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a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a) x = 4
b) x = -8
c) | x | - 9 = -2 + 17
| x | = 15 + 9
| x | = 24
x = 24 hoặc x = -24
d) |x – 9| = -2 + 17
|x – 9| = 15
x – 9 = 15 hoặc x – 9 = -15
x = 24 hoặc x = -6
a) -22<x<23
<=> x={-21,-20,-19,-18,....,19,20,21,22}
(-21)+(-20)+(-19)+(-18)+...+(-2)+(-1)+0+....+19+20+21+22
= 22+[(-21)+21]+[(-20)+20]+[(-19)+19]+[(-18)+18]+...+[(-2)+2]+[(-1)+1]+0
= 22+0+0+0+...+0+0
=22
b) -36<x<34
<=> x={-35,-34,-33,-32,...,32,33}
(-35)+(-34)+(-33)+...(-2)+(-1)+0+...+31+32+33
= (-35)+(-34)+[(-33)+33]+[(-32)+32]+...+[(-2)+2]+[(-1)+1]+0
=(-35)+(-34)+0+0+...+0+0
= -69
a) <=> 10 - 2x + 5 = 1 - 3x
<=> -2x + 3x = 1 - 10 - 5
<=> x = -14
b) \(\Leftrightarrow\frac{x-1}{x+1}-\frac{8}{9}=0\)
\(\Leftrightarrow\frac{9\left(x-1\right)-8\left(x+1\right)}{9\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{9x-9-8x-8}{9x+9}=0\)
\(\Leftrightarrow\frac{x-17}{9x+9}=0\)
ĐK: 9x + 9 \(\ne\)0 => x \(\ne\)-1
\(\Leftrightarrow x-17=0\)
\(\Leftrightarrow x=17\)
Ps: Câu b không chắc lắm.
phan b Vu Nhu Mai sai roi phai the nay moi dung
x-1/x+1=8/9
(x-1).9=(x+1).8
9x-9=8x+8
9x-8x=8+9
1x=17
x=17;1
x=17
a, \(\frac{5}{7}.3x-\frac{8}{5}=\frac{9}{35}\)
=> \(\frac{15}{7}x=\frac{9}{35}+\frac{8}{5}\)=> \(\frac{15}{7}x=\frac{9}{35}+\frac{56}{35}\)
=> \(\frac{15}{7}x=\frac{65}{35}=\frac{13}{7}\)=> \(x=\frac{13}{7}:\frac{15}{7}=\frac{13}{15}\)
vậy \(x=\frac{13}{15}\)
b, \(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)
=> \(\frac{10}{9}x+\frac{1}{2}=\frac{5}{36}+\frac{1}{18}\)=\(\frac{5}{36}+\frac{2}{36}=\frac{7}{36}\)
=> \(\frac{10}{9}x=\frac{7}{36}-\frac{1}{2}\)=\(\frac{7}{36}-\frac{18}{36}\)=\(\frac{-11}{36}\)=> \(x=\frac{-11}{36}:\frac{10}{9}\)=\(\frac{-11}{36}.\frac{9}{10}\)=\(\frac{-11}{40}\)
vậy x=\(\frac{-11}{40}\)
a, ta có : \(\frac{5}{7}.\frac{3x-8}{5}=\frac{9}{35}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}:\frac{5}{7}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}.\frac{7}{5}\)
\(\Leftrightarrow\frac{3x-8}{5}=\frac{9}{5.5}\Leftrightarrow3x-8=\frac{9}{25}.5\Leftrightarrow3x-8=\frac{9.5}{25}\)
\(\Leftrightarrow3x-8=\frac{9}{5}\Leftrightarrow3x-8=\frac{9}{5}+8\Leftrightarrow3x=\frac{9+8.5}{5}\)
\(\Leftrightarrow3x=\frac{49}{5}\Leftrightarrow x=\frac{49}{5}:3\Leftrightarrow x=\frac{49}{5}.\frac{1}{3}=\frac{49}{15}\)
~ Vậy, ta tìm được \(x=\frac{49}{15}\)
b, Ta có : \(\frac{2}{9}.\frac{5x+1}{2}-\frac{1}{18}=\frac{5}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5}{36}+\frac{1}{18}\)
