a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)

\(x=\dfrac{4}{-10}+\dfrac{3}{5}\)

\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)

\(x=\dfrac{1}{5}\)

b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)

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

\(\dfrac{3}{x}=\dfrac{4}{x}+4\)

\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)

\(3x=\left(4x+4\right)x\)

\(3x=5x\cdot x+4x\)

\(3x=x\left(5x+4\right)\)

\(3=5x+4\)

\(5x=-1\)

\(x=\dfrac{-1}{5}\)

Toán này lớp 6 á???

a. x = -4/10 + 3/5 = 1/5 

b.3 ( x + 4) = (x - 2).4 

=> 3x + 12 = 4x - 8 

=> -x = -20 

=> x = 20  

Câu c mình không hiểu ý bạn

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: \(x\in\left\{24;-6\right\}\)

\(3x^3+xy=3\)

\(\Leftrightarrow x\left(3x^2+y\right)=3\)

Với x = 1

\(\Rightarrow3x^2+y=3\Leftrightarrow y=0\)

Với x = 3

\(\Rightarrow3.3^2+y=3\)

\(\Leftrightarrow y=3-27=-24\)

Với x = -1

=> 3.(-1)^2+y=3

=>y=0

Với x = -3

=> y = -24

=18073

=18073

2009 * 2009 - 2001 * 2008

= 4036081 - 4018008

= 18073

bạn chi nhỏ câu hỏi ra nhé!!! 😅

Sorry nha, lộn rồi, x = 1

\(\frac{1}{6}< x< \frac{196}{98}\)

\(\frac{1}{6}< x< 2\)

Mà phân số \(\frac{1}{6}\)là phân số bé hơn 1

\(\Rightarrow x=2\)

\(\frac{1}{6}< x< 2\)

Vậy x phải = 1 nha bn nhớ tk dùm mk nha

X=5 y=21 nhé 

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