Đáp án:
Giải thích các bước giải:
Bài `1:`
`a) (3x-1)^3=\frac{-8}{27}`
`⇒(3x-1)^3=(\frac{-2}{3})^3`
`⇒3x-1=\frac{-2}{3}`
`⇒3x=\frac{1}{3}`
`⇒x=\frac{1}{9}`
`b) \frac{x^2+2}{9}=\frac{5}{12}+\frac{1}{4}`
`⇒\frac{x^2+2}{9}=\frac{2}{3}`
`⇒\frac{x^2+2}{9}=\frac{6}{9}`
`⇒x^2+2=6`
`⇒x^2=4`
`⇒x=±2`
`c) (2x+3)^2=\frac{9}{121}`
`⇒(2x+3)^2=(±\frac{3}{11})^2`
`+)2x+3=\frac{3}{11}`
`⇒2x=\frac{-30}{11}`
`⇒x=\frac{-15}{11}`
`+)2x+3=\frac{-3}{11}`
`⇒2x=\frac{-36}{11}`
`⇒x=\frac{-18}{11}`
Bài `2:`
a) Có `(2x-\frac{3}{5})^2≥0∀x⇒(2x-\frac{3}{5})^2-\frac{1}{7}≥\frac{-1}{7}∀x`
Dấu "`=`" xảy ra khi
`2x-\frac{3}{5}=0⇒2x=\frac{3}{5}⇒x=\frac{3}{10}`
Vậy Min `A=\frac{-1}{7}⇔x=\frac{3}{10}`
b) Có `-(6x+5)^2≤0∀x⇒-(6x+5)^2+\frac{3}{4}≤\frac{3}{4}∀x`
Dấu "`=`" xảy ra khi:
`6x+5=0⇔6x=-5⇔x=\frac{-5}{6}`
Vậy Max `B=\frac{3}{4}⇔x=\frac{-5}{6}`
