Giải pt (2x + 5)^2 = (3x – 1)^2 4x^2 (x – 1) – x + 1 = 0 9 (2x + 1)^2 = 4(x – 5)^2 17/10/2021 Bởi Eva Giải pt (2x + 5)^2 = (3x – 1)^2 4x^2 (x – 1) – x + 1 = 0 9 (2x + 1)^2 = 4(x – 5)^2
Đáp án: \((2x + 5)^2 = (3x – 1)^2\\\Leftrightarrow (2x+5)^2-(3x+1)^2=0\\\Leftrightarrow (2x+5+3x-1)(2x+5-3x+1)=0\\\Leftrightarrow (5x+4)(-x+6)=0\\\Leftrightarrow \left[ \begin{array}{l}5x+4=0\\-x=-6\end{array} \right.\\\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{4}{5}\\x=6\end{array} \right.\\4x^2 (x – 1) – x + 1 = 0\\\Leftrightarrow 4x^2(x-1)-(x-1)=0\\\Leftrightarrow (x-1)(4x^2-1)=0\\\Leftrightarrow (x-1)(2x-1)(2x+1)=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\2x-1=0\\2x+1=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\\9 (2x + 1)^2 = 4(x – 5)^2\\\Leftrightarrow 3^2(2x+1)^2-2^2(x-5)^2=0\\\Leftrightarrow [3(2x+1)]^2-[2(x-5)]^2=0\\\Leftrightarrow (6x+3)^2-(2x-10)^2=0\\\Leftrightarrow (6x+3-2x+10)(6x+3+2x-10)=0\\\Leftrightarrow (4x+13)(8x-7)=0\\\Leftrightarrow \left[ \begin{array}{l}4x+13 =0\\8x-7=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{13}{4}\\x=\dfrac{7}{8}\end{array} \right.\)\(\) Bình luận
Đáp án + Giải thích các bước giải: `a//(2x+5)^{2}=(3x-1)^{2}` `⇔(2x+5)^{2}-(3x-1)^{2}=0` `⇔(2x+5+3x-1)(2x+5-3x+1)=0` `⇔(5x+4)(-x+6)=0` `⇔` \(\left[ \begin{array}{l}5x+4=0\\-x+6=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=-\frac{4}{5}\\x=6\end{array} \right.\) Vậy `S={-\frac{4}{5};6}` `b//4x^{2}(x-1)-x+1=0` `⇔4x^{2}(x-1)-(x-1)=0` `⇔(x-1)(4x^{2}-1)=0` `⇔(x-1)(2x-1)(2x+1)=0` `⇔` \(\left[ \begin{array}{l}x-1=0\\2x-1=0\\2x+1=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=1\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array} \right.\) Vậy `S={1;\frac{1}{2};-\frac{1}{2}}` `c//9(2x+1)^{2}=4(x-5)^{2}` `⇔9(4x^{2}+4x+1)=4(x^{2}-10x+25)` `⇔36x^{2}+36x+9=4x^{2}-40x+100` `⇔36x^{2}-4x^{2}+36x+40x+9-100=0` `⇔32x^{2}+76x-91=0` `⇔32(x^{2}+(19)/(8)x-(91)/(32))=0` `⇔x^{2}+(19)/(8)x-(91)/(32)=0` `⇔x^{2}+2.x.(19)/(16)+((19)/(16))^{2}-(1089)/(256)=0` `⇔(x+(19)/(16))^{2}=(1089)/(256)` `⇔x+(19)/(16)=±(33)/(16)` `⇔` \(\left[ \begin{array}{l}x=\frac{7}{8}\\x=-\frac{13}{4}\end{array} \right.\) Vậy `S={\frac{7}{8};-\frac{13}{4}}` Bình luận
Đáp án:
\((2x + 5)^2 = (3x – 1)^2\\\Leftrightarrow (2x+5)^2-(3x+1)^2=0\\\Leftrightarrow (2x+5+3x-1)(2x+5-3x+1)=0\\\Leftrightarrow (5x+4)(-x+6)=0\\\Leftrightarrow \left[ \begin{array}{l}5x+4=0\\-x=-6\end{array} \right.\\\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{4}{5}\\x=6\end{array} \right.\\4x^2 (x – 1) – x + 1 = 0\\\Leftrightarrow 4x^2(x-1)-(x-1)=0\\\Leftrightarrow (x-1)(4x^2-1)=0\\\Leftrightarrow (x-1)(2x-1)(2x+1)=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\2x-1=0\\2x+1=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\\9 (2x + 1)^2 = 4(x – 5)^2\\\Leftrightarrow 3^2(2x+1)^2-2^2(x-5)^2=0\\\Leftrightarrow [3(2x+1)]^2-[2(x-5)]^2=0\\\Leftrightarrow (6x+3)^2-(2x-10)^2=0\\\Leftrightarrow (6x+3-2x+10)(6x+3+2x-10)=0\\\Leftrightarrow (4x+13)(8x-7)=0\\\Leftrightarrow \left[ \begin{array}{l}4x+13 =0\\8x-7=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=-\dfrac{13}{4}\\x=\dfrac{7}{8}\end{array} \right.\)\(\)
Đáp án + Giải thích các bước giải:
`a//(2x+5)^{2}=(3x-1)^{2}`
`⇔(2x+5)^{2}-(3x-1)^{2}=0`
`⇔(2x+5+3x-1)(2x+5-3x+1)=0`
`⇔(5x+4)(-x+6)=0`
`⇔` \(\left[ \begin{array}{l}5x+4=0\\-x+6=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\frac{4}{5}\\x=6\end{array} \right.\)
Vậy `S={-\frac{4}{5};6}`
`b//4x^{2}(x-1)-x+1=0`
`⇔4x^{2}(x-1)-(x-1)=0`
`⇔(x-1)(4x^{2}-1)=0`
`⇔(x-1)(2x-1)(2x+1)=0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\2x-1=0\\2x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array} \right.\)
Vậy `S={1;\frac{1}{2};-\frac{1}{2}}`
`c//9(2x+1)^{2}=4(x-5)^{2}`
`⇔9(4x^{2}+4x+1)=4(x^{2}-10x+25)`
`⇔36x^{2}+36x+9=4x^{2}-40x+100`
`⇔36x^{2}-4x^{2}+36x+40x+9-100=0`
`⇔32x^{2}+76x-91=0`
`⇔32(x^{2}+(19)/(8)x-(91)/(32))=0`
`⇔x^{2}+(19)/(8)x-(91)/(32)=0`
`⇔x^{2}+2.x.(19)/(16)+((19)/(16))^{2}-(1089)/(256)=0`
`⇔(x+(19)/(16))^{2}=(1089)/(256)`
`⇔x+(19)/(16)=±(33)/(16)`
`⇔` \(\left[ \begin{array}{l}x=\frac{7}{8}\\x=-\frac{13}{4}\end{array} \right.\)
Vậy `S={\frac{7}{8};-\frac{13}{4}}`