Tìm x, biết
a) x + 30%x = -1,3
b) ( 2x + 3/4 ) (
x
2
– 1/9 ) = 0
c) 7 ║ 3/5x – 40% ║ – 3/5 : 1/2 = 0,2
d) x-5/8 = 2/x-5
Tìm x, biết a) x + 30%x = -1,3 b) ( 2x + 3/4 ) ( x 2 – 1/9 ) = 0 c) 7 ║ 3/5x – 40% ║ – 3/5 : 1/2 = 0,2 d) x-5/8 = 2/x-5
By Melody
Đáp án:
a) x+30%x=−1,3x+30%x=-1,3
⇔1310x=−1,3⇔1310x=-1,3
⇔x=−1⇔x=-1
Vậy x=−1x=-1
b) (2x+34)(x2−19)=0(2x+34)(x2-19)=0
⇔⇔ ⎡⎢ ⎢⎣2x+34=0x2−19=0[2x+34=0x2−19=0
⇔⇔ ⎡⎢ ⎢⎣x=−38x=29[x=−38x=29
Vậy S=−38;29S=−38;29
c) 7∣∣∣35x−40%∣∣∣−35:12=0,27|35x-40%|-35:12=0,2
⇔7∣∣∣35x−40%∣∣∣−65=0,2⇔7|35x-40%|-65=0,2
⇔7∣∣∣35x−40%∣∣∣=75⇔7|35x-40%|=75
⇔∣∣∣35x−40%∣∣∣=15⇔|35x-40%|=15
⇔⇔⎡⎢ ⎢⎣35x−25=1535x−25=−15[35x−25=1535x−25=−15
⇔⇔ ⎡⎣x=1x=13[x=1x=13
Vậy S={1;13}S={1;13}
d) x−58=2x−5x-58=2x-5
⇔(x−5)2=16⇔(x-5)2=16
⇔x2−5x−5x+25=16⇔x2-5x-5x+25=16
⇔x2−10x+9=0⇔x2-10x+9=0
⇔x2−x−9x+9=0⇔x2-x-9x+9=0
⇔x(x−1)−9(x−1)=0⇔x(x-1)-9(x-1)=0
⇔(x−9)(x−1)=0⇔(x-9)(x-1)=0
⇔⇔ [x−9=0x−1=0[x−9=0x−1=0
⇔⇔ [x=9x=1[x=9x=1
Vậy S={9;1}
Đáp án:
Giải thích các bước giải:
a) `x + 30%x = -1,3`
`⇔ \frac{13}{10}x=-1,3`
`⇔ x=-1`
Vậy `x=-1`
b) `(2x+3/4)(x/2-1/9)=0`
`⇔` \(\left[ \begin{array}{l}2x+\dfrac{3}{4}=0\\\dfrac{x}{2}-\frac{1}{9}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{3}{8}\\x=\dfrac{2}{9}\end{array} \right.\)
Vậy \(S={-\dfrac{3}{8};\dfrac{2}{9}}\)
c) `7|3/5x-40%|-3/5:1/2=0,2`
`⇔ 7|3/5x-40%|-6/5=0,2`
`⇔ 7|3/5x-40%|=\frac{7}{5}`
`⇔ |3/5x-40%|=\frac{1}{5}`
`⇔`$\left[ \begin{array}{l} \dfrac{3}{5}x-\dfrac{2}{5}=\dfrac{1}{5} \\ \dfrac{3}{5}x-\dfrac{2}{5}=-\dfrac{1}{5}\end{array} \right.$
`⇔` \(\left[ \begin{array}{l}x=1\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={1;1/3}`
d) `\frac{x-5}{8}=\frac{2}{x-5}`
`⇔ (x-5)^2=16`
`⇔ x^2-5x-5x+25=16`
`⇔ x^2-10x+9=0`
`⇔ x^2-x-9x+9=0`
`⇔ x(x-1)-9(x-1)=0`
`⇔ (x-9)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x-9=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=9\\x=1\end{array} \right.\)
Vậy `S={9;1}`