Đáp án+Giải thích các bước giải:
Bài `4`:
`a)(x-1)^3=27`
`⇔(x-1)^3=3^3`
`⇔x-1=3`
`⇔x=3+1`
`⇔x=4`
Vậy `x=4`
`b)x^2+x=0`
`⇔x(x+1)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
Vậy `x=0` hoặc `x=1`
`c)(2x+1)^2=25`
`⇔(2x+1)^2=5^2`
`⇔`\(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
Vậy `x=2` hoặc `x=-3`
`d)(2x-3)^2=36`
`⇔(2x-3)^2=6^2`
`⇔`\(\left[ \begin{array}{l}2x-3=6\\2x-3=-6\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x=9\\2x=-3\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\frac{9}{2}\\x=\frac{-3}{2}\end{array} \right.\)
Vậy `x=9/2;x=-3/2`
`e)5^(x+2)=625`
`⇔5^(x+2)=5^4`
`⇔x+2=4`
`⇔x=4-2`
`⇔x=2`
Vậy `x=2`
`f)(x-1)^(x+2)=(x-1)^(x+4)`
`⇔x+2=x+4`
`⇔x-x=4-2`
`⇔0x=2`(vô lí)
không có giá trị `x` thỏa mãn
`g)(2x-1)^3=-8`
`⇔(2x-1)^3=-2^3`
`⇔2x-1=-2`
`⇔2x=-2+1`
`⇔2x=-1`
`⇔ x=-1/2`
Vậy `x=-1/2`
`h)1/4 . 2/6 . 3/8 . 4/(10) . 5/(12) ... 30/62 . 31/64=2^x`
`⇒1/(2.2) . 2/(2.3) . 3/(2.4) . 4/(2.5) .... (30)/(2.31) . (31)/(2.32)=2^x`
`⇒1/2 . (1/2 . 2/3 . 3/4 . 4/5 ... 30/31 . 31/32)=2^x`
`⇒1/2 . 1/(32)=2^x`
`⇒2^x=1/(64)`
`⇒2^x . 64=1`
`⇒2^x . 2^6=1`
`⇒2^(x+6)=1`
`⇒2^(x+6)=2^0`
`⇒x+6=0`
`⇒x=-6`
Vậy `x=-6`