\(\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5+2.1}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{7}{36}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}:\frac{2}{9}\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}.\frac{9}{2}\Leftrightarrow\frac{5x+1}{2}=\frac{7.9}{4.9.2}\)
\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{8}\Leftrightarrow5x+1=\frac{7}{8}.2\Leftrightarrow5x+1=\frac{7.2}{8}\)
\(\Leftrightarrow5x+1=\frac{7}{4}\Leftrightarrow5x=\frac{7}{4}-1\Leftrightarrow5x=\frac{7-1.4}{4}\)
\(\Leftrightarrow5x=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}:5\Leftrightarrow x=\frac{3}{4}.\frac{1}{5}\Leftrightarrow x=\frac{3}{20}\)
~ Vậy, ta tìm được \(x=\frac{3}{20}\)
a) \(55-4x-4\left(-x+3\right)=6-2\left(-8-3x\right)\)
\(55-4x+4x-12=6+16+6x\)
\(43-6-16=6x\)
\(6x=21\)
\(x=3,5\)
b) \(-5\left(-2x-6\right)-9\left(4-7x\right)=51-3x+6\left(x-9\right)\)
\(10x+30-36+63x=51-3x+6x-54\)
\(73x-6=-3+3x\)
\(73x-3x=-3+6\)
\(70x=3\)
\(x=\frac{3}{70}\)
c) \(93+\left|6-3x\right|-39=231\)
\(\left|6-3x\right|+54=231\)
\(\left|6-3x\right|=177\)
\(\Rightarrow\orbr{\begin{cases}6-3x=177\\6-3x=-177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6-177\\3x=6+177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-171\\3x=183\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-57\\x=61\end{cases}}\)
a, 2x -(-17) = 15
2x + 17 = 15
2x = 15 - 17
2x =(-2)
x = (-2) : 2
x = (-1)
a) Ta có: \(\frac{9}{25}=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)
TH1: \(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow2x+\frac{3}{5}=\frac{3}{5}\)
\(\Rightarrow2x=0\)
\(\Rightarrow x=0\)
TH2: \(2x+\frac{3}{5}=\frac{-3}{5}\Rightarrow2x=\frac{-6}{5}\Rightarrow x=\frac{-3}{5}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
Mà \(\frac{-1}{27}=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\Leftrightarrow3x=\frac{1}{6}\Rightarrow x=\frac{1}{18}\)
c) \(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{2}{3}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{2}{3}x+\frac{5}{6}=0\)
\(\Rightarrow-\frac{37}{6}x=\frac{-1}{6}\Rightarrow x=\frac{1}{37}\)
d) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3-x-\frac{1}{5}=0\)
\(\Rightarrow-3x=\frac{47}{10}\Rightarrow x=\frac{-47}{30}\)
| 3x + 2 | + 5 = 9 - x
=> | 3x + 2 | = 9 - x - 5
=> | 3x + 2 | = 4 - x
=> 3x + 2 = 4 + x hoặc 3x + 2 =4 - x
=> 3x - x = 4 - 2 hoặc 3x + x =4 - 2
=> 2x = 2 hoặc 4x = 2
=> x = 1 hoặc x = 0,5
| 3x + 2 | + 5 = 9 - x
=> | 3x + 2 | = 9 - x - 5
=> | 3x + 2 | = 4 - x
=> 3x + 2 = 4 + x hoặc 3x + 2 =4 - x
=> 3x - x = 4 - 2 hoặc 3x + x =4 - 2
=> 2x = 2 hoặc 4x = 2
=> x = 1 hoặc x = 0,5