\(\sqrt{x-3}-\sqrt{9x-27}+2\sqrt{16x-48}=6\)

\(\Leftrightarrow\sqrt{x-3}=1\)

hay x=4

lấy máy tính bỏ túi áấy bn  

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

a: =>5-x=-23

=>x=5+23=28

b: =>x-3-x+7-25+x=54

=>x-21=54

=>x=75

c: =>7-9x-2x+4=-5x-35+27-25=-5x-37

=>-11x+3=-5x-37

=>-6x=-40

=>x=20/3

a) \(10-x-5=\left(-5\right)-7-11\)

\(\Rightarrow5-x=-23\)

\(\Rightarrow x=5+23\)

\(\Rightarrow x=28\)

b) \(\left(x-3\right)-\left(x+17-24\right)-\left(25-x\right)=24-\left(-30\right)\)

\(\Rightarrow x-3-x-17+24-25+x=54\)

\(\Rightarrow x-21=54\)

\(\Rightarrow x=54+21\)

\(\Rightarrow x=75\)

c) \(\left(7-9x\right)-\left(2x-4\right)=-\left(5x+35\right)-\left(-27\right)-25\)

\(\Rightarrow7-9x-2x+4=-5x-35+27-25\)

\(\Rightarrow11-11x=-5x-33\)

\(\Rightarrow-11x+5x=-33-11\)

\(\Rightarrow-6x=-22\)

\(\Rightarrow x=\dfrac{22}{6}=\dfrac{11}{3}\)

a. 
10-x-5 = (-5) - 7 -11
=>5-x = 0
=>x=5
b
(x-3) - (x+17-24) - (25-x) = 24 - (-30)
=>x - 3 - x - 17 + 24 - 25 - x = 24 + 30
=>-x - 21 = 54
=>-x = 75
=>x = -75
c
(7 - 9x) - (2x - 4) = - (5x + 35) - (-27) - 25
=>7-9x - 2x + 4 = -5x - 35 + 27 - 35
=>11 - 11x + 5x = -43
=>16x = 11 + 43
=>16x = 54
=>x=4

Lời giải:
a.

$(x-15).27=0$

$x-15=0:27=0$

$x=15+0=15$

b.

$23(42-x)=0$

$42-x=0$

$x=42$

c.

$(9x+2).3=60$

$9x+2=60:3=20$

$9x=18$

$x=2$
d.

$71+(26-3x):5=75$

$(26-3x):5=75-71=4$

$26-3x=4.5=20$

$3x=26-20=6$

$x=6:2=3$

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

a, ( x + 1 ) = 0

<=> x = -1

b, x³ - 9x² + 27x - 27 = 0

<=> ( x - 3 )³ = 0 

<=> x - 3 = 0

<=> x = 3