bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

Câu 6: Khôg có cau nào đúng

Câu 7: C

Câu 8: B

Câu 9: B

Câu 10: D

6Không có đáp án nào đúng x=11/4

7C

8B

9B

10D

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\) 

\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[\left(x-2\right)-2\left(x-1\right)\right]\left[\left(x-2\right)+2\left(x-1\right)\right]=0\) 

\(\Leftrightarrow\left(x-2-2x+2\right)\left(x-2+2x-2\right)=0\) 

\(\Leftrightarrow\left(-x\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

cảm ơn mọi người nhìu nha!!!

bạn nào l giúp mình với

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

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

\(x=\dfrac{-1}{4}\)

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=9\)

\(\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)