a, |x^2 - 3x| = 0

=> x^2 - 3x = 0

=> x(x - 3) = 0

=> x = 0 hoặc x - 3 = 0 

=> x = 0 hoặc x = 3

vậy_

\(\left|a^2-3a\right|=0\)

\(\Rightarrow a^2-3a=0\)

\(\Rightarrow a\left(a-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)

Trả lời

Vì lx²-3xl=0

Nên x²-3x=0

=>x(x-3)=0

x=0.           x-3=0

                    x=3

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Giúp mình vs ạ. Cần gấp

\(\dfrac{3}{4}\)+\(\dfrac{1}{4}\): x=\(\dfrac{-2}{5}\)

⇒\(\dfrac{1}{4}\):x=\(\dfrac{-2}{5}\)-\(\dfrac{3}{4}\)

⇒ x=\(\dfrac{1}{4}\): \(\dfrac{-23}{20}\)

⇒ x=\(\dfrac{-5}{23}\)

\(\text{Phần a :}\)

\(\frac{2}{3}.x=\frac{-4}{27}\)

\(\Rightarrow x=\frac{-4}{27}:\frac{2}{3}=\frac{-2}{9}\)

\(\text{Phần b :}\)

\(-1\frac{1}{3}.x=1\frac{1}{15}\)

\(\Rightarrow\frac{-4}{3}.x=\frac{16}{15}\)

\(\Rightarrow x=\frac{16}{15}:\frac{-4}{3}=\frac{-4}{5}\)

\(=\left(\dfrac{3+8}{12}\right)\cdot\dfrac{4}{5}=\dfrac{11}{12}\cdot\dfrac{4}{5}=\dfrac{44}{60}=\dfrac{11}{15}\)

A) [124 - (20 - 4x)] = 12 . 20 

[124 - (20 - 4x)] = 240 

(20 - 4x)  = 240 - 124

(20 - 4x) = 116

4x = 116 + 20

4x = 136 

x = 136 : 4 

x = 34

B) (2x - 1) = 1/3 : -4/21

2x - 1 = -7/4

2x = -7/4 + 1 

2x = -3/4

x = -3/4 : 2 

x = -3/8

a) \(\frac{0,5}{0,2}=\frac{1,25}{0,1x}\Leftrightarrow0,1x.0,5=0,2.1,25\)

\(\Leftrightarrow0,1x.0,5=0,25\Leftrightarrow0,1x=0,5\Leftrightarrow x=5\)

b) \(x-\frac{3}{2}=2x-\frac{4}{3}\Leftrightarrow x-2x=\frac{-4}{3}+\frac{3}{2}\)
\(\Leftrightarrow x-2x=\frac{1}{6}\Leftrightarrow-x=\frac{1}{6}\Leftrightarrow x=\frac{-1}{6}\)

c) \(x+\frac{13}{14}=\frac{4}{7}\Rightarrow x=\frac{4}{7}-\frac{13}{14}\Rightarrow x=\frac{-5}{14}\)

d)\(-3\left(x-2\right)=2x+1\)

\(\Leftrightarrow-3x+6=2x+1\Leftrightarrow-3x-2x=1-6\)

\(\Leftrightarrow-5x=-5\Leftrightarrow x=1\)

e) \(\left(x-1\right)^2-4=0\Leftrightarrow\left(x-1\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

cậu có thể tham khảo bài trên ạ, nếu thấy đúng thì cho mk 1 t.i.c.k ạ, thank nhiều

\(d,-3\left(x-2\right)=2x+1\)

\(< =>-3x+6=2x+1\)

\(< =>-3x-2x+6-1=0\)

\(< =>5-5x=0\)

\(< =>5\left(1-x\right)=0< =>x=1\)

\(e,\left(x-1\right)^2-4=0\)

\(< =>\left(x-1+2\right)\left(x-1-2\right)=\left(x+1\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

Giải:

B=1,4 .15/49 - (4/5+2/3) :2 1/5

B=7/5.15/49 - 22/15 : 11/5

B=3/7 - 2/3

B=-5/21

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