`
⇒`4/(x+1)=0`
⇒`4=0`
⇒`x∈∅`
`c,(2x-8)/6-(3x+1)/4=(9x-2)/8+(3x-1)/12`
⇒`4(2x-8)-6(3x+1)=3(9x-2)+2(3x-1)`
⇒`8x-32-18x-6=27x-6+6x-2`
⇒`-43x=30`
⇒`x=-30/43`
`d,(x+2)(3-4x)=x^2+4x+4`
⇒`3x-4x^2+6-8x=x^2+4x+4`
⇒`-5x-4x^2+6=x^2+4x+4`
⇒`-5x-4x^2+6-x^2-4x-4=0`
⇒`(x+2)(5x-1)=0`
⇒\(\left[ \begin{array}{l}x=-2\\x=1/5\end{array} \right.\)
`e,(2x+1)(x+1)^2(2x+3)=0`
⇒\(\left[ \begin{array}{l}2x+1=0\\(x+1)^2=0\\2x+3=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-1/2\\x=-1\\x=-3/2\end{array} \right.\)
Đáp án:
5) \(\left[ \begin{array}{l}
x = – 2\\
x = \dfrac{1}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\left( {4x – 10} \right)\left( {20 + 5x} \right) = 0\\
\to \left[ \begin{array}{l}
4x – 10 = 0\\
20 + 5x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = – 4
\end{array} \right.\\
2)DK:x \ne \pm 1\\
\dfrac{{{{\left( {x + 1} \right)}^2} – {{\left( {x – 1} \right)}^2} – 4}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = 0\\
\to {x^2} + 2x + 1 – {x^2} + 2x – 1 – 4 = 0\\
\to 4x – 4 = 0\\
\to x = 1\left( l \right)\\
\to x \in \emptyset \\
3)\dfrac{{4\left( {2x – 8} \right) – 6\left( {3x + 1} \right) – 3\left( {9x – 2} \right) – 2\left( {3x – 1} \right)}}{{24}} = 0\\
\to 8x – 32 – 18x – 6 – 18x + 6 – 6x + 2 = 0\\
\to – 34x – 30 = 0\\
\to x = – \dfrac{{15}}{{17}}\\
5)(x + 2)(3 – 4x) = {x^2} + 4x + 4\\
\to – 4{x^2} – 5x + 6 = {x^2} + 4x + 4\\
\to 5{x^2} + 9x – 2 = 0\\
\to 5{x^2} + 10x – x – 2 = 0\\
\to 5x\left( {x + 2} \right) – \left( {x + 2} \right) = 0\\
\to \left( {x + 2} \right)\left( {5x – 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = – 2\\
x = \dfrac{1}{5}
\end{array} \right.\\
6)(2x + 1){(x + 1)^2}(2x + 3) = 0\\
\to \left[ \begin{array}{l}
2x + 1 = 0\\
x + 1 = 0\\
2x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{1}{2}\\
x = – 1\\
x = – \dfrac{3}{2}
\end{array} \right.
\end{array}\)