a) Ta có: \(3-\left(17-x\right)=-12\)

\(\Leftrightarrow3-17+x+12=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: x=2

b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)

\(\Leftrightarrow\left|x-9\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{24;-6\right\}\)

Lời giải:
$a+b+c=(a+b+b+c+a+c):2=(-21+49+10):2=19$

$a=(a+b+c)-(b+c)=19-49=-30$

$b=(a+b+c)-(a+c)=19-10=9$

$c=(a+b+c)-(a+b)=19-(-21)=40$

30 x ( x - 60 ) = 10 x x + 20

30x  - 180 = 10x + 20 ( vế trái nhân phá ngoặc ) 

30x -10x = 20 +180 ( chuyển vế đổi dấu ) 

20x = 200 

x = 10

vậy x = 10

=30*6-30*x=10*x+20

=180-30*x=10*x+20

=180+20=40*x

=200=40*x

x=200/40

x=5

bạn nguyễn đức minh sai rồi nhé! mình đang thử các bạn ấy

viết bài văn cho bn ko tra mạng ngày mai cx chx xong=))

thế tự viết chứ lên đây khác j chép mạng

a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x^2\right)^2}=2\)

⇔ \(\left|x^2\right|=2\)

⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)

⇔ x² - 2 = 0

⇔ ( x - √2 )( x + √2 ) = 0

⇔ x - √2 = 0 hoặc x + √2 = 0

⇔ x = ±√2 

b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )

⇔ \(3\sqrt{x+1}=8\)

⇔ \(\sqrt{x+1}=\frac{8}{3}\)

⇔ \(x+1=\frac{64}{9}\)

⇔ \(x=\frac{55}{9}\)( tm )

c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )

⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)

⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)

⇔ \(7\sqrt{x-3}=14\)

⇔ \(\sqrt{x-3}=2\)

⇔ \(x-3=4\)

⇔ \(x=7\)( tm )

d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )

⇔ \(\left|3x-1\right|=5\)

⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)

e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )

⇔ \(\sqrt{\left(x+2\right)^2}=6\)

⇔ \(\left|x+2\right|=6\)

⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)

\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)

​\(b)\)\(ĐK:x\ge0\)

\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)

Vậy \(x=\frac{64}{9}\)

\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)

Vậy \(x\in\hept{2;\frac{-4}{3}}\)

                -Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)

                -Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)

Vậy \(x=4;x=-8\)

x?1 hả bạn

Sửa : \(x+16⋮x+1\)

\(x+1+15⋮x+1\)

\(15⋮x+1\)hay \(x+1\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)

\(\left[{}\begin{matrix}\dfrac{x}{3}-\dfrac{12}{x}=0\\5:x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{x^2-36}{3x}=0\\\dfrac{5}{x}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}=>x=6và\left(-6\right)\\=>x=5\end{matrix}\right.\)

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)