(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

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

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

Chọn B nhé bạn

a: \(x^3=27\)

=>\(x^3=3^3\)

=>x=3

b: \(\left(2x-1\right)^3=8\)

=>\(\left(2x-1\right)^3=2^3\)

=>2x-1=2

=>2x=2+1=3

=>\(x=\frac32=1,5\)

c: \(\left(x-2\right)^2=16\)

=>\(\left[\begin{array}{l}x-2=4\\ x-2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4+2=6\\ x=-4+2=-2\end{array}\right.\)

d: \(\left(2x-3\right)^2=9\)

=>\(\left[\begin{array}{l}2x-3=3\\ 2x-3=-3\end{array}\right.\Longrightarrow\left[\begin{array}{l}2x=6\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=0\end{array}\right.\)

e: \(2x+5=3^4:3^2\)

=>\(2x+5=3^2=9\)

=>2x=9-5=4

=>\(x=\frac42=2\)

f: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

a: \(x^3=27\)

=>\(x^3=3^3\)

=>x=3

b: \(\left(2x-1\right)^3=8\)

=>\(\left(2x-1\right)^3=2^3\)

=>2x-1=2

=>2x=2+1=3

=>\(x=\frac32=1,5\)

c: \(\left(x-2\right)^2=16\)

=>\(\left[\begin{array}{l}x-2=4\\ x-2=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4+2=6\\ x=-4+2=-2\end{array}\right.\)

d: \(\left(2x-3\right)^2=9\)

=>\(\left[\begin{array}{l}2x-3=3\\ 2x-3=-3\end{array}\right.\Longrightarrow\left[\begin{array}{l}2x=6\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=0\end{array}\right.\)

e: \(2x+5=3^4:3^2\)

=>\(2x+5=3^2=9\)

=>2x=9-5=4

=>\(x=\frac42=2\)

f: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

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

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

Sao lm mỗi câu a, c thế

Câu c viết dấu hoặc đi

a, ĐKXĐ:\(x\ne-5\)

\(\dfrac{2x-5}{x+5}=3\\ \Rightarrow2x-5=3\left(x+5\right)\\ \Leftrightarrow3x+15-2x+5=0\\ \Leftrightarrow x+20=0\\ \Leftrightarrow x=-20\)

b, ĐKXĐ:\(x\ne3\)

\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\\ \Rightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x^2-x-6=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x\left(\dfrac{x+1}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4}{2\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x.\dfrac{x+1+x-3-4}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(2x-6\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x}{x+1}=0\\ \Rightarrow x=0\left(tm\right)\)