Question

[Trung Art]

Bài 3: Tìm x biết:

a) 4x(x+1) = 8(x+1);

c) 2x(x-2) - (2-x)² = 0;

d) (x-3)³ + 3 - x = 0;

g) 5x(x-2000) - x + 2000 = 0;

n) (x+1)² - 1 +x= 0;

k) (1-x)^2 - 1 + x = 0;

m) x + 6x^2 = 0;

h) x^2 - 4x = 0;

Mọi ng giải giúp mik vs ạ! Cảm ơn

Answer 1

[lê thị hương giang]

\(a,4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x^2+4x-8x-8=0\)

\(\Leftrightarrow4x^2-4x-8=0\)

\(\Leftrightarrow4\left(x^2-x-2\right)=0\)

\(\text{⇔}4\left(x^2-2x+x-2\right)=0\)

\(\text{⇔}4\left(x-2\right)\left(x+1\right)=0\)

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

\(c,2x\left(x-2\right)-\left(2-x\right)^2=0\)

\(\text{⇔}2x\left(x-2\right)-\left(x-2\right)^2=0\)

\(\text{⇔}\left(x-2\right)\left(2x-x+2\right)=0\)

\(\text{⇔}\left(x-2\right)\left(x+2\right)=0\)

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

\(d,\left(x-3\right)^3+\left(3-x\right)=0\)

\(\text{⇔}\left(x-3\right)^3-\left(x-3\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x^2-6x+9-1\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x^2-6x+8\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)

\(g,5x\left(x-2000\right)-x+2000=0\)

\(\text{⇔}\left(x-2000\right)\left(5x-1\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=2000\\x=\frac{1}{5}\end{matrix}\right.\)

\(n,\left(x+1\right)^2-1+x=0\)

\(\text{⇔}x^2+2x+1-1+x=0\)

\(\text{⇔}x^2+3x=0\)

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

\(k,\left(1-x\right)^2-1+x=0\)

\(\text{⇔}\left(1-x\right)^2-\left(1-x\right)=0\)

\(\text{⇔}\left(1-x\right)\left(1-x-1\right)=0\)

\(\text{⇔}\left(1-x\right).\left(-x\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

\(m,x+6x^2=0\)

\(\text{⇔}x\left(1+6x\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=0\\x=-\frac{1}{6}\end{matrix}\right.\)

\(h,x^2-4x=0\)

\(\text{⇔}x\left(x-4\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)