\(\left(x-1\right)\left(3x-15\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x-15=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

Vậy x = { 1; 5 }

( x - 1 ) ( 3x - 15 ) = 0

=> \(\orbr{\begin{cases}3x-15=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}3x=15\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy \(x\in\left\{5;1\right\}\)

( x - 1) ( 3x -15) = 0

=> x+ 1 = 0 => x =1

Hoặc 3x - 15 = 0 => 3x = 15 => x = 5

Vậy x = 1 hoặc x = 15

chiều nay tui giải đc hơm:)

tính cột dọc kiểu j?

18204

Chọn C và Chọn A

dài quá bạn ơi  môt 2 câu thôi

bn đang thi à

\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\left(đk:x>0,x\ne1\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)

\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\left(1-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{4}{\sqrt{x}+1}\)

565-13.x=370
13.x=565-370
13.x=195
x= 195:13
x= 15

565-13.x=370
13.x=565-370
13.x=195
x= 195:13
x= 15

  Vậy x = 15

15

mình làm câu dưới r nha

1) Ta có: \(\sqrt{2x+5}=\sqrt{3-x}\)

\(\Leftrightarrow2x+5=3-x\)

\(\Leftrightarrow2x+x=3-5\)

\(\Leftrightarrow3x=-2\)

hay \(x=-\dfrac{2}{3}\)

2) Ta có: \(\sqrt{2x-5}=\sqrt{x-1}\)

\(\Leftrightarrow2x-5=x-1\)

\(\Leftrightarrow2x-x=-1+5\)

\(\Leftrightarrow x=4\)

3 , \(PT\left(đk:\frac{16}{3}\ge x\ge3\right)< =>x^2-3x=16-3x\)

\(< =>x^2-16=0< =>\left(x-4\right)\left(x+4\right)=0< =>\orbr{\begin{cases}x=4\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)

4 , \(PT\left(đk:...\right)< =>2x^2-3=4x-3< =>2x^2-4x=0\)

\(< =>2x\left(x-2\right)=0< =>\orbr{\begin{cases}x=0\left(...\right)\\x=2\left(...\right)\end{cases}}\)

bạn tự tìm đk rồi đối chiếu nhé :P

5, \(PT\left(đk:...\right)< =>\left(2x-7\right)\left(x-2\right)=x-2\)

\(< =>\left(x-2\right)\left(2x-7-1\right)=0\)

\(< =>\orbr{\begin{cases}x=2\\2x-8=0\end{cases}< =>\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

6, \(PT\left(đk:...\right)< =>x^2-x-6=3x-1\)

\(< =>x^2-4x-5=0< =>\left(x-5\right)\left(x+1\right)=0\)

\(< =>\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

1) Ta có: \(\sqrt{4x}=\sqrt{5}\)

nên 4x=5

hay \(x=\dfrac{5}{4}\)

2) Ta có: \(\sqrt{16x}=8\)

nên 16x=64

hay x=4

3, \(2\sqrt{x}=\sqrt{9x}-3\left(đk:x\ge0\right)\)

\(< =>2\sqrt{x}-3\sqrt{x}+3=0\)

\(< =>3-\sqrt{x}=0< =>x=9\)(tmđk)

4, \(\sqrt{3x-1}=4\left(đk:x\ge\frac{1}{3}\right)\)

\(< =>3x-1=16< =>3x-17=0\)

\(< =>x=\frac{17}{3}\)(tmđk)

5, \(\sqrt{-3x+4}=12\left(đk:x\le\frac{4}{3}\right)\)

\(< =>-3x+4=144\)

\(< =>-3x-140< =>3x+140=0< =>x=-\frac{140}{3}\)(tmđk)

6, \(\sqrt{2-3x}=10\left(đk:x\le\frac{2}{3}\right)\)

\(< =>2-3x=100< =>3x+98=0< =>x=-\frac{98}{3}\)(tmđk)